Pinsky Problems in Physics Mir LIBRE
Pinsky Problems in Physics Mir LIBRE
Pinsky Problems in Physics Mir LIBRE
ITHHCKHA
~ ql1 no <I>1l3MRE
H3AATEJIbCTBO HAYHA MOCHBA
PROBLEMS
in PHYSICS
A.A. PINSKY
Translated from the Russian
by
Mark Samokhvalov, Cando Se. (Tech.)
MIR PUBLISHERS MOSCOW
First published 1980
Revised from the 1977 Russian edition
lIaAaTeJILCTBo Hayxa, 1977
English translation, Mir Publishers, 1980
PREFACE
The two volumes of the "Fundamentals of Physics",
published in two editions, have been translated into Polish
and English and have gained popularity among senior form
students of secondary schools where physics is studied at an
advanced level, among college freshmen and among instruc-
tors and teachers of physics. At the same time, reviews and
numerous letters from readers have stressed the need for a
system of problems adapted to the theoretical material
contained in the book which would enable the reader to
consolidate and to check his knowledge of the material stud-
ied, and to develop skills in the creative application of the
theory to specific physical problems.
,This book offers the reader over 750 problems concerning
the same subject matter as is treated in the two volumes of
the "Fundamentals of Physics". The order of presentation of
the theoretical material is also the same.
~ The availability of a great number of problem books based
on the traditional school physics curriculum prompted us
to enlarge those sections which are absent from traditional
problem books, namely the dynamics of a rotating rigid body,
the elements of the theory of relativity and of quantum and
statistical physics, of solid-state physics, wave optics, atom-
IC and nuclear physics, etc. Problems dealing with astrophys-
ics illustrate the application of the laws of physics to ce-
lestial bodies.
~ ~ I The book contains a few problems requiring elementary
skill in differentiating and integrating, as well as some prob-
lems to be solved with the aid of numerical methods, which
nowadays are being increasingly used.
B. M. Yavorsky and A. A. Pinsky. Fundamentals 0/ Physics,
v. I and II, Mir Publishers, Moscow, 1975.
6 Preface
As well as the practice exercises there are some rather
sophisticated problems requiring a deep knowledge of the
theory.
Most of the problems are provided with sufficiently de-
tailed solutions.
Whether a problem book should be provided with detailed
solutions, or only the answers should be supplied, is a con-
troversial question. True, the temptation to look into a
ready-made solution is quite strong. However we hope
that the reader wants to learn to solve the problems himself,
and so he will turn to read the solutions only in extreme
circumstances. On the other hand, having in mind that the
majority of readers will work with the book on their own, we
feel obliged to offer them help when they are unable to cope
with a problem. Note that the solutions provided are not
always the only ones possible. We shall be grateful to any
reader who suggests more elegant or original solutions.
Reference data required for the solution of the problems
is presented in the appendices. They augment the data con-
tained in the corresponding chapters of the series "Funda-
mentals of Physics".
The collection of problems may serve as an aid for students
preparing for examinations in physics. It may be used in
physico-mathematical schools or for extra-curricular work
in physics. The problems will be useful to students studying
to become physics teachers, to students at technical col-
leges, and to physics teachers in schools, technical schools
and secondary vocational schools.
The author expresses his sincere gratitude to Prof.
N. N. Malov and Prof. B. M. Yavorsky and also to Yu.
A. Selesnev, Ya. F. Lerner and M. M. Samokhvalov. Their
valuable remarks enabled the author to make corrections to
the manuscript.
A. A. Pinsky
CONTENTS
Preface
Some Practical Hints
5
. 10
Part One. Motion and Forces
1. Kinematics of a Particle
2. Force .
3. Particle Dynamics . . . . .
4. Gravitation. Electrical Forces
5. Friction
6. Theory of Relativity
Part Two. Conservation Laws
Prob-
lems
17
17
18
20
24
26
.28
.30
Sol u- Answers
tions and
Hints
123 306
123 306
125 306
127 306
132 307
135 308
144 309
148 309
7. The Law of Conservation of Momentum.
Centre of Mass . 30
8. Total and Kinetic Energy . .. . 32
9. Uncertainty Relation ..... . 33
10. Elementary Theory of Collisions . 33
11. Potential Energy. Potential . . ... 35
12. The Law of Conservation of Energy in
Newtonian Mechanics .. .. . . . . . . 36
13. The Law of Conservation of Energy 38
14. Rotational Dynamics of a Rigid Body 39
15. Non-inertial Frames of Reference and
Gravitation . 42
Part Three. Moleeular-kinetle Theory of Gases . 44
16. An Ideal Gas . . . . . .. . 44
17. The First Law of Thermodynamics . 48
1.8. The Second Law of Thermodynamics 50
19. Fundamentals of Fluid -Dynamics . 52
148 309
154 309
156 310
157 310'
163 310
165 310
170 311
172 311
172 312
183 312
183 312
188 313
192 313
198 313
8 Contents
Prob Solu- Answers
lems tions and
Hints
Part Four. Molecular Forces and States of Ag-
gregation of Matter . 55 201 314
20. Solids 55 201 314
21. Liquids 57 206 314
22. Vapours . . . . . 58 209 315
23. Phase Transitions 59 210 315
Part Five. Electrodynamics 61 211 315
24. A Field of Fixed Charges in a Vacuum 61
25. Dielectrics 64
26. Direct Current 65
27. A Magnetic Field in a Vacuum 69
28. Charges and Currents in a Magnetic
Field . . . . . . 71
29. Magnetic Materials 74
30. Electromagnetic Induction 77
31. Classical Electron Theory 80
32. Electrical Conductivity of Electrolytes 82
33. Electric Current in a Vacuum and in
~ s . ~
Part Six. Vibrations and Waves . 87
34. Harmonic Vibrations 87
35. Free Vibrations 88
36. Forced Vibrations. Alternating Current 91
37. Elastic Waves . . . . . . 94
"38. Interference and Diffraction 96
39. Electromagnetic Waves 97
40. Interference and Diffraction of Light 99
41. Dispersion and Absorption of Light 101
42. Polarization of Light . 103
43. Geometrical Optics . 104
44. Optical Instruments . 107
Part Seven. Fundamentals of Quantum Phy- .
sics . 110
45. Photons . 110
46. Elementary Quantum Mechanics . 112
47. Atomic and Molecular Structure . 113
"48. Quantum Properties of Metals and of
Semiconductors t.......... 117
211
217
220
226
229
234
237
240
241
242
245
245
248
251
256
257
258
263
266
270
271
277
281
281
286
290
296
315
316
316
317
317
318
318
319
319
319
320
320
320
320
321
322
322
322
323
323
324
324
324
324
325
326
327
Contents 9
Prob- Sol u- Answers
Iems tions and
Hints
Part Eight. Nuclear and Elementary Particle
Physics . . . . . . 119 299 327
49. Nuclear Structure . . 119 299 327
50. Nuclear Reactions . . . 121 302 328
Tables
1. Astronomical Data . . . . . .. . 329
2. Mechanical Properties of Solids . . . 329
3. Thermal Properties of Solids . . . . . ~
4. Properties of Liquids 330
5. Properties of Gases . . . . 331
6. Electrical Properties of Materials (20 C) 331
7. Velocity of Sound (Longitudinal Waves) 332
8. Refractive Indexes . . . . . . . . . . 332
9. Masses of Some Neutral Atoms (amu) 332
to. Fundamental Physical Constants . . . 333
SOME PRACTICAL HINTS
1. Before you attempt to solve the problems contained in
some chapter, study the corresponding chapters of the "Fun-
damentals of Physics". Bear in mind that the most frequent
reason that you cannot solve a problem is that your knowl-
edge of the theory is not profound enough or is too formal.
2. Think about assumptions which could simplify the
solution. For instance, when calculating forces in dynamics,
one usually assumes them to be constant, while in the theory
of oscillations they aretaken to be quasi-elastic. Processes
in gases are usually considered to be quasi-static, the ele-
ments of electrical circuits linear, the waves sinusoidal, etc.
When necessary, the violation of these conditions is specially
mentioned; in some cases it is evident from the particulars
of the problem (a solenoid with a ferromagnetic core, a modu-
lated wave, etc.).
3. Try to draw a schematic diagram or a sketch; this al-
ways makes consideration of the problem easier. Sometimes
it pays to show the evolution of one's thinking on the dia-
gram by partitioning it, or by introducing successive sim-
plifications (for instance, when determining' internal forces,
or when designing compound circuits). Remember, a good
diagram is half the success in solving a problem.
4. In most cases the problem should be solved in a gener-
al form with all the relevant quantities denoted by corre-
sponding symbols and the calculations made using symbols.
Don't let it trouble you if some of the quantities are not
specified in the statement of the problem-they will either
cancel out, or their values may be found in the appendices
to this book, or in the "Fundamentals of Physics". Do not
be scared by mathematical operations-the ability to per-
form them freely is an element of the mathematical knowl-
edge indispensible to the student of physics.
Note that it is not always convenient to solve the problem
in a general form. Sometimes the price of generality is an
Some Practical Hints 11
excessive volume of calculations. In such cases the problem
should be solved directly with numbers substituted for the
relevant physical quantities.
5. Having obtained the solution in a general form try to
make sure it is a sensible one. To do this, sometimes dimen-
sional analysis may be helpful, sometimes-the analysis
of particular or limiting cases, or a comparison with a sim-
ilar problem already solved is needed. For example, hav-
ing solved a problem in dynamics which takes account of
the forces of friction you may compare the result with that
of a similar problem without friction, a relativistic calcu-
lation can be compared with a similar calculation in New-
tonian mechanics" etc.
6. If the problem contains numerical values the final
answer should be numerical as well. Do not underestimate
calculations. In practice we are always interested in the
numerical values of the quantities sought and only rarely
in their expression in terms of other quantities.
All data, including those derived from the tables, should
be expressed in the same system of units (as a rule, in the
SI system) with the numerical data written in the standard
form, i.e. in the form of a X iOn where 1 -< a < 10. All
values should be specified to the same accuracy.
7. All calculations (including those in the majority of
problems involving the use of numerical methods) should
be performed with the aid of a slide rule, the use of which
guarantees reasonable accuracy. In the cases when the ini-
tial data are specified to two significant digits, the results
of the calculations should be rounded off to the same number
of digits.
<; Only a few problems on the theory of relativity, wave
optics, atomic spectra, etc. involve calculations requiring
an accuracy of four or five digits. For these mathematical
tables should be used.
8. Having obtained an answer compare it with the one
given at the end of the problem book. Do not be disappoint-
ed if your answer does not coincide with the author's. Both
answers might be two different forms of the same expression.
For instance, the expressions
sin ~ cos a and sin (a.-fP)
sin a+Jl. cos a, sin (a+ cp)
t2 Some Practical Hints
coincide if one puts IJ. = tan cp. If you are unable to find
ways of transforming one form of the answer into the other,
make a numerical calculation using both formulas, and if
the results coincide, your solution obviously may be taken
to be correct.
9. After you have obtained the correct answer it is a good
idea to look at the solution offered in the problem book.
Should the solutions turn out to be different, try to find out
which is the best. It will be all to the good if your solution
is simpler, shorter and more elegant than that of the author.
However, one should not preclude the possibility of errors
accidentally compensating each other, so that the answer
obtained is only formally correct. Such an analysis of the
solution is very helpful and instructive.
10. When you are unable to solve the problem right away,
don't hurry to read the solution supplied. Make a new study
of the respective theoretical material, paying attention to
the finer points. Experience shows that a repeated study of
the theory with a definite goal in mind is very effective and
ensures quick success when a second crack at the problem is
attempted.
However, repeated study of the theory still may not help
in solving the problem. This should not be a cause for gloom
and despair. This problem book contains many creative and
difficult problems, and it is not to be wondered at that you
will not be able to cope with them all single-handed. In
such a case try to make a detailed analysis of the solution
supplied by the author. Perhaps after that you will be able
to find a new solution. Experience shows that detailed study
of some of the ready-made solutions is also very instructive
and helps to raise the level of knowledge and to stimulate
creative abilities.
Here's a concrete example of the way the above recommen-
dations should be applied.
Problem. A small mirror with a mass of 9.0 mg is suspend-
ed from a thin quartz filament 4.0 em long. A powerful
laser flash is emitted in a direction perpendicular to the
mirror, so that the system is deflected from the vertical by
a certain angle. Calculate this angle knowing the energy of
the laser flash to be 1.0 X 10
2
J.
Solution will be performed by stages.
Some Practical Hints 13
1. The deflection of the system is caused by the momentum
the light transmits to the mirror upon reflection. Accelerat-
ed to a definite velocity the mirror rises to a definite height.
In this case the kinetic energy of the mirror is transformed
into its potential energy in the gravitational field. Conse-
quently, to solve the problem one must know the expression
for the momentum of the light flash, and the laws of conser-
vation of momentum and of energy.
2. To solve the problem, introduce several simplifying
assumptions. To begin with we shall neglect the mass and
E
YJ\Pc-
E'
let
I
I
I
I
I
I
I
I
I
I
f) __"/
1"""""'-'-
L...1 _
(t)
(a)
(b)
Fig. 1.
the elasticity of the filament, the friction at the suspension
and air resistance. Then we shall assume the mirror to be
perfect reflector (neglecting the absorption of light). Fi-
nally, because the mirror's velocity is small we shall per-
form all the calculations approximately, using Newtonian
mechanics.
3. It will be helpful to draw a schematic diagram depict-
Uig the dynamics of the process (Fig. 1). Here diagram (a)
shows the state of the system before the light falls on the
llil'fOr, (b) at the moment the light is reflected and (c) the
deflection of the mirror together with the filament by the
angle sought. The stages of the solution are also clear from
the diagrams.
4. Denote the energy of the laser flash bye, its momentum
p = ee/c, the velocity of light by c, the mirror's mass by m,
14 Some Practical Hints
its velocity at the moment of recoil by v, the height it rises
by h, the length of the filament by l and the angle of deflec-
tion by a.
It may be seen from the diagram that
h = l-l cos = l (1-cosa) = 2l sin
2
'7////////////7///7/////7///////7///////////a
Fig. 3.3.
Fig. 3.4.
mass M lying on a horizontal plate (Fig. 3.4). Neglecting
friction find the acceleration of both bodies (relative to the
plate) and the tension of the rope for the three cases:
Motion and Forces 21
(1) the monkey does not move with respect to the rope;
(2) the monkey moves upwards with respect to the rope
with acceleration b;
(3) the monkey moves downwards with respect to the rope
with an acceleration b.
3.5. A block of mass M lies on a plane inclined at an angle
a to the horizontal. A weight of mass m is connected to the
block by a thread slung over
a pulley (Fig. 3.5a). Find
the acceleration of the
weight and the tension of
the thread. Friction, the
mass of the pulley and of
the thread are to be neglect-
ed. m
3.6. A rod of mass m
2
"' rests
on a wedge of mass ml (Fig.
3.6a). Guides allow the Fig 3.5a.
rod to move only in the direction of the y-axis and the
wedge only in the direction of the z-axis. Find the accel-
Fig. 3.6a
erations of both bodies and the reaction of the wedge. Neg-
lect friction.
3.7. A block of mass m is placed on a wedge of mass M
(Fig. 3.7a). Find the accelerations of the block and the wedge
in the reference system fixed to the table, and the reaction.
Friction is to be neglected.
Analyse the limiting case when the wedge remains station-
ary.
22 Problems
3.8. Find the period of revolution of a conical simple pen-
dulum whose thread of length 1 makes an angle ex with the
vertical (Fig. 3.8).
3.9. An undeformed spring with the spring constant k has
length Zo. When the system (Fig. 3.9) rotates at an angular
Fig. 3.7a.
velocity 00, the weight with mass m causes an extension of
the spring. Find the length 1of the rotating spring.
Fig. 3.8.
m
Fig. 3.9.
3.10. A plane flies at a constant speed of 200 mls in a hori-
zontal path with radius of curvature equal to 5 km. What
is its angle of bank?
3.11. A plane flying at a constant speed of 300 mls makes a
wingover* in the vertical plane with the radius 1.3 km ,
* A flight manoeuvre in which a plane is put into a climbing
turn until nearly stalled after which the nose is allowed to fall while
the turn is continued until normal flight is attained in a direction
opposite to that in which the manoeuvre was entered. Also called
a Nesterov loop.
Motion and Forces 23
Find the change in weight in the upper and the lower points
of the loop.
3.12. A particle is thrown with an initial speed Vo at an
angle ex to the horizontal. Find the radius of curvature r
at the highest point of its path, and its ratios to the maximum
height H and to the distance L of the particle's flight.
3.13. The equation for a parabola is of the form x
2
= 2py,
where the parameter p > O. Find the radius of curvature of
the parabola at each point.
3.14. Prove that the tangent line to the parabola x
2
= 2py
at an arbitrary point makesan angle ex with the x-axis whose
H
x
l
Fig. 3.15. Fig. 3.f6.
tangent is equal to the x-coordinate of the point divided by
the parameter p (i.e. tan ex = x/p).
3.t5. The surface of a hill is inclined at an angle a to the
horizontal (Fig. 3.15). A stone is thrown from the summit of
the hill at an initial speed Vo at an angle ~ to the vertical.
How far from the summit will the stone strike the ground?
3.16. A body falls freely from some altitude H Fig. 3.16).
At the moment the first body starts falling another body is
thrown from the Earth's surface, which collides with the
first at an altitude h = H/2. The horizontal distance is l,
Find the initial velocity and the angle at which it was thrown.
3.17. Before the discoveries of Galileo the common opinion
was that the greater the mass of a body is the faster it falls.
Try to prove logically, using the fact of additivity of mass,
t ~ ~ t all bodies, independent of their mass, must fall in the
s ~ way. You will be repeating Galileo's reasoning. (Ga-
llleoused reductio ad absurdum).
Ie
}y--------
h
24 Problems
4. Gravitation. Electrical Forces
4.1. Find the mass of the Earth from its polar radius and
the free fall acceleration at the pole.
4.2. Find the mass of the Earth knowing the orbital period
and the radius of the Moon.
4.3. Find the mass of the Sun knowing the average distance
from the Earth to the Sun (the astronomical unit) and the
orbital period of the Earth.
4.4. Compare the forces with which the Sun and the Earth
act on the Moon.
How can you explain the fact that the Moon is a satellite
of the Earth, although the attraction of the Sun is stronger?
4.5. Find the distance from Venus to the Sun knowing its
orbital period and the orbital period of the Earth. .
4.6. At what altitude above a planet is the acceleration due
to gravity one half of that at its surface?
4.7. Find the acceleration due to gravity on the Venus, on
the Moon and on the Sun.
4.8. What should be the period of rotation of a planet about
its axis for the state of weightlessness to exist on the planet's
equator? Do the calculation for the case of the Earth.
4.9. Two small balls with masses of 0.5 g each hang on
threads 0.8 m long each tied to a common hook. What charge
has been acquired by the system if, as a result, an angle
2a = 12 has been established between the threads?
4.10. Two equal positive charges are placed at two corners
of a square, the other two corners having negative charges of
equal magnitude. Find the field intensity in the centre of
the square in the two cases.
4.11. A charge q is uniformly distributed over the surface
of a ring-shaped conductor of radius a. Find the field inten-
sity on the axis of the conductor at a distance .T, from the
plane of the conductor.
4.12. The molecule of water may, as a first approximation,
be considered as a dipole with an electric moment p = 6.1 X
X 10-
30
C m. Estimate the force of attraction between two
water molecules.
4.13. An electric field of intensity E is set up between two
parallel plates of length L. An electron beam enters the field
at an angle rx > 0 to the plates and leaves it at en n g l ~
Motion and Forces 25
p < 0 (Fig. 4.13). Find the initial velocity of the electrons.
The force of gravity is to be neglected.
4.14. Two plane-parallel plates l = 2 em long serve as the
control electrodes of a cathode-ray tube. The distance from
the control electrodes to the tube's screen is L = 30 em,
L
+ +
-I
Fig. 4.13.
An electron beam enters midway between the plates parallel
to them at a velocity of Vo = 2 X 10
7
m/s. What is the
+ +
h
Fig. 4.15.
electric field between the electrodes if the beam's displace-
ment on the screen is d == 12 em?
4.15. The length of plane-parallel electrodes is l, the dis-
tance between them is h. An electric field of intensity E
is set up between them. An electron enters the field close to
the lower plate at an initial speed Vo and at an angle e.t to
the plates (Fig. 4.15). What should the field intensity be for
tPe electron to pass between the electrodes without striking
~ t r of them? For what angles ex is this possible?
26 Problems
5. Friction
5.1. A stationary body of mass m is slowly lowered on to
a massive platform (M m) moving at a speed Do = 4 mls
(Fig. 5.1). How long will the body slide along the platform
and what distance will it travel during this time? The coef-
ficient of friction is I.t = 0.2.
5.2. Solve Problem 3.4 for conditions when the coefficient
of friction between the weight and the plate is f..l.
f 5
lJ
o
M
..
Fig. 5.1.
5.3. Find the acceleration of the block in Problem 3.5 for
conditions when the coefficient of friction between the block
and the inclined plane is f..L.
5.4. Find the reaction of the wedge in Problem 3.6, if the
coefficient of friction between the wedge and the table is
I.t, the friction between the rod and the wedge being
negligible.
5.5. Find the reaction in Problem 3.7, if the coefficient of
friction between the block and the wedge is f..L. Friction be-
tween the wedge and the table is to be neglected.
5.6. Find the reaction in Problem 3.7, if the coefficient of
friction between the wedge and the table is f..L. Friction be-
tween the block and the wedge is to be neglected.
5.7. A block lies on a wedge with the slope angle a. The
coefficient of static friction between the bar and the wedge is
I.t < tan cx. What should be the acceleration of the wedge to
prevent the bar from sliding down?
5.8. A disk rotates at 70 r.p.m. in a horizontal plane. How
would you place an object on the disk so that it would re-
main on it? The coefficient of static friction between the
object and the disk is f.t
st at
== 0.44.
5.9. In the motorcycle stunt called "the wall of death"
the track is a vertical cylindrical surface of 18 m diameter.
What should be the minimum speed of the motorcyclist to
Motion and Forces 27
I
I
Fig. 5.10a.
cp
I
I
I
I
I
I
~ R
I "'-
I
I
prevent him from sliding down? The coefficient of friction
is f..t -< 0.8. Take the motorcycle to be a point mass.
5.tO. A spherical bowl of radius R rotates about the verti-
cal diameter. The bowl contains a small object whose radius
vector in the course of rotation makes an angle a with the
vertical (Fig. 5.10a). What should be the minimum angular
velocity co of the bowl in order to
prevent the object from sliding
down, if the coefficient of static
friction is fJ-stat?
5.11. A motorcyclist rides at a speed
of 90 km/h. What is the radius of
curvature of the bend the motorcy-
clist can make, if the coefficient of
friction between the rubber tyre and
the asphalt is 0.65? What is the
motorcyclist's inclination to the hor-
izontal?
5.12. A glass ball of 4.0 mm dia-
meter falls in a glycerine solution
(Po = 1.21 X 10
3
kg/m", 11 =
5.02 X 10-
2
Pa -s), The density of the glass is p =
2.53 X 10
3
kg/m". Find the steady-state speed and the initial
acceleration. Make an approximate estimate of the time in
which the ball attains the steady-state speed, and the
distance the ball travels during this time.
5.t3. With the aid of numerical calculation find the in-
stantaneous values of the acceleration and the speed of the
falling ball in the previous problem and plot the graph.
Choose an interval of time equal to ~ = 0.02 s.
5.14. Estimate the steady-state speed of settling of dust
particles in a room l = 2.8 m high and the time of settling.
The minimum diameter of a dust particle is 2r === 0.06 mm.
The viscosity of air at 20 C is 11 === 1.8 X 10-
5
Pa -s, the
density of the dust particles is p = 2 X 10
3
kg/m",
5.15. Estimate the speed at which hail falls if the diameter
01 a hailstone is 2r = 5 mm, and its density is p === 8 X
X 10
2
kg/m",
5.16. When you have learned to integrate, find the depen-
dence on time of the instantaneous velocity of the ball in
Problem 5.13.
28 Problems
6. Theory of Relativity
6.1. Making use of the principle of relativity prove that the
lateral dimensions of an object do not change with a change
in the reference frame.
6.2. Estimate the relative error in calculations when the
classical law of addition of velocities is used instead of the
relativistic.
6.3. In a colliding-beam proton accelerator the protons of
both beams meet head on at a speed of O.99000c relative to
V
2
v,
-X.
Fig. 6.4.
the accelerator. What is the speed of a proton of one beam
relative to that of the other?
6.4. Here's one of the "disproofs" of the relativistic law of
addition of velocities. Suppose two objects initially at the
same point start moving relative to the Earth in opposite
directions (Fig. 6.4). The total distance covered by the objects
is
= - = v
t
t1t - ( - = (Vi +V2)
Hence, the speed of approach u = At = VI + V
2
We obtained the classical law of addition of velocities
and not the relativistic one. Where lies the error in such
reasoning?
6.5. The velocity of light in a stationary medium is u =
=== cln where c is the velocity of light in vacuum and n is the
refractive index of the material (see 63.1). Find the veloc-
ity of light in the medium moving at a constant speed
relative to the source of light.
6.6. In the Fizeau experiment two light beams travel head
OD, one along a stream of fluid, the other in the opposite
direction (Fig. 6.6). If the length of each tube is l, the veloc-
ity of the fluid v and its refractive index n, what will be
the difference between the travel times of the light beams?
6.7. What will be the distance a pion (pi-meson) travels
before it decays, if its speed is v == O.99c and its intrinsic
lifetime is 'to = 2.6 X 10-
8
5? What would be the transit
Motion and Forces 29
distance if there is no relativistic time dilation? The distance
is measured in the laboratory reference frame.
6.8. Find the intrinsic lifetime of a particle if its velocity
is below the light velocity in vacuum by 0.2% and the dis-
tance it travels before decaying is about 300 km.
6.9. At what speed must a particle move for its rest mass to
be trebled?
6.10. Find the expression for the density of a body in an ar-
bitrary inertial frame of reference.
6.t1. At two points of an iner-
tial reference frame separated by
a distance along the x-axis of
l = X
2
- Xl two events take place
simultaneously. Find the time
interval between the events in an
iarbitrary inertial reference frame.
;6."t2. An electron is accelerated
"in an electric field of intensity
E = 3.0 X 10
6
N/C. Find the
~ p e e of the electron after 1.0 ns .
t, --What would be the speed of
the electron, if its mass is inde-
pendent of speed? Fig. 6.6.
6.13. A force acts on a particle
moving at a relativistic speed in a direction perpendicular
to its path. How will the particle move? Express the force
in terms of -the speed and of the radius of curvature of the
path.
6.t4. When -you have learned to differentiate trigonometric
functions, try to prove that in the relativistic case the for-
mula F = ma does not hold, even if m is taken to be the
relativistic mass.
6.t5. Introduce the following definition: the length of a
moving rod is the product of its speed and the time interval
between the moments when its two ends pass a static clock.
~ h e proper length is determined in a similar way with the
bid of a clock moving at the same speed along a static rod.
Wind the relation between the length of a moving rod land
itS proper length lo.
1
c
..i'
Part Two
CONSERVATION LAWS
7. The Law of Conservation of Momentum. Centre of Mass
7.1. A wooden block of 5.0 kg mass lies on a horizontal
wooden table. A bullet of 9.0 g mass hits the block after
which it moves a distance 25 em across the table before stop-
ping. Find the bullet's speed.
7.2. A railway flatcar whose mass together with the artillery
gun is M moves at a speed V along the x-axis (Fig. 7.2).
Fig. 7.2.
The gun barrel makes an angle ex with this axis. A projectile
of mass m leaves the gun at speed v (relative to the gun)
in the direction of the flatcar's motion. Find the speed of
the flatcar after the gun has been fired. What should the
speed of the flatcar be for it to stop after the firing? Neglect
friction.
Conservation Laws 31
Put M = 10 tons, m = 120 kg, V = 6.0 mis, v = 900 mIs,
(X = 30.
7.3. The mass of a boat is M = 80 kg, the mass of a boy
is m = 36 kg. The boy moves from the stern to the bows of
the boat. What distance does the boat move, if its length is
l = 2.8 m? At such low speeds the water resistance may be
neglected.
7.4. Find the initial acceleration of a rocket if its initial
mass is 40 metric tons, the exhaust velocity of gases is
4 km/s and the fuel consumption 200 kg/s.
7.5. The initial mass of a missile is M 0 = 160 tons, the
exhaust velocity is 4 km/s. After 90 tons of fuel have burnt
out the first stage with a mass of 30 tons is detached. Next
additional 28 tons of fuel are burnt. What is the final speed
of the second stage?
>. What would be the speed of a single-stage missile with
the same mass of fuel?
7.6. When you have learned to integrate, derive the Tsiol-
kovsky formula.
7.7. Why do astronauts experience an increase in overload
as the spaceship is accelerated? Assume the fuel consumption
to be constant.
7.8. A carriage closed on all sides stands on rails. Can the
.passengers inside the carriage cause an oscillatory motion
of the carriage? Friction with the rails should be neglected.
.The mass of the carriage should be assumed to be commen-
surate with that of the passengers.
7.9. Prove that the centre of mass of a uniform triangular
plate coincides with the point of intersection of its medians.
~ t O Find the center of mass of a uniform plate shown in
-Fig. 7.10a.
7.11. Find the center of mass of a plate shown in Fig. 7.11.
.Put R = 5.00 em, r = 3.00 em.
'7.12. Using numerical methods find the center of mass of
.8. semicircle. To facilitate computation put R = 1.00.
i.13. Using numerical methods find .the center of mass of
a hemisphere.
,7.14. Using numerical methods find the center of mass of
-a right circular cone with the height h = 1 and its genera-
.\rix making an angle (L with the height.
7.15. Using integral calculus solve the Problems 7.12,
32
Problems
7.13, and 7.14 analytically.
7.16. The third Kepler law was derived in 9.4 for the case
of a planet mass being much smaller than that of the Sun,
6
J
Fig. 7.10a. Fig. 7..11.
so that the Sun could be considered to be stationary. Derive
this law for the case of two bodies rotating about their centre
of mass.
8. Total and Kinetic Energy
8.1. Find the rest energy (the proper energy) of an electron,
a proton, and a neutron.
8.2. Find the velocity of a particle whose kinetic energy is
equal to its rest energy.
8.3. Find the kinetic energy and the momentum of an elec-
tron whose speed is O.92c.
8.4. The kinetic energy of a proton is 10 GeV. Find its mo-
mentum and velocity.
8.5. The kinetic energy of electrons in the Kharkov and the
Erevan linear accelerators is 10 MeV. Find the speed of the
electrons.
8.6. What is the error when the classical expression for the
kinetic energy is substituted for the relativistic expression?
Calculate for Ul = O.1c; for U
2
= O.ge and for Us = O.99c.
8.7. The midship section of a launch is S = 4 m", the power
of its engine is P = 300 h.p., the efficiency is II = 25%.
What is the maximum speed of the launch? Put C = 0.5.
8.8. A winch powered by an engine of specified power P
pulls a weight up an inclined plane (see Fig. 3.5a, p. 21).
The plane makes an angle (X with the horizontal, and the
Conservation Laws
33
coefficient of friction is fl. For what angle of inclination will
the speed of the weight be a minimum?
J ~ 9 A hydraulic monitor emits a jet of water at a speed
.of100 m/s. The water flow rate is 144 m
3
/h. Find the power
of its pump if its efficiency is 75%.
8 ~ 10. An electron with zero initial velocity is accelerated
in an electric field of intensity E. Find the velocity of the
electron after it has travelled a distance l. Do the calcula-
tions for the classical and for the relativistic case. Show that
"for a weak field the relativistic formula becomes the same
as the classical formula.
'S.11. For an ultra-relativistic particle (pc ~ ~ o its total
energy may be assumed to be equal to the product of its
momentum and of the velocity of light in vacuum, i.e.
~ = pc. Determine the error arising from this assumption.
9. Uncertainty Relation
9.1. Assuming that in a hydrogen atom the electron rotates
about the nucleus in a circular orbit, estimate the radius of
-this orbit.
9.2. What kinetic energy must an electron have to be able
to penetrate the nucleus? The dimensions of a nucleus are
of the order of 10-
15
m.
9.3. Assess the kinetic energy of conduction electrons in
a .metal in which their concentration is of the order of
:f0
29
m:",
9.4. According to modern ideas, a pulsar is a star made up
'almost entirely of neutrons. * Assuming the mass of the pul-
'sar to be equal to that of the Sun (2 X 10
30
kg) and its radius
to be of the order of 10 kID, estimate the kinetic energy of
the neutrons.
10. Elementary Theory of Collisions
10.1. A block with the mass of 2.0 kg lies on a smooth hor-
.. izontal table. A bullet with the mass of 9.0 g flying at a
speed of 800 mls at an angle of 30
0to
the horizontal hits the
r * The possibility of such a state of matter was first suggested by
~ L. D. Landau, Member of the Academy of Sciences of the USSR,
. in 1932. See, for example, the paper "On the sources of stellar energy",
poklady Academii Nauk 88SR, v. 17, p. 301 (1937) (in Russian).
3-0360
34 Problems
block and sticks in it. What is the speed and the direction
of the resulting motion of the block?
Does the apparent loss of the vertical component of its
momentum contradict the law of conservation of momentum?
10.2. A radon nucleus with an atomic mass of 216 emits an
alpha-particle with an atomic mass 4 and a kinetic energy
8 MeV. What is the energy of the recoil nucleus?
10.3. A smooth ball hits a smooth wall at a certain angle.
The collision is elastic. Prove that the angle of reflection
is equal to the angle of incidence.
10.4. A ball moving parallel to the y-axis undergoes an
elastic collision with a parabolic mirror y2 = 2px. Prove
Fig. 10.7a. Fig. 10.8a.
that no matter where the point of impact lies, it will arrive
at the mirror's focus F. Find the position of the focus.
10.5. Prove that as a result of an elastic collision of two
nonrelativistic particles of equal mass the scattering angle
will be 90.
10.6. A relativistic proton with kinetic energy K collides
with a stationary proton. Assuming the collision to be elastic
and the energy to be partitioned equally between the parti-
cles, find the scattering angle. Calculate for the cases K =
= 500 MeV and K = 10 GeV.
10.7. A disk of radius r moving on perfectly smooth surface
at a speed v undergoes an elastic collision with an identical
stationary disk. Express the magnitude and the direction
of the velocity of each of the disks after the collision as a
function of the impact parameter d (Fig. 10.7a). Calcula-
tions to be made only for the non relativistic approximation.
Conservation Laws 35
to.8. Solve the previous problem assuming the mass of the
moving disk to be m, and its radius r
1
, and the corresponding
magnitudes of the stationary disk to be m
2
and r
2
(Fig. iO.8a).
to.9. Calculate the pressure exerted by a flux of particles
striking a wall at an angle ex, to its normal. Consider the case
of elastic collisions. The particle concentration is n.
10.10. Estimate the sail area of a sailing boat moving at a
constant speed in the direction of the wind assuming its
midship section to be So = 1.0 m", the coefficient C = 0.1,
-the boat's speed Vo = 3.0 mIs, and the wind velocity v ==
-= 6.0 m/s.
'10.11. A ball is thrown horizontally at a speed Vo from the
top of a hill whose slope is a (to the horizontal). Assuming
the ball's impact on the hill's surface to be elastic find the
,point where it will hit the hill the second time.
11. Potential Energy. Potential
it.1. Prove that in a uniform field the work is independent
of the path.
t1.2. When you have learned to integrate exponential func-
tions, try to derive formulas (18.6), (18.10), and (18.12).
11.3. Assume the potential energy of an object to be zero
if the object is infinitely distant from the Earth. Write the
expression for the potential energy of the object at an arbit-
rary point above the Earth.
..', What is its potential energy on the Earth's surface?
11.4. Assume the potential energy of an object to be zero
-if the object is on the Earth's surface. Write down the ex-
fression for the potential energy of the object at an arbitrary
point above the Earth.
What is its potential energy at an infinite distance?
tit.5. Calculate the energy of a dipole. What is the meaning
'of a minus sign?
11.6. The dipole moment of a hydrogen chloride molecule
!is 3.44 X 10-
30
Cm, the separation of the dipole is 1.01 X
X 10-
10
ID. Estimate the energy liberated in the course of
formation of 1 kg of hydrogen chloride from the starting ma-
if the number of molecules in 1 kg is 1.6 X 10
26
36 Problems
..
11.8. Find the sum of the kinetic and the potential energies
of an electron in the first Bohr orbit. Explain the meaning
of the sign of the total energy (see Problem 9.1).
11.9. Find the momentum and the velocity acquired by an
electrically charged particle which has travelled through a
potential difference fP = CPt - <P2. Take the initial velocity
of the particle to be zero. Do the calculation both for the
nonrelativistic and the relativistic cases.
11.10. Find the potential difference for which the error in
the value of the momentum for the previous problem calcu-
lated using the nonrelativistic approximation does not ex-
ceed 5%. Do the calculations both for the electron and for
the proton.
11.11. In the ultra-relativistic case the momentum of a
particle accelerated by a potential difference cp is found with
the aid of the formula p = eelc, this value being expressed
in units of MeVlc, where c is the velocity of light in a vacuum.
Express this unit in the SI system. Find out for what poten-
tial differences the use of this formula leads to an error of
less than 5%. Do the calculations both for the electron and
the proton.
12. The Law of Conservation of Energy in
Newtonian Mechanics
12.1. The ballistic pendulum is a block of 3.0 kg mass sus-
pended from a thread 2.5 m long. A bullet with the mass of
9.0 g hits the block and sticks in it, the result being a de-
flection of the system by an angle of 18
0
(Fig. 12.1). Find
the bullet '8 speed.
12.2. A body of 5 kg mass is raised vertically to a height
of 10 m by a force of 120 N. Find the final velocity of the
body using two methods: Newton's second law and the law
of conservation of energy. The initial speed is zero.
12.3. Solve Problem 5.1 using the law of conservation of
energy.
12.4. A weight is suspended from a thread of length l. What
is the initial speed that has to be imparted to it at the lowest
point to make it complete a full revolution? The mass of the
thread is to be neglected.
h
37
\
ex \
\
\
\
\
\
\
y,
Conservation Laws ~
~ .
ii 12.5. Solve the same problem for the case of a weight sus-
;;; pended from a thin rod of negligible mass.
12.6. A weight of mass m hangs on a thread. The thread is
deflected by an angle CXo and let go. Find the tension of the
~ ~ .
thread as a function of the angle cx .
o 12.7. A small disk of mass m lies on the highest point of a
: ; ~ sphere of radius R. A slight push makes the disk start slid-
ing down. Find the force of
pressure of the disk on the
sphere as a function of the
angle its radius vector makes
with the vertical. Where does
the disk lose contact with the
sphere? Friction is to be neg-
lected.
12.8. A cyclist rolls down a m v
"devil's loop" track from a ~ u
height H. Find the pressure of .
the cyclist on the track as a FIg. 12.1.
function of the angle the radius vector makes with the ver-
tical. Do the calculations also for the case when the cyclist
rolls down from the minimum height.
12.9. A small object loops a vertical loop in which a sym-
metrical section of'angle 2a has been removed! (Fig. 12.9).
Find the maximum and the minimum heights from whichthe
, object, after loosing contact with the loop at point A and
flying through the air, will reach point B. Find the correspon-
ding angles of the section removed for which this is possible.
12.10. The point of an elliptical orbit closest to the Sun is
called the perihelion, and the point most distant from it is
called aphelion (Fig. 12.10). Denoting the distance from the
perihelion to the Sun by r
o
, and the velocity of the planet at
the perihelion by Vo, find the radius of curvature of the orbit
at the perihelion and at the aphelion, the distance from the
aphelion to the SUD, and the velocity of the planet at the
aphelion.
Prove that the motion of a planet in an.ulliptical orbit
is only possible, if its total energy is negative.
12.1t. Prove that if a space vehicle travels along a para-
bolic path with the Earth (or some other planet) at its focus,
the total mechanical energy of the vehicle is zero.
38
Problems
12.12. Solve Problem 3.6 using the law of conservation of
energy.
12.13. Solve Problem 3.7 using the laws of conservation
of energy and momentum.
12.14. A space vehicle of 1 tonne mass is to take off from the
Moon and fly to the Earth. Find the amount of fuel required.
Compare the result with the amount of fuel needed to send
.
Find the decrease in the internal energy and the mass of the
Sun per second. How long will it take for the Sun's mass to
Conservation Laws 39
decrease due to radiation by 10%? The volume of the Sun
is to be assumed to remain constant.
13.4. A nonrelativistic particle collides inelastically with
an identical stationary particle. What is the kinetic' energy
of the body thus formed? What happened to the rest of the
kinetic energy?
13.5. A particle with rest mass M 0 splits up into two iden-
tical fragments which fly apart in opposite directions at
speeds of 0.90c. Find the rest mass of each fragment. 0
13.6. A relativistic particle collides inelastically with an
identical stationary particle. What are the internal and the
kinetic energies of the resulting object? The kinetic energy
of the particle before the collision is K = ecp where cp is
the potential of the accelerating electric field. Do the cal-
culations for protons with kinetic energies of 10 GeV and
76 GeV.
i3.7. Find the kinetic energy that must be imparted to a
positron for a proton-antiproton pair to be obtained as a
result of its collision with a stationary electron.
1'3.8. Solve the previous problem assuming the collision
to take place in a colliding-beam accelerator in which the
electrons and positrons meet head on with equal velocities.
13.9. Compare the efficiency of a colliding-beam accelerator
with that of an accelerator in which the particles strike a
target made up of identical stationary particles.
13.10. What should be the energy of a conventional acceler-
ator for it to be able to do the work of a colliding-beam accel-
erator of 200 MeV? Do the calculations both for electrons
and for protons.
c::
14. Rotational Dynamics of a Rigid Body
14.1. A force couple is the term used for a system of two equal
antiparallel forces; the arm of the couple is the shortest dis-
.tance between the forces. Prove that the torque is equal to
the product of the magnitude of the force and the arm no
'.Blatter what is the position of the point with respect to which
-the torque is determined.
,.14.2. Solve Problem 2.2, applying to the system 'two force
.:_uples having torques equal in magnitude and opposite in
-fin.'
40
Problems
A
Fig. 14.6.
14.3. Find the torque on the shaft of an electric motor of
20 kW power if its rotor turns at 1440 r.p.m,
14.4. The torsion modulus of a spiral
spring is 2 N -m/rad, The spring is tur-
ned 10 times. What is the work done?
14.5. Find the moment of inertia of
a disk about an axis passing through
a point on its circumference perpen-
dicular to its plane.
14.6. The mass of a disk with a cir-
cular hole cut in it (Fig. 14.6) is m.
Find its moment of inertia about an
axis passing through point A perpen-
dicularly to the disk's plane.
14.7. When you have learned to integrate, derive the
formula for the moment of inertia of a disk.
14.8. Derive, making use of an integral, the formula for the
moment of inertia of a sphere about its diameter.
14.9. Derive, making use of an integral, the formula for the
moment of inertia of a right circular cone about its height.
14.10. Solve Problem 14.8 using numerical methods.
v
Fig. 14.12. Fig. 14.13.
14.11. Solve Problem 14.9 using numerical methods.
14.12. A uniform rod of length l can rotate without friction
about an axis passing through its upper end (Fig. 14.12).
The rod is deflected by an angle ao and let go. Find the speed
of the lower" end of the rod as a function of the angle a.
14.13. A solid cylinder with base radius ]is placed on top
of an inclined plane of length l and slope angle a (Fig. 14.13).
Conservation Laws 41
The cylinder rolls down without slipping. Find the speed of
the centre of mass of the cylinder at the bottom of the plane,
if the coefficient of rolling friction is k. Can rolling friction
be neglected? Do the calculation for the following conditions:
l = 1 m, a = 30, r = 10 em, k = 5 X 10-
4
m.
What would be the speed if, in the absence of friction,
the cylinder slides down?
14.14. Solve Problem 14.13 for a thin-walled solid cylinder
of the same radius and mass.
14.15. A solid flywheel of 20 kg mass and 120 mm radius
revolves at 600 r.p.m. With what force must a brake lining
be pressed against it for the flywheel to
stop in 3 s, if the coefficient of friction
is O.1?
14.16. A flywheel with moment of iner-
tia 0.86 kg -m! and a cylinder of 5 em
radius of negligible mass are fixed to a
common shaft (Fig. 14.16). A thread is
wound around the cylinder, and a weight
of 6.0 kg mass is attached to it. What
time will the weight take to fall 1 m?
What will be its final speed? Assume the
initial speed to be zero.
14.t7. Solve Problem 3.2 assuming the
moment of inertia of the pulley to be I Fig. 14.16.
and its radius to be r.
14.t8. Solve Problem 12.7 assuming a ball of mass m and
of radius r to roll down from the top without slipping.
Neglect energy losses due to rolling friction.
f4.f9. A man stands in the centre of a Zhukovskii turntable
(8 rotating platform with frictionless bearings) and rotates
with it at 30 r.p.m. The moment of inertia of the man's
body with respect to the axis of rotation is about 1.2 kg -m",
the man holds in his outstretched hands two weights of
mass 3 kg each. The distance between the weights is 160 em.
What will be the change in the speed of rotation of the sys-
tem, if the man lets his hands fall so that the distance be-
... tween the weights becomes 40 cm? The moment of inertia
of the turntable is 0.6 kg -m", the change in the moment of
~ r t i of the man's hands and the friction are to be neglect-
42 Problems
14.20. A man of 80 kg mass is standing on the rim of a
circular platform rotating about its axis. The platform with
the man on it rotates at 12.0 r.p.m. How will the system
rotate, if the man moves to the platform's centre? What
work will the man perform in changing his position? The
mass of the platform is 200 kg and its radius is 1.2 m.
14.21. Suppose the Sun contracts (collapses) to a pulsar.
Estimate the minimum radius of the pulsar and its period of
rotation. The period of revolution of the Sun about its axis
is 25.38 days (1 day = 24 hours).
14.22. Compare the kinetic energies of rotation of the pul-
sar and of the Sun (see Problem 14.21). What is the source
of the increase in kinetic energy?
14.23. An electron has an intrinsic angular momentum
(spin) whose component in an arbitrary direction is one half
of the Planck's constant, i.e. L z = 11,/2 = 5.25 X 10-
36
.J s.
Making use of the fact that the speed of light in vacuum is
the maximum attainable, prove that a model in which the
spin of an electron is due to the rotation about its axis is
not feasible.
15. Non-inertial Frames of Reference and Gravitation
15.1. Solve Problem 5.7 in the frame of reference connected
with the wedge.
15.2. Solve Problem 5.8 in the rotating frame of reference
connected with the disk.
15.3. Solve Problem 5.9 in a rotating frame of reference.
15.4. Solve Problem 3.8 in a rotating frame of reference.
15.5. Solve Problem 3.9 in a rotating frame of reference.
15.6. What is the angular velocity of rotation of a star at
which the matter starts to escape from its equator? In cal-
culating make use of a reference frame fixed to the rotating
star. Compare with Problem 14.21.
15.7. A drop of fat in milk has a diameter of the orderkof
0.02 mm. Estimate the time it takes to separate cream in
a centrifuge at room temperature (t 20 C), if the depth
of the vessel is 20 em, the rotation radius 80 em and the speed
600 r.p.m. Compare with the time needed to separate cream
in the gravitational field.
15.8. A centrifugal governor is of the form shown in Fig. 15.8.
The mass of each weight is m, the spring constant is k. Will
Conservation Laws
43
i ~ 15.8.
this device work in conditions of weightlessness? What is
the dependence of the angle a on the speed of rotation of the
system? What is the maximum speed for which the device
is designed if the maximum
contraction of the spring is
10% of its original length?
15.9. Prove that the surface of
a liquid in a rotating vessel
assumes the form of a para bo-
loid of rotation.
15.10. Making use of the prin-
ciple of equivalence explain
the origin of weightlessness
in a spacecraft orbiting the
Earth (or some other planet).
15.11. What should be the
angular velocity of rotation of
a spacecraft about its axis for
the astronaut to feel effects similar to those of the Moon's
gravitational field, where the free fall acceleration is one
sixth of the Earth's? Take the spacecraft's diameter to
be 6 ID.
15.12. In October 1971 an atomic clock was placed on a
"Boeing 747" flying at an altitude of 10 km at a velocity of
1000 km/h eastwards. An identical clock, with time-keep-
ing accuracy of 1 ns (1 nanosecond = 10-
9
s) remained on the
Earth. The plane was in flight 60 h, after which a compari-
son was made of the clocks' readings. What was the differ-
ence in the readings of the clocks in the plane and on the
Earth? What were the contributions of the plane's elevation
and its speed of flight?
15.13. Find the gravitational shift in frequency on the Sun,
on a white dwarf, and on a pulsar. Assume the masses of
all three types of stars to be the same and equal to 2 X
X 10
30
kg; the radius of the Sun to be 7 X 10
6
km, of the
white dwarf 10
3
kID, and of a pulsar 10 krn.
15.14. Our solution of Problem 12.15 was incorrect; we
made use of the formula for the escape velocity derived from
nonrelativistic expressions for the kinetic and the potential
energies. Try to derive a formula for the radius of a black
hole from relativistic considerations.
Part Three
MOLECULAR-KINETIC THEORY OF GASES
16. An Ideal Gas
16.1. Prove that the magnitude of the hydrostatic pressure
is proportional to the height of the column of liquid (or
gas) and is independent of the vessel's shape.
16.2. Making use of the expression for the hydrostatic
pressure, derive an expression for the magnitude of the
Archimedes force.
16.3. What is the value of the unit of pressure 1 mmHg
(torr) on the Moon? On Venus? Use data from Problem 4.7.
16.4. In Stern's experiment (1920)
silver atoms emitted by a heated
filament passed through a slit and
were deposited on the cooled wall
of an outer cylinder (Fig. 16.4).
When the system was rotated. at
high speed there was a deflection
of the slit's image. The apparatus
was first rotated in one direction
and then in the opposite direction,
and the distance between the deflect-
ed images was measured. Find this
distance if the radius of the inter- Fig. 16.4.
nal cylinder is 2.0 em and of the
external one 8.0 em. The speed of rotation is 2700 r.p.m.
and the filament temperature 960C.
Estimate the errors of measurement, if the width of the
slit is 0.5 mm.
Molecular-kinetic Theory of Gases 45
16.5. At what speed must the rotor of Lammert's machine
rotate for gas molecules with velocities of 700 mls to pass
through the slits? What velocity spread will be recorded in
the experiment? Take the distance between the disks to be
40 em, the angle between the slits to be 20 and the angular
width of the slit to be 2.
Estimate the error in the experiment.
16.6. The temperature of the Sun's external layer (the pho...
tosphere) is about 6000 K. Why don't hydrogen atoms, the
main component of the photosphere, leave the Sun's sur-
face?
16.7. The height of the photosphere is much less than the
Sun's radius. Equating gravitational and pressure forces
try to estimate the height of the photosphere assuming it to
be made up entirely of atomic hydrogen.
16.8. The density of the photosphere assessed with the aid
of optical methods is 2 X 10-
4
kg/m", Find the average gas
pressure in the photosphere and the mean free path of hydro-
gen atoms.
16.9. Knowing the mass and the radius of the Sun one may
find the average density of the Sun's material. Estimate
the pressure and the temperature of the gas in the middle of
the radius assuming, for the sake of simplicity, that the
density is constant and that the acceleration due to gravity
at this point is one half its value at the surface. What is
the proton concentration at this point?
16.10. Explain the reason why the Moon cannot retain its
atmosphere. Take into account that during a lunar day its
temperature rises above 100C.
16.11. A vacuum has been created in a radio tube, i.e. a
state of gas where the mean free path of its particles exceeds
.the characteristic dimensions of the vessel. Assuming the
.tube's length to be 5 em and it to be filled with argon, esti-
mate the density and the pressure of the gas at room temper-
ature (20C).
16.12. Find the lifting force of a balloon of 2 X 10
4
rn
3
capacity filled with helium at the surface of the Earth and
.at an altitude of 10 km above sea level. The balloon's en-
velope is open underneath. For data on the properties of
the Earth's atmosphere see 26.10, Table 26.3.
46 Problems
Fig. 16.17.
u
16.13. Find the molecular formula of ammonia, if its den-
sity at the pressure of 780 mmHg and the temperature of
20C is 0.736 kg/m",
16.14. Dalton's law is formulated as follows: the total pres-
sure of a mixture of ideal gases is equal to the sum of the partial
pressures of these gases. The partial pressure is the pressure
of a given gas as if it alone occupied
the whole vessel. Prove this law.
16.15. A gas container of 20 I capa-
city contains a mixture of 10 g of
hydrogen and 48 g of oxygen. After
the mixture is ignited by a spark
the gas formed is heated to 300 "C,
Find the pressure of the gas.
16.16. Assuming air (M = 29
kg/mole) to be composed mainly
of oxygen and nitrogen, find the percentage composition of
these gases in the atmosphere.
16.17. In 1908-1910 Perrin determined the .Avogadro num-
ber. He did it by observing the distribution of tiny gumboge
gum balls in water with the aid of a short-focus microscope
(Fig. 16.17). By adjusting the focus of the microscope to
observe a definite layer he was able to count the number of
particles in each layer. In one of the experiments the fol-
lowing data were obtained:
Height of the layer above the
tray's bottom, urn 5 35 65 95
Number of particles in the layer 100 47 23 12
Knowing the ball's radius to be 0.212 urn, the density of
gumboge gum to be 1.252 X 10
3
kg/m", the density of water
at 27C to be 0.997 X 10
3
kg/m" find the Avogadro number.
16.18. A gas -rotates in a centrifuge. Taking into account
that the field of centrifugal forces of inertia is equivalent to
a gravitational field, write the expression for the baro-
metric distribution in the centrifuge.
16.19. Centrifuging may be used for the separation of iso-
topes. To do this a mixture of two gases is placed inside a
cylindrical vessel rotating at a high speed. Because of the
action of centrifugal forces, the isotope concentration near
the cylinder wall will be different from that in the centre.
Molecular-kinetic Theory of Gases 47
Compare the concentrations of the light and the heavy ura-
nium isotopes near the centrifuge walls, if the diameter of
the cylinder is 10 em, the rotation speed is 2.0 X ,10
3
f. p.s.,
the temperature of uranium hexafluoride is 27C. For con-
centrations in normal conditions see 25.6.
Find the enrichment factor in the mixture of the heavy
isotope near the walls of the vessel. The term enrichment
factor applies to the quotient obtained by dividing the
concentration ratio during rotation by the initial concen-
tration ratio:
x= ~ : ~
nl nOl
16.20. How many times in succession should the light frac-
tion he separated in the centrifuge to obtain a mixture con-
taining 80% of light uranium isotope?
16.21. We have obtained the barometric, distribution for
the case of an isothermal atmosphere; indeed, in 26.10 we
assumed the temperature to he the same at every point.
Actually, in the real atmosphere the temperature drops with
altitude. It may be demonstrated that if the decrease in the
'temperature with the altitude is linear, i.e. if T = To (1 -
- ah), the barometrical formula assumes the form
-L= ~ mg/akTo
Po To
Prove that if a is small this formula reduces to the for-
mula for the barometric distribution in an isothermal at-
mosphere.
16.22. Try to derive the barometrical formula for an atmos-
phere in which the temperature decreases linearly with the
altitude.
16.23. Measurements carried out by the Soviet "Venus"
space-probes with the aid of their landing modules have
shown that from an altitude of 50 km above the surface of
Venus the temperature of the planet's atmosphere changes
linearly as the altitude decreases. Using the data given below
prove that this layer of the atmosphere consists mainly of
carbon dioxide gas.
Altitude above surface h; km 50 42 37 15 0
Pressure p, atm 1 3.3 6 37 90
Temperature t, C 80 160 200 360 485
T, K 353 433 473 633 758
48
Problems
t 7. The First Law of Thermodynamics
Fig. 17.6.
0.2 0.5
Fig. 17.2.
------71
---/ i
I I
I I
I I V,m
3
17.1. A vessel contains helium, which expands at a constant
pressure when 15 kJ of heat is supplied to it. What will be
the variation of the internal energy of the gas? What is the
work performed in the expansion? )K
17.2. A cylinder contains 0.15 kg of hydrogen. 'The cylinder
is closed by a piston supporting a weight of 74 kg (Fig. 17.2).
What amount of heat should be supplied
to lift the weight by 0.6 m? The process
should be assumed isobaric, the heat capa-
city of the vessel and the external pres-
sure should be neglected.
17.4. For most diatomic gases at room
temperatures 'V = + 1.40 + 0.01. Find the
specific heat of nitrogen in these conditions.
17.5. A cylindrical vessel of 28 em diameter contains 20 g
of nitrogen compressed by a piston supporting a weight of
75 kg. The temperature of the gas is 17 C. What work will
the gas do, if it is heated to a
temperature of 250 C? What p,105Pa
amount of heat should be sup- 8
plied to it? What distance will
the weight be raised? The process
should be assumed isobaric; the 4
heating of the vessel and the
external pressure should be neg-
lected.
17.6. Upon expansion, the pres-
sure of a gas rose linearly (Fig.
17.6). What work did the gas
perform? By how much did its
internal energy increase? What quantity of heat has been
supplied to it? The gas was monoatomic.
What was the molar heat of the gas in this process? Com-
pare with the specific heats at constant pressure and at con-
stant volume. .
Molecular-kinetic Theory of Gases 49
17.7. The initial gas pressure is 6 X 10
5
Pa and the volume
1 m", Expansion at constant temperature leads to its volume
being increased two-fold. Using numerical methods calculate
the work of expansion of the gas.
Compare with the formula in 27.6 and estimate the error.
17.8. When you have learned to integrate derive the for-
mula to calculate the work of expansion of a gas at constant
'temperature.
17.9. A gas has been subjected to an isochoric-isobaric cycle
1-2-3-4-1 (Fig. 17. 9a). Plot the graph of this cycle in the
p-p, V-T and p-T coordinates.
P
2
V
2 J
:::.
.4 I
1
..::
4
!
I
I
I
I
V
7i
T
Eig. 17.9a.
A gas has been subjected to isothermal-isochoric
-cycle 1-2-3-4-1 (Fig. 17.10a). Plot the graph of this cycle
in p-V, p-p, and p-T coordinates.
17.11. When you have learned to integrate derive the Poiss-
r
on formula for an adiabatic process.
. . 17.12. Express the relations between the pressure and the
temperature and between the volume and the temperature
in an adiabatic process.
17.13. The initial pressure of air is 4.0 X 10
5
Pa, the ini-
,tial volume is 2.0 m". The gaswas compressed adiabatically
:80' that its volume decreased ,to a quarter' of its original vol-
ume. Find the final pressure. Compare with the, pressure
would result from a similar compression of the gas at
'a, constant temperature.
, Which process requires the greater work to be performed
in compressing the gas?
6-0360
50 Problems
17.14. The initial pressure of neon is 2.0 X 10
6
Pa, the
initial volume is 0.4 m". The gas expanded adiabatically so
that its volume increased three times. Find the final pres-
sure. Compare with the pressure that would result from an
expansion at constant temperature. In which process does
the gas perform more work upon expansion?
17.15. Find the degree of compression of air if its tempera-
ture rises from 15C to 700C upon compression. Assume the
compression process to be adiabatic.
17.16. The distance between the atomic centres in a nitro-
gen molecule is 1.094 X 10-
10
ID. Find the moment of inertia
of the molecule and the temperature at which molecu-
lar collisions cause the state of the rotational motion to
change.
17.17. The natural frequency of vibrations of a nitrogen
molecule is 4.4 X 10
14
rad/s. Find the temperature at which
vibrations of the nitrogen molecules are excited.
18. The Second Law of Thermodynamics
18.1. What is the probability of extracting from a pack
of 36 cards (a normal pack with all the 2 ~ 3s, 4's, and
5"s removed) (a) a spade card; (b) a red card; (c) a queen
of any suit?
18.2. What is the probability of extracting from a pack
of cards (a) a court-card; (b) a red court-card?
18.3. What is the probability of extracting two aces in
succession from a pack as in Problem 18.1 (a) if the ace
~
o GTI EE [[J rn
o 0 0 0, 0 0
00 00 00 00
Fig. 18.4.
extracted first is returned to the pack; (b) if the ace extracted
first is not returned?
18.4. Find the mathematical and the thermodynamical prob-
abilities of the five possible distributions of four balls
in two halves of a vessel (Fig. 18.4), assuming them to be
distinguishable.
Molecular-kinetic Theory of Gases
51
4IIl
"-
L..".I
J
1---
'-J.'rt
l .
jL.o'
I
-==
.)
"
l)C
J
....
/
\.
.....,. ...
IP
18.5. Try to generalize the result of the previous problem
to include the case when one part of the vessel contains k
out of n balls (k n) in conditions when
(a) the probabilities of a ball being in the left-hand and
the right-hand parts of the vessel are different;
(b) the probabilities of a ball being in either part of the
vessel are equal.
18.6. Plot the graphs of the functions C: and (Choose
the scale of the z-axis so that the graphs can be conveniently
compared. For instance, for n = 6
you can use the scale 1 : 13 mm
and for n=8 the scale 1 : 10 mm.)
"18.7. A vessel of capacity V
o
con-
tains n molecules. Calculate the
probability of all the molecules
assembling in a part of the ves-
lsel V < yo.
f8.8. Prove the theorem converse
:00the one of 28.8: if in the course
"of the heat exchange between two
bodies contained in a closed and an
isolated system the
i; .entropy rises, then the heat tran-
-sfer will be in the direction from Fig. is.9.
. the heated body to the cold one. "
:.t8.9. Figure 18.9 shows the results of a series of observations
of a migrating Brownian particle. The observations were
at intervals of 30 s, the temperature of water was 25 C,
the radius of a Brownian particle is 4.4 X 10-
7
ID. Measur-
:ing the "steps" of the particle in the scale specified, find the
t aquare of the r.ID.S. displacement for a given time, and cal-
culate the Boltzmann constant and the Avogadro number.
scale is as follows: 1 mm on the graph corresponds to a
of 1.25 um,
Plot the T-S diagram (i.e. the entropy vs tempera-
(a) for an adiabatic process; (b) for an
.. process.
"::,f.;:.----
In this book, as in "Fundamentals of Physics" by B. Yavorsky
P.insky, the is used for (1:), i.e. the number of
of k objects from n,
((fA 4*
52 Problems
Fig. iB.i?
.3
2 ~ ~ ~
I
I
I
18.11. How can you calculate the amount of heat received
(or delivered) by a system, using the T-S diagram?
18.12. Express the amount of heat received by a system in
the course of isothermic expan-
sion in terms of temperature and
entropy. P
18.13. When you have learned
to integrate, calculate the change
in entropy in the course of an
arbitrary quasi-static process.
18.14. Solve the previous prob-
lem for the cases of an isochoric,
an isobaric and an isothermic
process.
18.15. Find the work per cycle
in Problems 17.9 and 17.10.
18.16. Plot the Carnot cycle in the T-S coordinates and
calculate its efficiency.
18.17. Figure 18.17 depicts an idealized cycle of a petrol
internal combustion engine. The segment 1-2 corresponds to
the adiabatic compression of the combustible mixture;
segment 2-3, to the isochoric combustion of fuel in the course
of which the working fluid receives an amount of heat Q;
segment 3-4 corresponds to the adiabatic expansion of the
working fluid; segment 4-1, to the isochoric exhaust of spent
gases. Express the engine's efficiency in terms of the gas
compression ratio x = V
2/V1
Fig. 20.4.
20.6. The theoretical values for the rupture strength obtained
in the previous problem exceed by tens of times the rup-
ture strength of good steels and by many thousands. of times
the rupture strength of real ionic crystals. What is the ex-
planation for this?
20.7'. A steel flywheel is made in the shape of a solid ring
of 40 em external and 30 cm internal diameter. What is
its maximum design speed? At what speed will it fly apart?
20.8. What pressure can a spherical steel container with-
stand, if its internal radius is R and the wall thickness is
d? Do calculations for R == 50 em, d == 5 mm.
20.9. Prove that in similar conditions a cylindrical contain-
er will withstand a pressure half as great.
20.10. A copper rod is fixed between two supports. Its tem-
perature was raised by 50C. What is .the resulting stress
in the rod?
20. t 1. A steel cylinder was cooled in liquid nitrogen (72 K)
and fitted without play into a nickel-chrome steel shell at
room temperature (20C). The internal radius of the shell
is 25 .mm and the external radius is 35 mID. Neglecting defor-
mation of the cylinder find the stress in the shell and the
nature of its deformation.
20.12. Water penetrated into a crack in a rock and froze
there. What is the resulting pressure?
20.13. To determine the volume expansion coefficient of
kerosene, one end of a V-tube filled with it was held at 10C,
Molecular Forces and States of Aggregation 57
and the other at 80 C. The level of liquid in one tube was
280 mm and in the other 300 mm. Find the coefficient.
20.14. What is the number of atoms in an elementary cell
ef a simple cubic lattice?
20.15. What is the number of atoms in an elementary cell
of a face-centered cubic lattice?
20.16. What is the number of atoms in an elementary cell
of a closely packed hexagonal lattice?
21. Liquids
21.1. The viscosity of mercury decreases with the rise in
temperature (Table 21.1a). Check whether relation (34.10)
is valid for mercury. Calculate the activation energy.
Table 21.1a
.. Tempera ture Viscosity n. Temperature t ; Viscosity T},
t , C mPas
c
mPas
0
1.681 50 1.407
10 1.621 60 1.367
20 1.552 70 1.327
30 1.499 100 1.232
40 1.450
2t.2. To what height will water in a capillary of 0.8 mm
diameter rise? Assume the contact angle to be zero.
2t.3. A capillary of 0.8 mm diameter is immersed in water,
and rises 2 em above the water. To what height will the
water rise in it? How can the result be made consistent with
the result of the previous problem?
21.4. There were 100 droplets of mercury of 1 mm diameter
on a glass plate. Subsequently they merged into one big
drop. How will the energy of the surface layer change? The
process is isothermal.
21.5. To pump liquid out of a vessel which is not wetted by
it into a vessel wetted by it one may make use of surface
tension forces (a capillary pump). What will be the speed of
flow of petrol in a capillary of 2 mm diameter and of 10 em
length? The experiment is conducted in conditions of weight-
lessness.
58 Problems
21.6. Compare the effectiveness of a capillary pump for
water at low and high temperatures.
21.7. Liquid in a capillary rises to a height h. What column
of liquid will remain in the capillary, if it is filled in a hor--
izontal position and then
placed in a vertical po-
sition? Assume thecapil-
lary to be sufficiently
long.
21.8. Find the height to
which a liquid rises bet-
ween two long parallel
plates, a distance d apart.
21.9. A drop of water of
0.2 g mass is placed bet- Fig. 21.10.
ween two well cleaned
glass plates, the distance between which is 0.01 cm. Find
the force of attraction between the plates.
21.10. Two soap bubbles with radii of curvature R
1
and R
2
,
where Rcz < R
1
are brought 1 into contact as shown in
Fig. 21.10. What is the radius of curvature of the film be...
tween them? What is the contact angle of the films?
22. Vapours
22.1. Making use of Table 35.1 check the validity of the
Mendeleev-Clapeyron equation for the case of saturated
'Yater vapour. Can saturated vapour be assumed to be an
ideal gas?
22.2. Isn't the result of the previous problem in contradic-
tion with the fact that the ideal gas isochore in p-T coordi-
nates is represented by a linear graph, while the isochore of
saturated vapour is nonlinear (see 35.3, Fig. 35.2)?
22.3. A cylinder closed by a piston contains 8 g of water
vapour at a temperature of 55C. The vapour is compressed
isothermally. What will be its volume when dew begins to
appear?
22.4. A cylinder closed by a piston contains 3.5 g of water
and 2.9 g of water vapour at a temperature of 40 C. The gas
expands isothermally. What will be the volume correspond..
ing to complete evaporation of water?
Molecular Forces and States of Aggregation 59
22.5. The air temperature is 18C, the dew point is 7C.
Find the absolute and the relative humidities of the air.
22.6. During the day the air temperature was 25C, the
relative humidity was 68%. At night the temperature fell
to 11 "C. Will dew precipitate? If the answer is positive,
what will be the precipitation per cubic metre of air?
22.7. 5 m
S
of air with a relative humidity of 22% at 15C
and 3 m
3
of air with a relative humidity of 46% at 28C have
been mixed together. The total volume of the mixture is
8 m", Find the relative humidity of the mixture.
22.8. Making use of the values of the critical parameters of
water ( 35.5) check whether those parameters satisfy the
_ ideal gas law. Explain the result.
22.9. Table 22.9 contains the values of the density of liquid
Table 22.9
Temperature
t, C
I
Densi ty of liquid I
p, kg/m
3
Density
of vapour P.
kg/m
a
Vapour pressure
p, MPa
0 914 96 3.47
10 856 133 4.48
20 766 190 5.70
25 703 240 6.41
30 598 334 7.16
31 536 392 7.32
31.25 497 422 7.38
31.35 464 464 7.39
carbon dioxide, as well as the pressure and the density of its
saturated vapour. Find the critical parameters of this sub-
stance, Plot the density vs temperature graphs.
23. Phase Transitions
23.1. What amount of work is performed when 1 kg of water
turns into steam at 100C? How much energy is spent to
break the bonds between the molecules?
If you sling a thin wire loop around a block of ice and
attach to it a weight of several kilograms then after some time
the wire will pass through the block of ice, but the block
remains intact (Fig. 23.2). Explain this phenomenon.
23.3. 0.2 kg of water vapour at 100C is admitted into a
60 Problems
mixture consisting of 5 kg of water and 3 kg of ice. What
will happen? Neglect radiative losses.
23.4. Solve Problem 23.3 assuming that 1.1 kg of water
vapour was admitted into the mixture.
23.5. 0.5 kg of ice at -15C is thrown into a litre of water
at room temperature
(20 C). What will hap-
pen? Neglect losses.
23.6. Solve Problem
23.5 assuming the
amount of water to
be 31.
23.7. Pure water can
be supercooled down
to -10C. If a small
ice crystal is thrown . Fig...23.2._
into, it immediately
freezes. What fraction of water will freeze? The system is
adiabatically isolated.'
23.8. Water is boiling in a kettle on an electric hot-plate of
800 W power. Find the steam outflow velocity, if the cross
section ~ f the spout is 0.9 cm
2
and the pressure at the output
is normal. The efficiency of the hot plate is 72%.
23.9. Ice at 0 C is enclosed in an adiabatic shell and is com-
pressed to a pressure of 600 atm. It is known thatan increase
in the pressure of 138 atm causes the melting point of ice to
drop by 1 K. Assuming the phase diagram in this _part. to
be linear, find the fraction of the ice that is going to melt.
~ 3 1 To determine the quality of thermal insulation of a
Dewar vessel, it is filled with ice at 0 C. 42 g of ice have
melted in 24 h. Usually liquid nitrogen at 78 K is kept in
this flask. Assuming the quantity of heat entering the flask
to be proportional to the difference in the internal and the
external temperatures of the vessel, find the amount of liq-
uid nitrogen that is going to evaporate in 24 h. The. am-
bient temperature is 20C, the heat of vapourization of Iiq-
uid nitrogen at normal pressure is 1.8 X 10
6
JIK.
23. t t. The triple point of carbon dioxide (C0
2
) corresponds
to a pressure of 5.18 X 10
6
Pa and a temperature of 216.5 K.
In what temperature range can liquid carbon dioxide be
obtained? In what conditions does sublimation take place?
Part Five
ELECTRODYNAMICS
24. A Field of Fixed Charges in a Vacuum
-24.1. Estimate the upper limit of the error made in calcu-
'Iating the force of interaction between charged spherical
'conductors with the aid of the Coulomb law. The radii of
the spheres are r
o
, the distance between their centres is r,
'Carry out the calculations for r 20r
o
Electrodynamics
75
29.2. Solve a similar problem for the case of a tungsten ball.
The magnetic susceptibility of tungsten is Xm = 1.76 X
X 10-
4
-I-
--
7
-,
t
o
500
Fig. 29.9a.
1000 1500 H,A/m
cive force and saturation induction from the gr aph. Calculate
the saturation magnetization and remnant magnetization
Mr.
29.11. For several practical applications the so-called
"dif t i 1" ti hil i I 1 dB d t
I eren ra magne ic permea I tty =-= dH an no
the usual magnetic permeability f.1 =:=: B/ftolI is the para-
meter of interest. Here is the derivative of the field indue-
, Electrodynamics 77
Table 29.10
Magnetic induction Magnetic induction
Magnetic
B, T
Magnetic
B, T
ttcld strength
lower I upper
field strength
lower I upper H, Aim
branch of branch of
H, Aim
branch of branch of
the loop the loop the loop the loop
0 -0.23 0.23 500 0.92 1.15
100 0 0.46 600 1.10 1.19
200 0.23 0.69 700 1.20 1.24
:100 0.46 0.92 800 1.26 1.26
1,.00 0.69 1.08
tion with respect to the field intensity, i.e. the slope of the
graph in Fig. 29.9a. For the purpose of practical calcula-
tions one can assume f.t' = , where IJ.B and IJ.H are
chosen so small that the respective segment of the graph
may be regarded as a straight line. Find the approximate
values of the differential magnetic permeability for the same
values of the magnetic field strength as in Problem 29.9.
30. Electromagnetic Induction
30.1. A plane with a wing span of 18 m flies horizontally at
a speed of 800 km/h. The vertical component of the Earth's
magnetic field strength is about 40 A/m. Find the voltage
across the tips of the wings.
Will a light bulb connected to the wing tips glow?
30.2. A conductor of length 1and mass m can slide without
friction, but with an ideal electrical contact, along two ver-
tical conductors AB and' CD connected through a capacitor
(Fig. 30.2). Perpendicular to the plane of the figure a uni-
form magnetic field of induction B is set up. Find the voltage
across the capacitor plates as a function of h.
30.3. What will be the motion of the conductor of the pre-
vious problem, if a resistor of resistance R is connected into
the circuit instead of the capacitor? Neglect the resistance of
the conductors.
78 Problems
H
h
c
]) _ ~
II
lr
Fig. 30.2.
F
30.4. A rod of length l is perpendicular to the lines of induc-
tion of a uniform magnetic field of induction B. The rod re-
volves at an angular speed ro about an axis passing through
the rod '8 end parallel to the lines of
induction. Find the voltage across A _--11 .........--.. _
the rod '8 ends.
30.5. The length of the conductor in
the diagram shown in Fig. 30.5 is l =
= 20 em, its speed is v = 1 mls and
the resistance of the bulb is R = 1
ohm. A magnetic field with induction
B = 0.5 T is set up perpendicular to
the plane of the diagram. What force
should be applied to the conductor to B
make it move at the speed specified?
30.6. A horizontal flat coil of radius
a made of w turns of wire carrying a
current i sets up a magnetic field. A horizontal conducting
ring of radius r is placed at a distance Xo from the centre of
the coil (Fig. 30.6). The ring is dropped. What e.m.f. will
be established in it? Express
the e.m.f. in terms of the speed.
30.7. A magnet was inserted
into a wire ring connected to
Fig. 30.5.
Fig. 30.6.
a ballistic galvanometer of 30 ohm resistance, this causing
a 20 division deflection of the galvanometer's pointer. What
magnetic flux passes through the pole piece of the magnet,
if the galvanometer constant is 3 X 10-
5
C/div? Neglect the
resistance of the ring and of the leads.
30.8. To find the magnetic field induction in the gap be-
tween the pole pieces of an electromagnet, a coil of 3.2 ern"
Electrodynamics 79
160
Fig. 30.14.
area made of 50 turns of thin wire connected to a ballistic
galvanometer of 100 ohm resistance with a constant of
2 X 10-
0
C/div is inserted into it. When the coil is withdrawn
from the field, the galvanometer pointer moves 20 divisions.
What is the field induction?
30.9. A ring is made from a dielectric with polar molecules.
What will happen to the dielectric, if a magnet is inserted
into it?
30.10. A flat circular coil of 10 em radius has 200 turns of
wire. The coil is connected to a capacitor of 20 1iFand placed
in a uniform magnetic field whose in-
duction decreases at a constant rate
of 10-
2
,Tis. Find the capacitor's charge.
The plane of the coil is perpendic-
ular to the lines of induction of the
field.
30.11. An electric motor works from
an accumulator battery with an e.m.f.
of 12 V. With the rotor stalled the
current in the"circuit is 10 A. What is
the motor's power at nominal load, if
the respective current is 3 A?
3 t l 2 ~ 1200 turns of copper wire are
wound onto a cardboard cylinder 60 em
long and 5 em diameter. What is the
inductance of the coil?
30.13. A current of 500 rnA flows in the coil of the previous
problem. When the current is switched off it vanishes after
a time of 10-
4
s. Supposing the current to decrease linearly',
find the e.rn.f. of self-induction.
30.14. A core with the shape and dimensions-in millimeters
shown in Fig. 30.14 has been manufactured of a ferromagnet-
ic material whose magnetization curve is shown in Fig. 29.9a
(p. 76). One layer of wire of 0.6 mm diameter (including
insulation) was closely wound on the core. Find the induc-
tance of the coil fora current of 2CO rnA flowing in it. Find
the energy of the magnetic field and the energy density.
30.15. What would be the energy of the magnetic field of
the coil of the previous problem, if its core were of a non-
ferromagnetic material? What is the source of excess energy
in the case of a ferromagnetic core?
80 Problems
Fig. 30.16.
~
60
~ J f J ~
30.16. The core and the armature of an electromagnet with
dimensions in millimeters as shown in Fig. 30.16 have been
manufactured from a ferromagnetic material whose proper-
ties have been described in Problem 29.10. What is the force
with which the core attracts the arma-
ture, if the material has been magne-
tized to saturation? What force will
remain active after the current is
switched off?
30.17. A bulb with 1.2 ohm resistance
is connected to an accumulator in se-
ries with a choke. Estimate the in-
ductance of the choke, if the bulb
starts to burn brightly 2.5 s after the
circuit has been closed.
30.18. When you have learned to in-
tegrate, try to analyze the process of
shorting a circuit made up of a coil
and a resistor connected to a power
supply with constant e.m.f., i.e. the dependence of the
current on time. Assume the coil to be without a ferromag-
netic core.
30.19. What time does it take for the current in a circuit
made up of a coil and a resistor to reach 0.9 of its stationary
value?
30.20. According to the formula obtained in Problem 30.18
the stationary value of the current can be reached only after
infinite time. How can this be made consistent with the fact
that actually the stationary value is reached in a finite time?
What are the limits in which the formula obtained in Prob-
lem 30.18 is applicable?
31. Classical Electron Theory
31.1. The coil employed in an experiment similar to that
of Stewart and Tolman has a diameter d = 500 mm and
N == 400 turns of copper wire. The moment the coil stops
it is connected through a pair of sliding contacts to a bal-
listic galvanometer (Fig. 31.1). The total circuit resistance is
R == 50 ohm. The coil is rotated at a constant speed of
n = 6000 r.p.m, and quickly brought to a halt, the charge
Electrodynamics 81
passing through the galvanometer being Q == 1.1 X 10-
8
c.
Find the specific charge of the charge carriers in copper.
31.2. A copper disk of r
o
== 20 em radius revolves in a
vertical plane at 3000 r.p.m. One contact from a sensitive
galvanometer is connected to the disk's axis, the other by
means of a mercury contact to the outer
edge of the disk (Fig. 31.2a). Find the +
voltage.
Will the galvanometer pointer point
in another direction, if the direction of
rotation of the disk is changed?
The Earth's magnetic field is compen-
sated.
3f.3. The dimensions of a copper plate
are as shown in the diagram (Fig. 31.3).
When the longitudinal voltage is L\<p, a
eurrent i flows in the conductor. If a
magnetic field with induction B perpen-
dicular to the plate is established with
the current still flowing, a Hall voltage
of d<PH appears between the upper
and the lower faces of the plate. Find
\he concentration and the mobility of Fig. 31.1.
conduction electrons in copper, if I =
=.60 mm, h == 20 mm, d == 1.0 mm, mY,
... 55 nY, i == 10 A, B == 0.1 T.
31.4. Calculate the Hall constant for silver knowing its
density and atomic mass.
3t.5. The resistivity of indium arsenide is 2.5 X
'x 10-
3
ohm -rn, its Hall constant is 10-
2
m
3/C.
Assuming the
conductivity to be due to carriers of one sign only, find the
-ecncentration .and the mobility of the charge carriers. Com-
pare with Problem 31.3.
3t.6. The safe current in an' insulated aluminium wire of
.1 mm" cross section is 8 A. Find the average drift velocity
of the conduction electrons.
31.7. Find the mean free transit time and the mean free path
the electrons in copper (at room temperature).
The constant of a constantan-copper thermocouple
:-is, 4.3 X 10-
2
mVIK. The resistance of the thermocouple is
'ohm, that of the galvanometer 100 ohm. One junction
'-0360
82 Problems
of the thermocouple is immersediin -melting ice, the. other
in a hot liquid. What is its temperature if the CUrrent' in
the circuit is 56!1A? .
31.9.. The constant of a thermocouple is7.6 (lV/K, the tem-
perature of its cold junction is -80C (dry ice), and that
Jt1-----r
h 1000.........,iJI.
l ~ _ r
I
I" -I
Fig. 31.2a. Fig. 31.3.
of the hot 327C (molten lead). What charge will flow through
the thermocouple, if the hot junction absorbs quantity of
heat equal to one joule? The efficiency' of the thermocouple
is 20%.
31.10. The' Debye otemperature for silver is 213 K, the lat-
tice constant is 2 A. Find the velocity of sound.
31.11. Find the heat flux through a copperplate 5 cm thick,
if a temperature difference of 100 K is maintained between
its ends.
31.12. The dimensions of a brick wall of a living room are:
thickness 40 em, width 5 m and height 2.8 m. A temperature
of- 20C is maintained inside the room, the temperature
outside being -15C. How, much heat is lost through this
wall in 24 hours? -
31.13. 'Find the mean free path of phonons in carbon tetra-
chloride.
31.14. Find the heat conductivity of silver and of mercury"
at room temperature (20C).
32. Electric Conductivity of Electrolytes
32.1. Find the dissociation coefficient of an aqueous potas-
sium' chloride solution with a concentration of 0.1 g/cm'',
if the' resistivity of this solution at 18C is 7.36 X
X 1 ~ 2 ohm-rn. The mobility of the potassium ions .is
Electrodynamics 83
6.7 .X 10-
8
m
2
/(V. s), that of the chlorine ions 6.8 X
X 10-
8
m
2
/ (V. s).
32.2. Find the thickness of a nickel layer deposited on an
article with surface area of ~ O O cm
2
in the course of a 6-hour.
electrolysis at a current of 10.5 A.
32.3. flow much copper will be deposited from a vitriol
solution in 3 minutes, if the current through the electrolyte
changes in accordance with the law i = 6 - 0.03t? All
quantities are expressed in the SI system.
32.4. An electrolytic bath containing a vitriol solution is
connected to a d.c, power supply with an e.m.f. of 4 V and
an internal resistance of 0.1 ohm. The resistance of the solu-
tion is 0.5 ohm, the polarization e.m.I. is 1.5 V. How much
copper will be deposited in one hour?
32.5. Find the minimum e.m.f, of the power supply at
which the electrolysis of acidified water can take place, if
the combustion of 1 g of hydrogen liberates 1.45 X 10
2
kJ.
32.6. A 10 flF capacitor is charged to a voltage of 600 V.
Suppose it is discharged through an electrolytic bath con-
taining acidified water. How much hydrogen will be libe-
rated? What energy can be gained by burning this hydrogen?
How can it be made consistent with the energy conserva-
tion law?
32.7. What energy should be spent to fill a balloon with a
lifting force of 3000 N with hydrogen under normal condi-
tions! How much..will it cost at the price of 4 copecks per
kW. h? Ignore the heating of the solution in the course of
electrolysis.
'33. Electric Current in a Vaeuumand in Gases
33.1. Find the saturation current in a diode with a tungsten
cathode at cathode temperature of 2700 K, if the length of
the cathode filament is 3 em and its diameter is 0.1 mm. The
constant B ~ 6 X 10
5
A/(m
2
K2).
33.2. How will the' saturation current" in' a diode with a
caesium-coated tungsten cathode change, if the cathode tem-
perature is raised from 1000 K to 1200 K? "
3 3 3 ~ In modern diodes the anode is often brought very close
to the cathode so that" their areas are approximately equal.
Assuming the electrons; to leave the cathode with zero velo-
6*
84 Problems
city find the force with which they act on the anode. The
current in the tube is i
sat
= 500 rnA, the anode voltage is
CPa = 600 V.
33.4. Compare the emissivities of a caesium-coated tungsten
cathode at 1000 K and that of a pure tungsten cathode at
360
330
300
270
-.30
-20 -10
90
60
30
o
Fig. 33.5.
10 20
2700 K. Assume the constant B in the Richardson-Dush-
mann equation to be the same for both cathodes.
33.5. Figure 33.5 depicts the grid characteristics of a triode
plotted at anode voltages of 450 V and 600 V. Find the
triode's internal resistance R i in the linear section of the
characteristic and its amplification factor u, i.e. the ratio
of the change in the anode voltage to the change in the grid
Electrodynamics
.voltage which causes a given change in the anode current'.
'33.6. Inside an ionization chamber there are two planar
.alectrodes of 300 cm'' area, 2 em apart. The chamber is
filled with air under normal conditions. At a voltage of
200 V the current is equal to 1.8 flA which is far below the
saturation value.
Find the ion concentration and the ionization
coefficient of air. The mobilities of the ions are:
b; === 1.37 X 10-
4
m
2
/(V. s), b_ = 1.89 X 10-
4
m
2
/(Y. s).
.. 33.7. Oxygen is ionized by gamma-radiation, the ion con-
: centration being 10
15
m-3. Find the conductivity of the gas
; in these conditions. The ion mobilities are: b, === 1.32 X
X 10-
4
m
2
/(Y. S), b_ === 1.81 X 10-
4
m
2
/ (V. s).
33.8. What changes in the current in the range below sat-
uration will take place, if the electrodes of an ionization
chamber are brought closer? How will the saturation current
change? Plot the current-voltage characteristics for some
inter-electrode distance d
1
and for < d
1
" Assume the other
parameters to remain constant.
33.9. The saturation current in an ionization chamber of
0.5 I capacity is 0.02 l-lA. Find the ion generation rate per
second.
33.10. The ionization energy of a hydrogen atom is Cion ==
. = 13.6 eY. Yet the ionization of hydrogen atoms is observed
at temperatures for which the average kinetic energy is
Lmuch less. How can this fact be explained?
\33.11. A high-temperature hydrogen plasma with a tempera-
(. ture of 10
5
K is placed in a magnetic field with induction
',of 0.1 T. Find the cyclotron radii of the ions and electrons
;;(i.e. the radii of the orbits in which the positive ions and
the electrons move).
Mercury flows at a speed of 20 cmls in a pipe with
,: conducting walls of 5 em diameter. The pipe is in the gap
between the pole pieces of an electromagnet, the magnetic
in the gap having induction of 0.6 T. Will the magnetic
f:fteld affect the hydraulic friction coefficient? The conductiv-
ity of mercury is 10
6
ohm-1 m-1"
r33.13. Estimate the effect of the magnetic field in the COI1-
ditions of the previous problem, if a 30% solution of sulphur-
:,'ie acid flows in the pipe. The conductivity of the solution is
74 ohm-1 -m-1.
86 Problems
33.14. Estimate the induction of the magnetic' field of a
pulsar taking into account that for ordinary stars the field
induction is of the order of 10-
6
to 10-
4
T (see Problem 14.21).
33.15. Find the pressure of a pulsar's magnetic field and
compare it with the pressure of gravitational forces (see
Problem 16.9).
Part Six
VIBRATIONS AND WAVES
34. Harmonic Vibrations
34.1. Harmonic vibrations comply to the law
s = 0.20 cos (300t +2)
Find the amplitude, the frequency, the period and the
initial phase of the vibrations *.
34.2. A particle with a mass of 0.2 kg moves according to
the law
s = 0 .08 cos (20nt + )
Find the velocity of the particle, its acceleration and the
acting force, as well as the amplitudes of the respective quan-
tities.
In the conditions of the previous problem find the
kinetic, the potential and the total energy of the oscillator.
34.4. In conditions of the previous problem find the fre-
quency and the period of the variation of kinetic energy.
34.5. A particle vibrates harmonically at a frequency of
0.5 Hz. At the initial moment it is in an equilibrium posi-
tion moving at a speed of 20 cm/s. Write down the equation
of the vibrations.
;,; 34.6. At the initial moment a particle's displacement is
. 4.3 em and its velocity is -3.2 m/s. The particle's mass
is 4 kg and its total energy 79.5 J. Write down the
: Here and below the units used are the 51 units, i.e, the displace-
ment amplitude is expressed in meters, the time in seconds, the
frequency in hertz, the phase in radians.
88 Problems
of the vibrations and find the distance travelled by the
particle in 0.4 s.
34.7. Add up two vibratory motions analytically and using
a vector diagram:
81 = 3 sin (6t + ~ and 82 = 4 sin (6t - ~ )
Find the amplitude of the velocity of the resulting vibrations.
34.8. Find the resulting amplitude and phase of the vibra-
tions
s=Acoswt+ ~ cos(wt+ ~ 1cos(wt+n)+
A ( 3n \
+8 cos rot +-2-)
34.9. Beats result from two vibratory motions:
8
1
= cos 4999nt and 52 = cos 5001nt
Find the period of beat and the "conventional" period of
the almost sinusoidal vibrations.
34. to. A particle oscillates according to the law
s = 4 (cos" O.5t) (sin 1000t)
Expand this motion into a harmonics series and plot its
spectrum.
34. t 1. A particle vibrates according to the law
s == (1 + cos" t + sin1t) sin 500t
Expand this motion into its harmonic components and plot
its spectrum.
34.12. A particle oscillates according to the law
s = (1 + cos
2
t + cos" t) sin 500t
Expand this motion into its harmonic components and plot
its spectrum.
35. Free Vibrations
35.1. A weight of mass m is attached to a spring hanging
vertically, which causes an extension 1. Subsequently the
weight is pulled down a little and let go. What is the natural
frequency of the vibrations?
Vibrations and Waves
89
'35.2. A spherical copper weight of 3.0 em radius submerged
in olive oil hangs from a spring whose elasticity coefficient
'(force constant) is 1.0 X 10
2
N/m (Fig. 35.2). Find the
natural frequency of the oscillatory system, its Q-factor
and the time the oscillations will take to practically damp
-out.
,35.3. A weight of mass 1 kg attached to a spring with
a force constant of 20 N1m is able to oscillate on a horizontal
Fig. 35.2.
r--f
If ====::D
Fig. 35.3.
steel rod (Fig. 35.3). The initial displacement from the posi-
tion of equilibrium is 30 em, Find how many swings the
weight will make before stopping completely. One swing
is the movement from maximum displacement to the equi-
librium position (or back). For numerical calculation put
g = 10 m/5
2
and coefficient of friction JJ, == 0.05.
35.4. A piston of mass m divides a cylinder containing
gas into two equal parts. Suppose the piston is displaced
to the left to a distance x and let go (Fig. 35.4). Assuming
the process to take place at a constant temperature, find
the frequency of the piston's oscillations.
35.5. Solve Problem 35.4 assuming that the process is
adiabatic.
35.6. Mercury fills a glass tube (Fig. 35.6) so that the total
column is 20 em long. The tube is then rocked, so that the
mercury begins to oscillate. Find the frequency and the
period of the vibrations.
35.7. A block of solid oak with dimensions 10 em X 20 em X20
em floats in water (Fig. 35.7). The block is submerged
90 Problems
a little' and let go. Find the frequency and the period 'of
vibrations.
35.8. A pendulum clock is accurate on the Earth's surface.
How slow will it be, if it is lifted to the hundredth floor
of a sky-scraper? The height of a storey is 3 ID.
35.9. The period of a pendulum whose suspension is sta-
tionary relative to the Earth's surface is 1.50 s. What will
J
I
I
I
I
I
I
Fig. 35.4. Fig. 35.6.
~ c m ~
Fig. 35.7.
1--
its period be, if it is placed in a car moving horizontally
with an acceleration of 4.9 m/s
2
? What will be the change
in the pendulum's angle of
equilibrium?
35.10. A mathematical pen-
dulum 1 m long is deflected
from the vertical by an angle
of 40 and let go. Find the
period of oscillations using
numerical methods.
What will be the error, if in
this case we use the formula
for small oscillations?
35.11. The period of a simple
pendulum for large deflection angles may be determined
from the approximate formula
T=2n V; (1 ++sin
2
~ o )
Compare with the result of numerical calculations for the
previous problem.
Vibrations and Waves
35.12. A uniform rod of length l oscillates about an axis
passing through its end.' Find the oscillation period and the
reduced length of this pendulum. .
35.13. A physical pendulum shown in Fig. 35.13 is made
up of a rod 60 cm long with, mass 0..50 kg and a disk of
3.0' em radius with mass 0.60 kg. Find the pe-
riod of this pendulum.
35.14. An oscillatory circuit is made up of a 100
~ l capacitor and a 64 fJ.H coil with resistance
of 1.0 ohm. Find the natural frequency, the pe-,'
riod of oscillations and the Q-factor of the cir-
cuit. .
36. Forced Vibrations. Alternating Current
36.1. An iron ball of 0.8 kg mass hangs on a
spring with a force constant of 10
3
N/m. An al-
ternating magnetic field acts on the ball with a
sinusoidal force whose amplitude is 2.0 N. The
Q-factor of the system is 30. Find the amplitude Fig. 35. 13.
of forced vibrations for to == roo/2, co = roo, ro =
:;= 2{t)o.
36.2. Plot the resonance curve for the amplitude of the
velocity in the previous problem.
36.3. A weight of 0.5 kg mass is suspended from a spring,
causing it to extend by 5 mm. When the system is dis-
Fig. 36.4.
placed from its equilibrium position and then set free, its
natural vibrations continue for 3.5 s. Find the amplitude
of the system at resonance. What will happen in the case of
resonance?
36.4. A radio receiver receives radio telegraph signals in
Morse code in the form of sinusoidal wave packets (Fig. 3.6.4).
The inductance of the circuit is 100 uH, the capacitance
92 Problems
R
c
Fig. 36.11a.
is 250 ~ l and the resistance is 0.2 ohm. Find the interval
between the impulses 't'sp needed to prevent two adjacent
signals from merging.
Assuming the duration of the "dot" signal to be l"dot ==
== 1.51'sp and that of the "dash" 'rdash == 4.51'sp find the
maximum amount of information that can be transmitted
per unit time.
36.5. Derive the expression for the inductive reactance
and for the phase shift in an a.c. circuit containing a coil
of zero resistance.
36.6. Derive the expression for the
capacitive reactance and for the phase
shift in an a.c. circuit containing a
capacitor.
36.7. Plot a vector diagram for a cir-
cuit containing a coil and a resistor
connected in series, and find the im-
pedence of this circuit. Find the
phase shift.
36.8. Do the same for a capacitor connected in series with
a resistor.
36.9. Do the same for a capacitor connected in parallel
with a resistor.
36.10. Express the inductance of a series connected circuit
made up of a resistor, a coil and a capaci tor in terms of
its Q-factor and the frequency ratio y == 00/000.
36.11. Plot the vector diagram for currents in the circuit
shown in Fig. 36.11a and find the current in the unbranched
section of the circuit. What is the condition for the current
in the unbranched section to be a minimum? What is the
phase shift between the voltage and the current in general,
and at resonance?
36.12. The capacitance in the circuit shown in Fig. 36.113
is 20 flF, the inductance is 0.2 H and the resistance is 5 ohm.
What power is consumed in this circuit, if the voltage
at its terminals is II = 312 cos 314t?
36.13. What is the frequency in the circuit with parameters
as specified in the previous problem when the current in
the unbranched section of the circuit is a minimum? What
power will be consumed, if the voltage amplitude remains
the same?
Vibrations and Waves 93
36.14. Prove that an electrodynamic wattmeter measures
active power P == IV cos cp in an a.c, circuit.
36.15. The reading of the wattmeter on a panel is power
12 kW. The reading of the voltmeter is voltage 380 V, of
the ammeter, current 36 A. What is the phase shift in the
circuit? What are the impedance and the ohmic resistance
of the load?
36.16. The starting voltage of a neon glow lamp is 80 V,
the quenching potential is 70 V. A voltmeter in an a.c.
circuit measures a voltage of 60 V. Will
the lamp glow in this circuit?
36.17. The breakdown voltage indicated
on. a capacitor is 300 V. Can it be used
in a 220 V a.c, circuit?
36.18. A two-wire power line transmits
100 MW power. Its power factor is 0.87
and its resistance 8 ohm. What is the
transmission voltage, if the power loss
is 2%?
36.19. The primary of an arc welding
transformer has 120turns of wire of 20 mm"
cross section; its resistance is 8 X 10-
2
Fig. 36.21.
ohm. The current at nominal load is 40 A.
How many turns are there in the secondary and what is the
cross section of the wire in the secondary, if the transforma-
tion ratio is k==220/60?
Assuming the windings to be of a single-layer type, find
the resistance of the secondary.
Neglecting losses due to magnetic reversal and Foucault
currents (i. e. losses in the steel core), find the power loss
due to the heating of the windings and the efficiency of
t.he transformer. The transformer power rating is 8 kW.
96.20. Show that the "losses in the steel core" are practically
equal to the open-circuit power consumption of the transformer.
36.21. Explain, why a copper ring "floats" in air when
an a.c, current is supplied to the winding (Fig. 36.21).
36.22. The primary of a transformer is at a voltage of
220 V drawing a current of 1.5 A. The secondary feeds an
incandescent lamp with a current of 20 A at a voltage of 12 V.
The transformer's efficiency is 91%. Find the power
f4lctor at this load.
94 Problems
37. Elastic Waves
37.1.. Compare the velocity of sound in a gas with the root-
mean-square velocity of its molecules. Do the calculations
for a diatomic gas.
37.2. The sound velocity in a duralumin rod is 5.1 X 10
3
mls
while the density of the material is 2.7 X 10
3
kg/m", Find
the Young modulus.
37.3. An 'observer at a distance of 800 m from a sound
source heard first the sound signal which travelled through
water and 1.78 s later-the signal which travelled through
air. Find the velocity of sound in water and the compres-
sibility of water. The air temperature is 17C.
37.4.- The velocity of sound in oxygen in normal 'conditions
is 317.2 m/s. Find the Poisson ratio.
37.5. Awave with a frequency of 440 Hz travels in a cylin-
drical .tube containing' air. The wave intensity is 1.2 X
X 10-
2
W/m
2
Find the energy density and the amplitude
of oscillations, .if the air temperature is 27C and its pres-
sure is 780 mmHg: .
37.6. A sound source of small dimensions radiates waves
at frequency of 500 Hz. The power of the source is 5 W, the
air temperature is 0 c, the pressure is 1.01 X 10
5
Pat
What are the amplitudes of the sound wave 10 m and 15 m
away from the source? .Ncglect attenuation.
37.7. Compare the intensity levels of the sound wave in
the previous .problem.
37.8. The intensity of a sound wave 20 m away from the
sound source is 3.0, nW1m
2
Find the' intensity' of the wave
32 ill away from the source, if the half-thickness for .sound
of this frequency is 120 m,
Compare. the' intensity levels.
37.9. Find the relation between the linear absorption coef-
ficient of .sound wave f.1, and the half-thickness L.
37.10. Sound travels in 'acylindrical tube 80 cm long.
The linear absorption coefficient is 1.2 X 10-
2
m-1. Com-
pare the sound intensity levels at the entrance and the exit
of the tube.
37.11. Two tuning forks with natural frequencies of 340 Hz
move relative to a stationary observer. One fork moves away
from the observer, while the other moves towards him at the
Vibrations' and' VVaves 95
same speed. The observer hears beats of frequency 3 Hz.
Find the speed of the tuning forks assuming the velocity of
sound in air to be 340 m/s.
37.12. Two trains move towards each other at a speed of
80 km/h relative to the Earth's surface. One radiates a
520 Hz signal. What frequency will the observer on the
other train hear?
How will this frequency change when the trains pass one
another? .
37.13. The equation of a plane sound wave is
s = 6.0 X 10-
6
cos (1900t +5. 72x)
Find the frequency, the wavelength and the velocity of
the wave. Compare the wavelength with the amplitude of the
oscillations and the wav"e velocity with the amplitude 'of
the velocity of the oscillations.
37.14. Under the conditions of the previous problem, find
the distance between any two nearest points of the wave
oscillating in opposite phase. What is the phase shift be-
tween the oscillations of two points 37 em apart in the direc-
tion of the sound ray?
37.15. Find the minimum and the maximum wavelengths
,,'of sound in air that a man is able to hear (see 58.1). How
will this range change, if the sound travels in water? ..
. 37.16. A ceramic ultrasonic transducer is immersed in castor
', oil. What fraction of the energy is transmitted to -the .oil?
The density of ceramic material is 2.8 X 10
3
kg/m", the
: velocity of sound in it is 6.2 X 10
3
m/s. .
. 37.17. Solve Problem 37.16 for a magnetostrictive nickel
:. 'j .transducer working in water.
. .tS.Why when operating in ultrasonic defectoscope is
j. care always taken to see that there is a' film of oil between
the transducer and the part under inspection?
37.19. An ultrasonic defectoscope operates at a frequency
.of 1.2 MHz and radiates pulses of the order of 60 periods of
oscillations. What is the resolution of 'the instrument? We
Shall define resolution as the minimum distance of the defect
r from .the part's. surface . which may be determined: with.
the aid of the Instrument.
96 Problems
38. Interference and Diffraction
Fig. 38.6.
a-
1
I
38.1. Prove that when a wave is reflected by a medium of
higher characteristic acoustic impedance, a standing wave
displacement node is formed at the boundary, and that
when it is reflected by a medium of lower characteristic
acoustic impedance, an antinode is formed.
38.2. A quartz plate (the X-cut) 7 mm thick serves as the
radiating element of an ultrasonic transducer (see 58.3,
Fig. 58.4). What is the fundamentalop-
erating frequency of the ultrasonic gen-
erator?
Will the transducer's frequency change,
if the air gap is filled with oil?
38.3. A magnetostrictive transducer op-
erates at a frequency of 25 kHz. Find the
thickness of the pack of nickel plates
of the transducer.
38.4. An organ pipe 17 em long open
at one end radiates a tone of 1.5 kHz
at an air temperature of 16C. What
harmonic is this? What is the funda-
mental frequency of the oscillations?
38.5. Two organ pipes closed at both
ends serve as sound sources, the result-
ing. beat frequency being 2 Hz. The length
of both pipes is 24.0 em, the temperature of air in one tube
is 17 "C. What is the temperature of air in the other?
38.6. To measure the speed of sound in air the apparatus
shown. in Fig. 38.6 is used. A sound source with a frequency
of 1.20 kHz is placed close to the upper end of a narrow
tube. By moving the left-hand vessel we cause the level of
the liquid in the narrow tube to change. Acoustic resonance
is observed when the height of the air column is 6.8 em, 20.6
em and 34.8 em. Find the velocity of sound and estimate
the error of the value obtained.
38.7. The transducer of an ultrasonic defectoscope of 12 mm
diameter operates on a frequency of 1.2 MHz. What is the
angular width of the principal diffraction maximum, if
the ultrasonic wave travels in castor oil?
Vibrations and Waves 97
38.8. What is the diameter of the transducer of an echo-
sounder operating on a frequency of 50 kHz in sea water, if
the angular width' of the principal maximum is about 60?
39. Electromagnetic Waves
39.1. Find the wavelengt.hs in air and in transformer oil,
if the transmitter frequency is 60 MHz.
39.2. Find the fundamental frequency radiated by a half-
wave antenna and the frequencies of the harmonics.
39.3. A half-wave antenna 0.5 m long is immersed in ethyl
alcohol. What is the wavelength of the electromagnetic
waves outside the vessel (in air)?
39.4. A plane electromagnetic wave
E, = 200 cos (6.28 X 10
B
t +4 .55x)
is completely absorbed by the surface of an object" perpendic-
ular to the x-axis. In what medium does the wave propagate?
What pressure does it exert on the object? What energy is
absorbed by 1 m
2
of the surface per second? .
39.5. The current amplitude in a half-wave antenna is
0.5 A. What is the radiation power? What is the equivalent
resistance of this vibrator? For simplicity of calculations
assume the current to be the same in every' point.
39.6. An electron bunch circulates in the storage ring of a col-
. liding-beam accelerator. The current is 500 rnA, the speed
of the electrons is 0.99 of the speed of light. What is the
power of synchrotron emission?
39.7. An oscillator radiating electromagnetic waves 25 m
long is required to transmit with minimum distortion sound
signals with frequencies "up to 2 kHz. Find the parameters
of the resonant circuit.
39.8. Derive the relation between the frequencies of a wave
in two reference frames (the Doppler effect, see 59.8)
and the relation between the values of the cosines of the
angles the ray makes with the direction of the source's
motion (in both reference frames).
39.9. Try to derive the expression for the relativistic longi-
tudinal Doppler. effect from the principle of relativity' and
the classical Doppler effect, without using the Lorentz trans-
formation.
7-03GO
98
Problems
39.10. Find the Dopplerbroadening of spectral lines in the
spectrum of a "white dwarf" (surface temperature about
10 000 K). Compare with the gravitational red shift of the
spectral lines, assuming the mass of the "white dwarf" to
be equal to that of the Sun and its radius to be 0.01 of the
Sun's radius.
39.11. The spectrum of excited singly ionized helium atoms
includes a line of 410 nm wavelength. A beam of such ions
leaves a cyclotron with an energy of 40.0 MeV. Find the
Doppler shift of this line, if it is observed at an angle of
30 to the beam's direction.
39.12. The observations of the spectral line of hydrogen
H Ii with a wavelength of 4861.33 1\ in the solar spectrum
lead to the discovery of a difference in the wavelength of
this line on the opposite fringes of the solar disk equal to
0.065 1\. Find the period of rotation of the Sun about its axis.
39.13. In astrophysics frequent use is made of the quantity
Z == (A - Ao)/A
o,
equal to the relative variation of a spec-
tral line wavelength. Here "'-0 is the wavelength emitted by
the source and A is the wavelength observed. Express this
quantity in terms of the radial velocity of the source in
the reference frame of the observer.
39.14. The relative variations of spectral line wavelengths
measured for an optical galaxy, for the radiogalaxy 3C295
and for a quasar (quasi-stellar radio source) 3C9 were Zl ==
=== 0.034, Z2 == 0.46 and Z3 == 2. Find the ratios of the radial
velocities of these sources to the velocity of light; find
the velocities of the sources. .
39.15. The Doppler effect made possible the discovery of the
so-called spectroscopic binaries. The spectral lines of these
stars are periodically doubled. This leads to the conclusion
that the source is constituted of two stars revolving about
a common centre of mass. The maximum distance between
the components of a periodically splitting hydrogen line
of 4 3 4 ~ 7 A wavelength in the spectrum of one such star
is 0.53 A. Find the projection of the orbital speeds of the
two stars on the line of sight.
39.16. The maximum relative shift of spectral lines of
a binary star is 2.08 X 10-
3
, the' period of line-splitting
being 3 days 2 hours and 46 minutes. Both stars being
identical find their masses and the distance between them.
Vibrations and Waves 99
40. Interference and Diffraction of Light
40.1. In Young's arrangement (see 61.5, Fig. 61.5) the
distance between the slits is 1.5 mnl and the distance of the
screen from the slits is 2 Ill. The slits are Illuminated by
a source provided with a red filter (A ::::::: 687 nm). Find the
distance between the interference fringes .
. How will the distance between the fringes change, if the
red filter is replaced by a green one (Iv == 527 nrn)?
40.2. How many interference maxima will be observed, if
Young's arrangement described in the previous problem is
illuminated with white light?
The wavelength limits are Iv
r
==
= 690 nm, Iv
v
== 420 nm. What
is the distance on the screen
between the red and the violet
maxima?
40.3. A thin wire of 0.05 mm Fig. 40.3a.
diameter is placed between the
edges of two well-polished planar plates; the opposite
edges of the plates are tightly pressed together (Fig. 40.3a).
Light falls perpendicularly onto the surface. An observer
sees interference fringes on the 10 em long plate, the dis-
tance between them being 0.6 mID. Find the wavelength.
40.4. When the mirror in the Michelson interferometer is
displaced, the interference pattern is shifted py 100 bands.
The light used in the experiment is of 5460 A wavelength.
What is the mirror's displacement?
40.5. Cylindrical tubes 10 em long, each closed at both
ends by transparent plane-parallel plates, were placed in the
path of the two light rays in the Michelson interferometer.
First the tubes were evacuated, then one was filled with
hydrogen, and the interference pattern was observed to shift
by 47.5 fringes. What is the refractive index of hydrogen?
The light in the experiment was of wavelength 590 nm,
40.6. During the quality control of surface finish with
the aid of the Linnik interference microscope, a scratch was
observed on the surface which produced a curvature of 2.3
fringes in the pattern of interference fringes. Green light
of 530 nrn wavelength was used in the test. Find the depth
of the scratch.
7*
100 Problems
40.7. The yellow sodium line consists of twg components
with the wavelengths of 5890 A and 5896 A. When- the
Michelson interferometer is illuminated with this light and
the mobile mirror is moved, the interference pattern period-
ically vanishes and then appears again. What is the cause
of this phenomenon?
40.8. Green light of 500 nm wavelength falls on a slit 8 um
wide. Find the angles at which the first- and the second-order
minima are observed.
40.9. A diffraction grating has 400 slits per 1 mm. A mono-
chromatic red light of 650 nm wavelength falls on the grating.
What is the angle at which the first-order maximum is vis-
ible? What is the total number of maxima produced by this
grating?
40.10. Find the wavelength of monochromatic light falling
normally on a diffraction grating with a grating constant of
2.20 f.LID, if the angle between the directions of the first-
and the second-order maxima is 15.0.
40.11. Light of 530 nm wavelength falls on a grating with
a grating constant of 1.50 J.1m, and with a total length of
12.0 mm. Find the angular width of a principal maximum
and- the resolving power of the grating.
40.12. What should be the length of a diffraction grating
with a density of ruling of 3QO slits per 1 D!m to resolve
two spectral lines of 6000.00 A and 6000.50 A wavelength
in the second-order spectrum? What should be the length
in the spectrum of the highest order?
40.13. The grating constant of a diffraction grating is
0.01 mm, the total number of slits is 990. Are we able to
see in the first-order spectrum the two components of thoe
sodium ye!low line doublet with wavelengths of 5890 A
and 5896 A? What is the angular spacing between these
maxima in the second-order spectrum?
40.14. A plane wave falls on a diffraction grating with
a grating constant do at a glancing angle a. Prove that the
resulting diffraction will be the same as when the wave
falls normally on a grating with a grating constant of d =
= do sin t'%.
40.15. A narrow beam of X-rays falls at a glancing angle
of 20 on a diffraction grating with a constant of 2.0 um.
The first diffraction maximum is observed at an angle of
Vibrations and Waves 101
12' to the direction of the beam. Find the wavelength of
the X-rays.
40.16. A parallel beam of X-rays with a wavelength of
1.47 Afalls at a glancing angle of 313' on a face of a rocksalt
crystal. Find the spacing between the atomic planes of the
crystal, if the second-order diffraction maximum is observed
at this angle.
41. Dispersion and Absorption of Light
41.1. A proton beam with an-energy of 10.0 GeV enters a Ce-
renkov counter made of rock salt. Find the deflection angle
of the threshold red (0.67 urn) and of the violet (0.40 urn)
rays from the cone axis.
41.2. A beam of relativistic electrons in a Cerenkov counter
filled with water radiates in the violet spectral range inside
a cone with an aperture of 8220'. Find the kinetic energy
of the electrons.
41.3. Assuming the free electron concentration in a plasma
to be no, find the dependence of the plasma's dielectric
constant on the frequency of an electromagnetic wave, ne-
glecting the interaction of the electromagnetic wave with
the positive ions.
4t.4. Express the group velocity of light in terms of the
velocity of light in a vacuum, the refractive index and the
derivative of the refractive index with respect to frequency.
41.5. Prove that in the normal dispersion range the group
velocity is less than the velocity of light in a vacuum.
41.6. Find in the optical range the refractive index of a plas-
ma and the phase and the group velocities of a wave in
a plasma.
41.7. The phase velocity of light in a plasma exceeds the
velocity of light in a vacuum. Is this not in contradiction
with the fundamental principle of the theory of relativity
that the velocity of light in a vacuum is the maximum pos-
sible velocity?
41.8. Can Cerenkov radiation be induced in a plasma?
41.9. Find the free electron concentration in the ionosphere,
if the refractive index of the ionosphere for radio waves
of 3.0 m wavelength is 0.90.
102 Problems
41.10. In discussing the action of X-rays of sufficient energy
the bonding energy of the electrons in the lattice can, be
neglected, and the electrons may be considered free. Cal-
culate in this approximation \he refractive index of alu-
minium for X-rays of 0.50 A wavelength.
41. t 1. Find the reflection coefficient for optical waves at
the vacuum-plasma interface. (Take into account that in
the optical range the approximate equality n + 1 ~ 2 is
valid to a high degree of accuracy.)
41.12. Compare the reflection coefficients for red and for
violet light at the air-molten quartz interface. The light is
perpendicular to the interface.
41.13. The distance between the toothed wheel with 720
teeth and the mirror in the Fizeau experiment is 7.0 km.
Two consecutive rotation speeds of the wheel for which the
light disappeared were 283 r.p.s. and 313 r.p.s. Find the
velocity of light.
41.14. Prove that in a plasma the relation uU === c
2
holds,
where u is the phase and U the group velocity of an electro-
magnetic wave.
41.15. Find the group and the phase velocitiy of l ~ t
in sylvite for the wavelength of 5086 A in the 5461-4861 A
spectral interval.
41.16. Two plates, one 3.8 mm thick and the other 9.0 mm
thick, were manufactured from the same material. The plates
are placed in succession in a narrow beam of monochromatic
light with the result that the first plate transmits 0.84
and the second 0.70 of the light flux. Find the absorption
coefficient and the half-thickness of the material. Neglect
secondary reflections.
41.17. A point light source is in the centre of a spherical
layer of a substance, with an internal radius r
1
and an exter-
nal radius r
2
The refractive index and the absorption coef-
ficient of the substance are known. Find the transmittance
of the layer of substance. Neglect secondary reflec-
tions.
41.18. A light filter 5 mm thick has a variable absorption
coefficient which depends on the wavelength according to
the law f.t ~ ~ o + ex (Ao - "-)2, where ex == 5.6 X 10
20
m:",
Ao=== 5000 A, fJ-o === 4 m". Find the transmittance of the
filter for the wavelength A
o
and its transmission band width
Vibrations and Waves 103
The transmission band width includes all wavelengths for
which the transparency of the 'filter is not less than half of
the resonant transparency. Neglect the reflection of light
from the surfaces.
How many half-thicknesses are there in a plate which
. reduces the intensity of a beam 60 times? .
.u.m"
,I""',
i ..
A,A
46
50
54
S8
Fig. 41.20.
41.20. Figure 41.20 shows the dependence on the wavelength
of the absorption coefficient of lead for gamma-rays from
a radio-active source. What is the maximum half-thickness of
lead for gamma-rays?
42. Polarization of Light
42.1. Natural light falls on a system of two polaroids the
J.. 'angle between whose optical axes is 45. By how much
will the intensity of the light be reduced? 10% of the light
is lost in each polaroid. Neglect losses due to reflection.
42.2. If a third polaroid with its optical axis at an angle
of a to the optical axis of the analyzer is placed between
two crossed polaroids, the field of view brightens. Find
the intensity of the transmitted light. Neglect losses due
to reflection and absorption. What is the angle ex of maximum
brightness?
42.3. The ordinary and the extraordinary rays are obtained
by decomposition of a given beam of natural light. Will
104 Problems
there be an interference pattern of minima and maxima,
if both rays are combined?
42.4. Find the thickness of a calcite plate which creates
in yellow light a phase difference of n/2 between the ordinary
and the extraordinary rays of light at a wavelength of 5893 A
(the quarter-wave plate). What will be the phase shift for
violet light (4047 A) passing through this plate?
42.5. To compensate the phase shift introduced bya calcite
quarter-wave plate a quartz quarter-wave plate was placed in
the path of a light beam. Compare the thicknesses of both
plates. The light used in the experiment was in the green
part of the spectrum (5086 A).
42.6. A glucose solution with a concentration of 2.8 X
X 10
2
kg/m" in a glass tube rotates the polarization plane
of light passing through it by 64. Another solution in the
same tube rotates the polarization plane by 48. Find the
concentration of the second solution.
42.7. What must the thickness of a quartz plate placed be-
tween crossed polaroids be to make the field of view turn
red? blue? The polarizer is illuminated with white light.
43. Geometrical Optics
43.1. A. plate of 4.0 em thickness was cut out .of calcite
perpendicular to the optical axis. A narrpw beam of natural
yellow light with a wavelength of 5893 A falls on the plate
at an angle of 60. Find the spacing between the ordinary
and the extraordinary rays at the point of exit from the
plate to the air.
43.2. A ray of white light falls on a prism of crown glass
perpendicular to its face. Find the angle of refraction of
the prism for which the red rays pass out of it into air while
the violet rays experience total internal reflection.
43.3. The sides of an isosceles right prism are coated with
a reflecting coating.' A ray of light falls on the hypotenuse
at an arbitrary .angle. Prove that the ray leaving the prism
is parallel to the incident ray.
43.4. A point source of light is placed on the bed of a pond
80 cm deep. Find the diameter of the illuminated circle on
the surface of water.
Vibrations and Waves 105
Fig. 43.10.
- --
-- -- --
43.5. A prism of flint glass with an angle of refraction of
30 is placed in water. At what angle should a ray of light
fallon the face of the prism so that. inside the prism the ray
is perpendicular to the bisector of the angle of refraction?
Through what angle will the ray turn after passing through
both faces of the prism?
43.6. A lens made of crown glass has a focal power of 8 diop-
ters in air. What will its focal power be in water? In hydro-
gen sulphide (n = 1.63)?
43.7. A system is made up of two thin converging lenses
arranged perpendicularly to their common axis. Where is the
image of the anterior focus
of the left-hand lens? Trace
the rays.
43.8. Prove that the lens
focal power of a system made
up of two tightly packed
thin lenses is equal to the
sum of the focal powers of
each of them.
43.9. What .is the experimental procedure to find the focal
power of a diverging lens?
43.10. A convexo-concave lens of crown glass has the radii
of curvature equal to 1 m and 12 em. What is its focal
power? The lens is placed horizontally and filled with water
,,(Fig. 43.10). How will its focal power change?
::. 43.11. Derive the formula for the focal power of a plano-
convex lens by tracing the rays passing through it.
#43.12. Two thin lenses with focal .lengths of-_u/t-.=- 7 em
Iand /2 == 6 em are placed at a distance d = 3 em apart.
is the distance of the focus of the system from the
lens? The system is a centred one.
Two thin converging lenses are placed on a common
;:: axis so that the centre of one of them coincides with the
r'l'::.focus of the other. An object is placed at a distance twice
focal length from the left-hand lens. Where will its
>image be? What is the lateral magnification? The focal
.power of each lens is Q). .
A converging bundle of light rays in the shape of
ia cone with the vertex angle of 40 falls on a circular dia-
i,phragm of 20em diameter. Alens with a focal power of 5 diop-
t06
Problems
ters is fixed in the diaphragm. What will the new cone angle
be?
43.15. Compare the longitudinal and the lateral magnifica-
tions of a thin lens. Consider the case of small longitudinal
dimensions of the object.
43.16. A ball is placed at a distance double the focal length
from the lens on its axis. What will be the shape of its image?
M
o
a F
Fig. 43.19a. Fig. 43.20a.
43.17. Find the magnitude of the chromatic aberrations
of a lens made of flint glass, if the radii of curvature of both
its surfaces are 0.5 m. We define the chromatic aberration
as the difference between the focal lengths for the red and
oA'
"A
Fig. 43.22a.
the violet rays. Find the ratio of the chromatic aberration
to the average focal length of the lens. .
43.18. A concave spherical mirror with a radius of curvature
of ~ m is filled with water. What is the focal power of
this system?
43.19. A ray of light falls on a concave spherical mirror,
as shown in Fig. 43.19a. Trace the path of the ray further.
43.20. A ray of light falls on a convex mirror, as shown
in Fig. 43.20a. Trace the path of the ray further.
Vibrations and Waves 107
Figure 43.21a shows the optical axis of a lens, the
<point source of Iight A and its- virtual image A'. Trace the
to find the position of the lens and of its focuses. What
r, type of lens is it?
.22. Solve., the problem similar to the previous one using
43.22a.
-43.23. Prove that a parabolic mirror is free from spherical
"liberrations.
Can a 'strictly parallel bundle of light rays be ob-
tained with a parabolic mirror?
44. Optical Instruments
'44.1. A monochromatic light source of 555 nrn wavelength
radiates a total light flux of 1200 Im. What is the radiant
power? What should be the radiant power in order to obtain
the Same light flux at a wavelength of 480 nm? 600 nm?
,44.2. A monochromatic point source of 520 nm wavelength
has a luminous intensity of 20 cd. What are the amplitudes
of the electric field intensity and of magnetic field induction
at a distance of 50 cm from the source?
44.3. A cylindrical hall of diameter D is illuminated by
Ii lam-p fixed at centre of the ceiling. Compare the minimum
illuminance of the wall and of the floor. The height of each
wall is h.
44.4,. A round table of radius r is illuminated by a lamp of
luminous intensity (candlepower) I hanging above its centre.
What should be the height of the lamp above the table for
the illuminance of the table fringe to be maximum? What
is Its value? What is the illuminance at the centre of the
table in these conditions?
A point light source illuminates a screen, the maximum
illuminance being Eo. How will the illuminance of this point
change' if a large plane ideally reflecting mirror is placed
behind the source half way between the lamp and the screen?
44.6. Street lamps of 10 em diameter and of 1.8 X 10
5
cd/m''
brightness are suspended at a height of 12 ID, the distance
between them being 40 m. Find the illuminance under each
lamp and midway between them.
44.7. A point light source of 10 cd luminous intensity is
in the centre of a concave mirror with a radius of curvature
108 Problems
of 40 em and of 20 em diameter. It illuminates a screen
at a distance of 2 m from the source. What is the maximum
illuminance of the screen? How will the illuminance change,
if the mirror is withdrawn?
44.8. A lens of diameter D and a focal length f projects
a small object placed at a great distance from the lens onto
a screen. Show that the illuminance of the image on the
screen will be proportional to the luminance and to the
aperture ratio of the lens. (The aperture ratio is the square
of the ratio of the lens' diameter to its focal length.)
44.9. A screen is 1 m away from a light source. A diverging
lens of -5 diopters focal power is placed between the source
and the screen so that the position of the light source coin-
cides with that of the virtual focus. How will the illuminance
of the screen on the optical axis of the system change?
44.10. The distance of maximum visual activity for a near-
sighted eye is 9 em. What glasses should be worn to correct
the eyesight?
44.11. Show that the estimate of the resolving power of
the eye based on diffraction from a single slit practically
coincides with the minimum angle of view estimated on the
basis of the minimum distance between nearest elements of
the retina. The diameter of the pupil in conditions of good
illuminance is 2-3 mm.
44.12. The length of a microscope tube is 16 em, the focal
power of the objective is 185 diopters and of the eyepiece
50 diopters. Find the angular magnification of the instru-
ment.
44.13. The numerical aperture of a microscope in air is
0.46. What is the minimum .distance this instrument can
resolve?
44.14. A telescope objective of 2 diopters focal power
and of 10 cm diameter is part of a Keplerian telescope whose
magnification is 12. Find the focal power and the diameter
of the eyepiece, as well as the minimum angle this telescope
can resolve.
What are the dimensions of objects on the, SUD this
telescope can resolve?
44.15. Galileo's telescope is a telescopic system in which
a long-focus convex lens serves as the objective, and a short-
focus concave lens serves as the eyepiece. The posterior vir-
Vibrations and Waves 109
tual focus of the eyepiece coincides with the posterior focus
of the objective. Trace the path of rays in this system and
find the angular magnification.
44.16. Compute the minimum distances between two points
on the Moon and on Mars which can be resolved with the aid
of a reflecting telescope having a mirror of 6 m diameter.
The shortest distance between the Earth and Mars (the
"great opposition") is 5.6 X 10
10
m.
44.17. A radio telescope of about half a kilometer diameter
operates in the centimeter wave range of the hydrogen
spectrum (21 em). Estimate its resolving power. Compare
with the resolving power of an optical telescope with a three
meter mirror.
44.18. At present the best sprinters run 100 m in 10 s.
What is the appropriate exposure in making snapshots, if the
blurring on the negative must not exceed x = 0.5 mm? The
snapshots are made at a distance of ~ 6 ill, the focal power
of the photographic objective is (J) == 20 diopters.
44.19. Landscape shots are made with a camera having an
objective of focal power 7.7 diopters. The camera is focused
on objects 12 m away from it. It is desired to obtain a suffi-
ciently clear image of objects within a distance of 3 m in
front and behind-their blurring on the negative should
not exceed 0.2 mm. What should be the setting (i.e. diameter)
of the diaphragm? What will the aperture of the objective
be at this setting?
44.20. A camera with an objective of 8 diopters focal power
is used to take photographs of an object lying on the bottom
of a pond 1.2 m deep. What is the distance of the film from
the objective lens? The lens is placed close to the surface
of the water.
Part Seven
FUNDAMENTALS OF QUANTUM PHYSICS
45. Photons
45.1. Solar radiation with an intensity of 1.36 kW/m
2
per
second falls on the illuminated surface of the Earth. As-
suming the Sun's radiation to be similar to that of an abso-
lute black body, find the temperature of the photosphere.
45.2. The Sun radiates maximum energy at a wavelength
of 470 nm. Assuming the Sun's radiation to be similar to
that of an absolute black body, find the temperature of the
photosphere.
45.3. The surface temperature of "white dwarfs" is of the
order of 10
4
K. To what spectral band does the maximum
value of their radiation belong?
45.4. The sensitivity of the human eye in darkness is high:
at a wavelength of 555 nm it detects a light signal of no less
than 60 photons per second. What is the wave intensity?
What is the power of the light source, if its distance from
the eye is 10 km? The diameter of the pupil in darkness
is 8 mm.
45.5. Find the threshold frequency of the photoemissive
effect for lithium, zinc, and tungsten.
45.6. Find the maximum kinetic energy and the velocity
of photo-electrons emitted from a metal irradiated with
'V-rays of 0.3 A wavelength.
45.7. Find the cut-off voltage which stops the emission of
electrons from a caesium cathode irradiated with light
of 600 nm wavelength.
Fundamentalsof Quantum Physics 1"11
45.8. When a cathode was irradiated with light first of
440 nm wavelength and then of 680 nm, the cut-off voltage
was changed by a factor of 3.3. Find the electron work func-
tion.
45.9. There are types of photographic paper which can be
processed in red light of over 680 nm wavelength. Find the
activation energy of the chemical reaction.
45.10. Find the energy, the mass and the momentum of a pho-
ton of ultraviolet radiation of 280 nm wavelength.
45.11. Find the wavelength of X-ray radiation for which
the photon energy is equal to the intrinsic electron energy.
45.12. A moving source emits photons. Find the expressions
for the energy and the momentum of a photon in the labo-
ratory reference frame.
45.13. Making use of the photon concept compute the light
pressure on a reflecting surface, if the angle of incidence is cx.
45.14. A small ideally absorbing plate of 10 mg mass is
suspended from a practically weightless quartz filament
20 mm long. A light flash from a laser falls on its surface
perpendicularly to it, causing the filament with the plate
to deflect from the vertical by an angle of 0.6. Estimate
the energy of the laser flash.
45.15. Estimate the dimensions of a particle, if the pres-
sure of the light from the Sun on it compensates the gravita-
tional force. Assume the particle to be an absolute black
body, its density to be 2.0 X 10
3
kg/m", and the solar con-
stant to be 1.36 kW/m
2
R2
r 1
= ym
2
T2j4n
2
, R2
r 2
=: "(1n
1
T2j4rt
2
Adding both equalities, we obtain
R3 = 'Y (ml +m2) T2
4n
2
This is the desired expression for the generalized third Kepler law.
8. Total and Kinetic Energy
8.2. It follows from K == that CS == 0' or y 1 - v
2/c2
== 1/2.
Hence
v= c y3/2= 2.6 X 10
8
mls
8.3. The electron mass is
m= mo mo 10mo 2.
56m
o
1/ 1- 0. 92
2
Y1.92XO.08 y15.3
Therefore the total energy is = 0' and the kinetic energy
K ==
The momentum is
p == mu= 2.56 X 9.1 X 10-
31
X 0.92 X 3.0 X 10
8
=== 6.4 X 10-
22
kg -m/s
8.4. The total energy is = K + 0 === 10.94 GeV. The momentum
may be found from the relation
p= == .-!.-
c c
3.0; 108 V 10 X 11.88 X 1.6
2
X 10-
20
10.9 X 1.6 X 10-
10
3.0 X 108 5.8 X 10-
18
kg. m/s
If K 0 one may with sufficient accuracy assume that
K 10 X 1.6 X 10-
1
()
P=-c-= 3.0 X 10
8
5.3 X 10-
18
kg-rn/s
The error will be
E=={5.8-5.3)/5.8 5%
The velocity may be found from the relation
.!i.-=10.0=O.92
c mc
2
10.9
8.6. The relative error is
E=Krel-Kcl 1- m
gu
2
Krel 2 (mcs - m
oc
2
)
Conservation Laws 155
Denoting == ul c and noting that = V1- we obtain
v42 1
V1- t\2)
8.7. v = t/2P'll/CSp.
8.8. As the weight moves along an inclined plane at constant speed,
the power is
P = (F +T) v = lngv (sin a+ cos ex) == mgu (sin a+tan cp cos a) =
lngv sin (a+ <p)
cos cp
where tan cP == fl. The speed
p cos fP
v==------.;.....--
mg sin (a+<fJ)
will be at its minimum if sin (a + cp) = 1, i.e. if a + <p == . Hence
n 1
a ==-2-arctan Il == arctan-
8.10. Noting that K
o
== 0 we see that the work of the electrical forces
is equal to the ultimate kinetic energy:
K == A == eEL
(1) The nonrelativistic case: 1/2mou2 == eEL, hence
u == 1/2eEl/rn
o
(2) The relativistic case: If + p2c2 - <tb 0 == eEL. Hence
2 2 2
J
J2
C
2 = eEL eEL), or mou c == eEL (2$ I eEL)
t-u
2jc2
OT.
After some simple transformations \ve obtain
c VeEL +eEL)
u= 'Ho+eEl
(3) If eEl then
u= V
r
21noc
2
eEl - }I/ 2eEl
Ill"
i.e. the nonrelativistic formula is valid.
8.1 f. The absolute error is
156 Solutions
9. Uncertainty Relation
is called the first
9.1. Assuming the electron moves in a circular orbit about the nucleus
(proton), we can write the equation of motion in the form
e
2
l11V
2
= -r-
Hence the kinetic energy of the electron is
rnv
2
e
2
K=--=--
2 8JlE
o
r
But if the electron is localized in a region of characteristic dimension
r, its kinetic energy
Comparing the two expressions, we obtain r >
. 4ne h
2
The quantity a
o
= = 5.24 X 10-
11
m
mt
Bohr radius (see i .15).
9.2. Let's first estimate the momentum and the velocity of the elec-
tron. We have
r >hl a 10-
19
kg-rn/s
From the formula p == moBc ,where == ul c, we obtain
.r: == 360
V n!oc
Hence 1, i.e. the electron is an ultra-relativistic one. Its kinetic
energy is
K == pc== 3 X 10-
11
J == 200 MeV
9.3. Estimate first the region of localization of the electron, its mo-
mentum and its velocity... We have .
ID, p>n!a=5xl0-
25
kg-rn/s
The velocity may be found in the same way, as was done in the PW
vious problem:
Since 2 X 10-
3
, the velocity' of the electron is much lower than
the velocity of light, i.e, the conduction electrons in a metal are non-
Conservation Laws 157
relati vistic particles. Their kinetic energy is
n,2
n
2
/
3
K 1.3X 10-
19
i ev
9.4. First find the number of neutrons and their concentration:
'N -- 2 X 10
3 0
_ -03 1 44 -3
-- 1.67 X 1U-27 1.2 X 10
57
, 11, -- 4j3rrra - X () ill
The region of localization of a neutron is a. = n-
1
; 3 = 1.5 X 10-
15
m.
The momentum of a neutron is p 1Ia 7 X 10-
20
kg .m/s; the
relativistic factor is
0.14
7 X .10-
2
0 p
moc 1.67x1u-
27
x 3LJ x 10
8
from which it follows that 0.14, i.e. the neutron is a nonrela-
ti vistic particle. I ts kinetic energy is
!l2
n
2
/
3 o
K 1.3 X 1r,-12J 9l\fpV
10. Elementary Theory of Collisions
oX
A
I
I
I
I
,
I
t
I
I
N
F
o
B
to.t. The result is not in contradiction with the law of conservation
of momentum: the vertical component of the momentum is transmitt-
ed to the Earth.
10.3. Direct the x-axis along Y
the wall and the y-axis normal
to j t (see Fig. 10.3). Since the
Fig.
Fig. 10.4.
wall is smooth, the momentum component along the x-axis will
not change. Assuming the mass of the wall to be infinitely greater
than that of the ball, we find that; the momentum component along
.the y-axis changes sign while its magnitude remains constant (see
158
17.3). Thus
Solutions
Hence a' == a (we are not interested in the sign of the angle).
The ball, after striking the surface of the parabolic mirror at
point M (Fig. 10.4), will be reflected to point F called the focus.
The angle of incidence ex is the angle between the direction of the
velocity of the ball and the normal MN; it is equal to the angle be-
(J
..
Fig. 10.5a.
\
\
\ P
Fig. 10.5h.
./
,/
./
tween the tangent ME and the x-axis. According to Problem 3.14,
we have tan a == xlp , where x == OK == MB is the x-coordinate of
the ball before impact. As may be seen from the figure, L Blvl F =
= = 2a - 11/2 and BF = BM tan = x tan ( 2a - =
x2- p2
= -x cot 2a == --.
2p
The focal length
X
2_
p2 p2 =L
j==OF=OB-BF=y- 2p 2p 2
Thus, the focus of a parabola is on its axis of symmetry (the y-axis)
at a distance p/2 from the origin. A particle moving parallel to the
axis of symmetry after an elastic reflection arrives at the focus no
matter at what distance from the axis it was moving.
10.5. In the given reference frame one of the particles is at rest before
the impact, the other moves at a speed v. After the collision their
velocities are VI and v
2
, respectively, the scattering angle is a
(Fig. 10.5a). Construct a triangle of momenta (Fig. 10.5b). Taking
account of the fact that the total momentum and the total kinetic
energy are retained after an elastic collision, we obtain
L-...EL 0-
- 2 -r 2
2m m m
Hence cos = 0 and a == = 90.
10.6. Since the energy is equipartitiened between the particles, the
same will happen to the momenta. Consequently, after the collisions
the protons scatter at equal angles to the original direction of the
Conservation Laws 159
proton projectile (see Fig. 10.6). From the laws of conservation of
total momentum and of total kinetic energy we obtain
ex.
2PI cosT= P, 2K
1
=K
From the relation between the energy and the momentum =
m,p
o
Fig. 10.6.
+ p2C2 we obtain, noting that = + K,
K2+ 2K'H
o
==p
2
c
2
, Ki+ 2K
1
Cfl
o
= pfc
2
Eliminating the momenta and the proton's kinetic energy after
collision, K
h
from these equations, we obtain
2 a
cos T=
whence
cos a = 2 cos- -.:x--1
2
K
Note that in the nonrelativistic case, when K 0, we shall
have cos a 0 and ex. :rr/2 (compare with Problem 10.5).
For K = 500 MeV we obtain, noting that 0 = 938 MeV, (see
Problem 7.1) cos ex = 0.117, a = 0.46n.
For K = 10 GeV we obtain cos a = 0.728, a = O.24n. We see
that as the kinetic energy of the projectile particle rises, the scattering
angle decreases, approaching zero for ultra-relativistic particles
(K '
10.7. One of the disks is at rest before the impact; after the impact its
velocity will be in the direction of the centre line at the moment
of contact (Fig. 10.7b) for this is the direction in which the force
acted on it. Thus, sin a2 = dl2r, al + (l2 = 1(/2 (see Problem 10.5).
Since the masses of both disks are equal, the triangle of momenta
turns into the triangle of velocities (see figure). We have
. VI = V cos al = v sin a
2
= vd/2r
V
2E:::VCOS
cx
2
= v V1- d
2
f4r
2
160 Solutions
to.8. Direct tho coordinate axes as shown in Fig. to.8b. The direction
of velocity of the larger disk may be found, as in the previous prob-
lem, from the condition
sin (X2
1
= d/(rl + T2)
The remaining three unknowns, the velocities VI and V
2
and the
angle aI, will be found if we wri te down the equations for the conser-
y
Fig. 10.7b.
vation of the x- and y-components of the momenta and the equation
for the conservation of kinetic energy
Ply - P2y ==0, PIX 1- P2X == P, K
1
+!(2 == K
.or in the form
PI si n at - P2 sin Cl
2
::= 0, PI cos al -+- P2 cos (1,2 = p,
__ p2 ~ p2
_1_+__= __
2ml 2m2 2m,!
After some transformations we obtain
2m
2
Pcos a2
p 2 = ----.:;;
nll + nl
2
'
PI = P V (tnl + f1l
2
)2- 4 m
ltrl 2
cos
2
(1,2
m
1+m2
sin at == (P2 sin (X2)lpl
Note that for ml == m2 and rl = r2 we shall arrive at the results
of the previous problem.
10.9. The number of particles striking the wall during the time ~
(Fig. 10.9) is
N == nuS1. ~ == noS ~ cos (1,
(see 17.5). Since only the normal component of the velocity changes
after impact, the force is
2mu cos a
/!it
and the pressure
Conservation Laws
The force of pressure is
F = N] = 2nSmv
2
cos
2
ex.
161
p = 2nmv
2
cos
2
ex.
10.10. The problem stipulates that the boat moves over water at
a constant speed, and this means that the drag and the force acting
on the sails are equal in magnitude.
To find the formula for the drag of the water compute the Rey-
nolds number Re = PulJolo/110' It is given that Vo = 3 mls and that
o
o
o.
v
m '" r-,
~
v.1
Fig. 10.9.
the characteristic dimension is lo ~ 1 m. Thus, Re = 3 X 10
6
, and
so the pressure drag R = CSoPov'512 plays a basic role.
To be definite let us assume that the interaction of the air with
the sails is elastic. Noting also that nm = P is the density of air, we
obtain ~ C S o P o v ~ = 2pSv2. The area of the sails is
S - C S o P o v ~ 0.1 X 1.0 X 10S X 9.0 5 m
2
- 4pv
2
4 Xi. 3 X 36.0
10.11. Arrange the coordinate axes as shown in Fig. 10.11. Then com
ponents of the initial velocity along the axes will be Vox = Vo cos a,
vo
y
= Vo sin a and the acceleration components will be ax = g sin a,
all = -g cos CL. The equation of motion for the first part of the tra-
jectory should be written thus:
V
x
= vox+axt=Vo cos a+gt sin ex.
v
y
= voy+ayt = Vo sin ex.-gt cos ex.
ax
t 2
gt
2
sin a.
x=xo+vox
t
+ -2-=votcosa+ 2
a t
2
gt
2
cos a
Y==Yo+voyt+ +=votslna- 2
t 1-0380
162 Solutions
Since at the point A
h
the ordinate Yl = 0, it follows that
t _ 2vo tan a x _ sin ex
1 - g , 1 - g cos2ex, ,
v1y==-vosina
cos ex,
The longitudinal velocity component does not change after an
elastic impact, but the lateral component changes sign. Therefore
A ----
I - ................. ,
""-
"
"
,
,
,
\.A
2
Fig. 10.11.
for the second part of the trajectory
Vox==Vtx== voy=-v1y==vosina
cos ex
The equation of motion, by analogy with. the first case, will he
+ +
axt2 sin ex. + vot (1+sin
2
a) +gt
2
sin ex.
x=x v t --=-.....;----
lOX 2 g cos2 ex. cos ex. 2
--l + a
y
t
2
8
The increase in the kinetic energy of rotation of a star in the process
of its contraction (collapse) is at the expense of the work of gravita-
tional forces.
14.23. The moment of the ball's momentum is L =-= It =
=; mr
2
-;- = mvr, where v is the orbital velocity on the equa-
tor. Since v < c, L < ; mer, whence
5L 5n
r '> 2n2c' or r > 4mc==4.8x10-
13
m
This dimension does not agree wi th experimental data, according
to which the effective electron radius is two orders of magnitude less.
15. Non-inertial Frames of Reference and Gravitation
15.1. The reference frame fixed to the wedge moves with an acceler-
ation a in the x-direction and therefore is non-inertial. There are
four forces acting on the block: the force of gravity mg, the reaction Q,
mg
Fig. 15.1a.
mg
Fig. 1S.1h.
the friction force T, and the inertial force I = -rna. To solve the prob-
lem one should consider the same two cases as in Problem 5.7 (see
Fig. 15.1a and 15.1b). The block is not accelerated with respect to
the non-inertial frame, and for this reason the sum of the projections
of forces on both coordinate axes is zero. We have for both cases
12*
130 .Solutions
y-axis:
Qcos a+T sin a-mg=O, Qcos a-Tsin a-mg=O
z-axls:
-I1-Tcos a+Qsin ct.= 0, -1
2
+T cos a+Qsin a== 0
The equations obtained are equivalent to the equations of Problem 5.7,
therefore we shall obtain the same answer.
15.2. Four forces act on the body in the rotating coordinate system:
the force of gravity mg, the reaction Q, the centrifugal inertial force
Q
Jef
Q
Fig. 15.2. Fig. 15.3.
mg
mg
Fig. 15.4.
let = mw
2r
and the force of friction T (Fig. 15.2). To prevent the
disk from sliding the following condition should be satisfied:
I cr or nlw
2
r Ilstatmg
hence r -< f.t
stat
glro'i..
15.3. The forces acting on the motorcyclist in the rotating reference
frame are mg, Q, T and let (Fig. 15.3).
Evidently
Q= I cr== mw
2r,
T ==flQ == = flmv
2Jr
The motorcyclist will not slip off the
wall, if T > mg. We have
(!mv
2/r
> mg; v VgrIll
15.4. Three forces constituting a
closed triangle act on the weight in
the rotating reference frame (Fig.
15.4). We have
I cr -=mg tan c, or mw
2r
= mg tan a
Noting that r == 1 sin ct. and ro == 2rr/T, we obtain
/r l
T=2n}/ -cosa
g
15.5. From the condition that the centrifugal force of inertia is equal
to the elastic force, we obtain the same result as in Problem 3.9.
Conservation Laws 181
15.6. The forces acting on material particles on the star's equator
in the rotating reference frame are the force of gravity F = y ; ; ~ and
the centrifugal force of inertia I cf = mCJ)2R. Matter will start to escape,
if I cr ~ F, i.e. CJ) >- 1/ rMt R3.
15.7. The time needed for the oil droplets to rise in the gravitational
field is (see 11.9)
't' -- 9l1] 9 X o. 2 X 10-
3
9 X t ()3 s == 2. 5 h
-- 2r
2
g (Pt - p) 2 X 10-
10
X 9.8 X 10
2
In the centrifuge the part of the gravitational field, in accordance
with the principle of equivalence ( 24.5), is played by the field of
centrifugal forces of inertia. Consequently, in the above formula one
must simply substitute w
2
R for g. We obtain
9lTl 9 X 0.2 X 10-
3
't ~ 2r
2
CJ) 2R (Pt -p) 2 X 10-
10
X 4 X 10
2
X 11;2 X 0.8 X 102 = 28 s
The centrifuge is (U2R/g = 320 times more efficient.
15.8. When the system is at rest, the spring is undeformed and its
length is maximum, lo == 2a. When the system rotates, the weights
move away from the rotation axis, and the length of the spring becomes
1 = 2a cos a. The change in length is
~ I := lo-1==2a (1-cos ex) == 4a sin
2
~
2
To relate the deflection angle to the rotation speed, we employ
a rotating reference frame. In compressing the spring, work is per-
formed, equal to
Wei == ~ k { ~ 1 2 == 8ka
2
sin
2
5!:...
2 2
This work was performed by the centrifugal forces of inertia, which
displaced each weight by a distance r = a sin ct.. The work of the
inertial forces is
Wet = 2 X -}- I etr= mw
2a2
sin
2
a
Equating the two expressions for the work, we obtain
a, ct. "1;- mw
2
8ka
2
sin" 2 == m,w
2a2
sin
2
a whence tan 2 = V 2k
Obviously the governor will work in conditions of weightlessness,
for the presence of gravi ty was mentioned nowhere.
The maximum rotation speed may be found from the condition
!:J.l -< 0.1 X 2a. From this 2a (1 - cos ex) -< O.2a, or 0.9 -< cos a <
< 1. Expressing the cosine in terms of the tangent of half the angle
and substituting the value obtained into the condition of equilibrium,
we obtain
(Omax ~ -V 2k/19m
182 Solutions
15.9. Fix the reference frame to the rotating liquid. Then there will be
three forces acting on every particle: the force of gravity mg, the
reaction of the liquid N and the centrifugal force of inertia let =
= moo
2
x. (Fig. 15.9). The angle between the reaction and the y-axis
is equal to the angle between the tangent and the x-axis (since the
edges of the angles are mutually perpendicular). We have
tan a= lef = w
2
x
mg g
It was shown in Problem 3.14 that the tangent to a parabola x
2
=
= 2py makes an angle of tan a = xl p with the x-axis. Therefore we
may conclude that the surface of a liquid
in a rotating vessel is a paraboloid formed
by the parabola x
2
= y rotating about
ro
the y-axis.
1.5.10. A spacecraft orbiting a planet
forms a non-inertial reference frame
moving with an acceleration g = lef
= yM/ R2 directed towards the centre of
the planet. Two forces act on an object
inside the spacecraft: the force of gravi-
ty F = ymM/R2, directed towards the
centre of the planet, and the oppositely 0
directed centrifugal force of inertia let =
= mv2/R. The resultant of these forces Fig. 15.9.
is zero, therefore the object is weightless
with respect to the spacecraft.
15.12. The slowing down of time measured by the "plane" clock is
due to two effects: to the decrease in gravitation and to the speed of
flight. We have
/ 2q> /' v
2
t p ] =:: tEarth J. 1--- JI 1--:..;.=::
r c
2
c
2
=tEarth V
1
- V1-
(cp is the potential of the gravitational field). Since g: and 1,
c c
the expression for the time assumes the form
tpi = tEarth ( 1- ) ( 1- ;c: )=
=tEarth
c
2
2c
2
The time dilation is
&t = [Earth- tpi = tEarth ( + )
Molecular-kinetic Theory of Gases
Here
eb 9.81 X 10
4
7== 9.0 X 10
16
1.09 X 10-
1 2
,
183
0.43 X 10-
12
v
2
10
12
2c
2
== 2 X 3.6
2
X 10
6
X 9.0 X 10
16
Since tEarth == 2.5 days = 2.5 X 8.64 X 10
4
s, the time dilation is
~ 2.5 X 8.64 X 10
4
X 1.52 X 10-
12
==3.3 X 10-
6
s ==33) DS
The effect of the general theory of relativity (of the gravitational
field) turns out to be 2.5 times more pronounced than that of the
special theory (of the speed).
15.13. For the Sun and for the white dwarf i\v = ..9:
2
= V
2
M.
For the
veeR
pulsar we can make use of the precise formula
i\v = 1- .. / 1- 2VM
v V c
2
R
15.14. Light cannot escape from the star's gravitational field, if the
gravitational potential on the surface is so high that 2cp/c
2
~ 1.
Hence the radius of a black hole is R < 2-yM/c
2
f kT
, = exp { Eo 2) }
Hence we obtain after taking the logarithms
Eo (T t - T2) = In
kTIT2 111
or
kT 1T2 In ("2/111) 2.3kT1T2 log (T12/ rh )
Eo =
T
t-T2
T
1-T2
21.3. Water will start rising and will reach the upper end of the capil-
lary. Here the radius of curvature will decrease until the pressure
of the curved surface becomes equal to the hydrostatic pressure of
the water column. Then the water will stop rising. The condition
for equilibrium is
== cos e = pgh
r
For the contact angle we obtain
cos e= = 0.544, 8= 57
0
20
208 Solutions
21.4. The total surface area of all the droplets is 8
0
= 400nT
2
=
= 100nd
2
, their total volume is V0-. == 1001lt
3/6.
After the droplets
merge, the volume remains unchanged, but the surface area decreases:
V = nD3/6 == V
o
, 8 == 1tD2. From the condition of equality of the
volumes, find the diameter of the large drop:
100nd
3
nD3 whence D= dV"100
-6-=-6-'
The surface area of the large drop is 8 == nd2 V10
4
The decrease
in the surface layer energy corresponding to the decrease in the sur-
face area of ~ s == So - S == nd
2
(10
2
- 10
4
/
3
) is
L\esur=aLlS ~ oL\S=1tod
2
(10
2-10
4
/
3
)
21.5. Two forces act on a fluid in conditions of weightlessness: the
force of surface tension Ssur = 1tad and the Iorce of hydraulic resis-
1 pv
2
nd
2
tance F
res
= Ad T T (see 30.17). Since the velocity
of the fluid is small, for small Reynolds numbers A=
=6
R4
= 64
11
d
Equating the forces, we obtain after some
e pv
simplifications
v = (1d/8f1l
Compute the velocity to find the Reynolds number,
and make sure that it is much less than 2320.
21.7. When the fluid reaches the lower end of the tube,
a convex meniscus will be formed there wi th a shape
identical to that of the upper meniscus (Fig. 21.7). The
excess pressure is tlp = 2 X 20/T. But 20/T = pgh,
h-p = pgH, from which H = 2h.
21.9. Suppose the drop spreads evenly and as seen from
above has the shape of a circle of radius R (Fig. 21.9).
The area of this circle is S = vt == m/pd. The force of Fig. 21.7.
attraction between the plates is F = SpS; where!1p ==
= 2(1/d is the excess pressure under the curved surface. Therefore
F == 2(Jm/pd
2
21.10. The pressures from the left and from the right are equal, i.e.
AP1+L\p=dP2' or ~ 4cr ~
R
1
R R
2
Here Sp == 40/ R, for the film has two surfaces-the external and
the internal. Hence
Molecular Forces and' States of Aggregation 209
Since at the point of contact of the three films we have a system
of three forces of equal magnitude in equilibrium in a plane, the angle
----==---
---
Fig. 21.9.
between the forces is found from the condition that they form a closed
triangle.
22. Vapours
22.2. As distinct from an ideal gas, whose molecular concentration
(and density) does not change in an isochoric process, the molecular
concentration (and density) of a saturated vapour rises with tempera-
ture because of additional evaporation of the liquid.
22.3. The density of saturated vapour at 55C is 104.3 g/m
3
Therefore
at this temperature 8 g of saturated vapour occupies a volume of
= 7.6 X 10-
3
rn
3
= 7.6 Iitres. In a smaller volume there will
be a precipitation of dew.
22.7. Find first the mass of moisture in each volume of air, i.e. the
absolute humidity of the volumes which are mixed. We have in 5 rn
B
of air
ml = /1VI = VI = 12.8 X 0.22 X 5= 14.1 g
In 3 rn
s
of air
m
2
= X 0.46 X 3=37.5 g
Next find the absolute humidity of the mixture:
t = (ml + m
2
)/ (V
1
+ V
2
) = 6.45 g/m
8
To find the relative humidity we must find the temperature of
the mixture. Neglecting the vapour mass we may write the equation
of heat balance in the form
POV1Co POV2Co (28-t)
where the subscripts 0 show that the density and the specific heat
refer to air. We have t = 20"C. Now it is easy to compute the relative
humidity.
14-0380
210 Solutions
" R' .
'M PerTer = X 10
7
Pa; at the same time Per= 218 atm =
= 2.2 X 10
7
Pat
Hence the Mendeleev-Clapeyron equation is inapplicable to the
critical state...1'4Js is because of the large part played in this case by
3031.35 t,Oc 25 "20
ioq
:0 __.....&.._---.............
900
600
500
464
400
300-
200'
700
800
f ._ .: s . . Fig., 22.9.
molecular interactions, which are neglected in the case of an ideal gas.
22.9. See 22,.9.
23. Phase Transitions
. ..
23.1. Since evaporation takes place at a constant. pressure, W:::;::
= P (vX
ap
- where Vo = tIp is the specific volume, I.e. the
volume of 1 kg of the substance. Hence
':.r1! or .
' ':. n : W = p (P:ap pil
q
)
The density of steam' at ioo-c may befound from Table 35.1 (see
l the. specific volume of vapour is almost a thousand times
.. liquid.. it follows with. better
than t % accuracy that W = plPvapo .
Electrodynamics 211
The energy spent on breaking the bonds between the molecules
can be found from the first law of thermodynamics: I!1U = L - W
where L is the specific heat of evaporation,
23.2. Ice melts under the pressure of the wire, and the wire sinks;
the water formed above the wire immediately freezes again. .
23.3. The heat liberated in the processes of condensation of water
vapour and of cooling of water down to the melting point of ice is
Q= ma (L + ~ t = 5.2 X 10
5
J. This is not enough to melt all
the ice. For this the. required heat is Qm = m2A = 10
6
J. Therefore
the ice will only partially melt.
23.7. The heat of fusion liberated in the process of freezing of water
will be spent to heat the remaining water to ooe. Let the total mass
of supercooled water be m, the mass of ice formed nlt, the mass of
remaining water m2 = m - m1.. From the heat balance equation we
have ml'" = (m - ml) cSt; whence
x =.!!!:l. = cAt
m A+cf1t
23.8. The outflow velocity is v = pipS, where fJ. is the amount of water
that evaporates per second. Obviously, I.l. = Z= f, where P is
the power of the hot-plate and" is its efficiency. Hence
v = TlPlpLS
23.9. First find the melting point: t = -Aplk = -4.35e. As the
ice is cooled to this temperature, the heat liberated is Q= mcSt =
= me I t I. This heat will be spent to melt the ice: Qm = m1A. Hence
the fraction of the ice that will melt will be x = ::: = ci
f
In the calculation assume that the specific heat of ice and the
heat of melting remain constant. .
23.10. The heat flowing to the Dewar vessel is- Q = a (T
a
l r - T),
where (J, is a certain coefficient, and T is the temperature inside the
Bask. For ice and liquid nitrogen we obtain the ratio
Ql Talr-T
1
Q;=Tslr-T
s
But for nitrogen Q2 = m2L, where L is its heat of evaporation, the
respective value for ice being Ql = miA. Hence
mt
A
Tafr -
T
l from which m1A(T
a
l
r
- T
2
)
m
2
L =Talr-T
a
ms= L(T
a
l r - Tl )
24. A Field of Fixed Charges in a Vacuum
24.1. It is evident from Fig. 24.1 for the case of charges of different
signs that the actual force of interaction of the charges is greater than
it would have been, if the charges were concentrated in the centres of
the spheres, and less than if the charges were. concentrated at-the
14*
212 Solutions
nearest points of the spheres:
q2 < F < q2
4ne
or
2
411:8
0
(r- 2ro)2
The absolute error is less than the difference between the bounding
values, the relative error being less than the ratio of this difference
r
Fig. 24.1.
to the minimum force. Hence,
8 < F
2-F1
= r
l-(r-2r
o
) 2 4r
o
(r-r
o
>
F1 (r-2r
o
)2 (r-2r
o
)2
24.2. Direct the coordinate axes as shown in Fig. 24.2. Let M ba'one
of the points where the potential is zero: I(
q > = q > q > 2 = q ~ = O f
4n8
0r
1 411:8
0
' 2
Hence r2 = 2rl.
Substituting rl = V(3a - X)2 + y2 and r2 = V(3a + X)2 ~ y2,
we obtain after some manipulations
(x - 5a)' + y2 = 16a
2
I
This is the equation of a circle of radius 4a, with centre at (5a, 0).
r
Fig. 24.2.
24.4. A force of gravity directed downwards and equal to rrig ~
= 1/6nD3pg acts on the droplet. I t is counterbalanced by an. electric
force F = gE = qcp/d (Fig. 24.4). The charge of the droplet IS found
from the balance of forces.
Electrodynamics 213
24.5. In this case equipotential surfaces are spheres with a common
centre at the source, and radii normal to the surfaces. If the potential
is q> = q/4'ItEor, the field strength is
d<p d ( q) q
E=-Cir= -a;: 41tB
o
r = 41tB
o
r2
and just this was to be 'proved.
24.6. q>= _ r Edr= __q_ r
J 41tB
o
J r
2
41tB
o
r
To find the potential, divide the ring into small segments and
add up the potentials of these segments. The field strength is
(x
2+a2
) - 1/ 2
dx 411:8
0
dx
= i.. _q_ (x
2
+a
2
)- 3/ 2 . 2x = qx
2 4"';80 41tB
o
Y(x
t
+a
2
)3
24.8. (a) It is evident from Fig. 24.8a that in this case a force couple
acts on the dipole. The resultant of the couple is zero. To obtain the
torque, multiply the force by the + + +. +
dipole separation.
(b) It is evident from Fig. 24.8b tF
that in this case two forces act on the
dipole in the direction of the radius
vector, Hence it is clear that in the
lat-ter case the torque is zero, and mg
the' resultant acts in the direction of
the radius to the source. The re-
sultant can be found by two methods. Fig. 24.4.
One may use the Coulomb law:
F=F_+F = - QQ + Qq
+ 4nB
o
(r-l/2)2 41tB
o
(r+ lj2)2
2QrQl
+
t:;t
Fig. 24.8b.
o-------.......
Q
t
0=======
Q Fet-
Fig. 24.8a.
Noting that the problem stipulates that I r, we obtain
F= _ 2Qrql = _ 2QPe
4:t8
0
r
4
4n8
0
r
3
214 Solutions
c,
+ -
The same result may be obtained from formula (37.15), if the
derivative is substituted for the ratio of increments. We have
F=Pe dE =Pe-!!.. (_Q_) = _ 2QPe
dr dr 4ne
o
r
2
4aB
o
r
3
the capacitors are connected in parallel (Fig. 24.9), all
their Interconnected plates have the same potential, therefore qt =
= C
t
(cpt - CP2) and q2 = C
2
(j)l - <J>2)' where ql and q2 are the charges
of the respective capacitors.
The charge of the system is q=
= qt + q2 = (C
1
+C
2
) (cpt - CP2)
On the other hand, q = C (cpt -
- <J>2)' where C is the equiva-
lent capacitance. Hence the for-
'It
+
-
+ - +
-:-
II
0
II
C
f
"
C
2
C
2
Fig. 24.9. Fig. 24.10.
mula sought.
24.10. When the capacitors are connected in series (Fig. 24.10), their
charges are redistributed so that the charges of all the capacitors are
Fig. 24.11a.
Fig. 24.11b.
equal. To make sure that this is so, consider using Fig. 24.10, what
will happen if the system is discharged by shortcircuiting the points
at potentials <PI and CJ'2"
We have lpt - <p' = qlC, <p' - <J>2 = q1C
2
For the system as
a whole, CPt - q>2 = c. Adaing the two former equations and com-
paring the result with the third, we obtain the formula sought. .
24.1t. The solution is obvious from Figs. 24.11a and 24.11b, which
give the diagrams of the parallel and series connections of four iden-
tical capacitors. The energy in the case of parallel connection is
W
C
par = 2 ::= -2-
Electrodynamics
215
and of series connection
W
ser= 2 -
(Coin) nC
o
q>3 W
2 == -2- == par
24.12. A possible circuit diagram is shown in Fig. 24.12 for the case
of a five-capacitor system. It follows from the diagram that a ten-
position switch is required. When the knob "is in the upper position,
--.----t-...-..........
In parallel
\ \\ \\ \\ \\ \
In series
I I r r I I
Fig. 24.12.
Fig. 24.14.
the capacitors are connected in parallel; when the knob is in the
lower position, they are connected in series. ,
24.13. Suppose that there is a field inside the sphere. It is obvious
from considerations of symmetry that in this case equipotential
surfaces must be spherical surfaces concentric with the charged sphere.
Accordingly, the lines of force coincide with the radii, Le. they must
either begin or end at the centre of the sphere. This would have been
possible, if there were a positive or a negative charge at the centre
of the sphere. But since there is no charge inside the sphere, the lines
of" force cannot begin or end there. Consequently there is no field
inside the sphere.
24.t4. Construct a second sphere around the sphere under considera-
tion and suppose it carries a charge equal in magnitude, but opposite
in sig-n (Pig. 24.14). According to the result obtained in the previous
problem, the charge of the outer sphere does not create a field inside
it. Therefore the field between the spheres is created only by the charge
of the internal sphere. If there is little difference between the radii
of these spheres (i .e. if R 1 - R <t R), the field in between will be
almost homogeneous, and its strength will be (see 37.5)
E--!!.--!L- Q
- eo - eoS - 4tte
o
R2
24.15. Let the volume density of the charge be p = where
R is the radius of the sphere. Choose a point M inside the sphere at
216 Solutions
r
R
pR Q
E
sur
= __= 4 2
3e
o
rtoR
This dependence is plotted in Fig. 24.1.5.
24.16. The field potentials on the surfaces
of both spheres are cp = q/(4ne
o
R }, CPt =
=q/(4ne
o
R
t
}, respectively. The potential
difference is
a distance r < R from its centre and consider a concentric sphere
through it. From the results obtained in Problem 24.13 it is evident
that the spherical layer lying outside point M does not create a field
at this point. The field is created solely by the charge q = { nrp
contained inside the smaller sphere. By the result of the previous
problem, the field at the surface of this
sphere, i.e. at point M, will be
q 4J'tr
sp
pr
E = 4n80r2 - 3 X 41t8
0
r
2
= 3e
o
We see that the field inside the sphere
increases in proportion to the radius; at the
centre its i tensity IS zero, and on the sur-
face it is
Fig. 24.15.
q>-q>l=-q- = q(R1-R)
4nB
o
R R
1
41toRR
t
The capacitance is
C _ q _ 4n8
oRRl
q>-q>1 - R
1-R
If d = R:i - R R, we obtain the approximation
C= 4nEoR2 = 8 0S
d d
which is the expression for the capacitance of a plane capacitor. The
error is
l)- _ Rt-R _...!!:-.
- Csp - R - R
24. t 7. A ball of radius a carrying a surface charge may be regarded
as a spherical capacitor whose external sphere is infinitely far away
(i.e, R = a, R
1
00). Making use of the result ,of the previous prob-
lem, we obtain
C= lim 4n8
n
aR1 = lim 4neoa = 4nB
oa
R1 -. 00 R1 - a R1-' 00 1-(a/ R1)
The energy of the field is
- 2C - 8ne
oa
Electrodynamics 217
Equating it to the rest energy of an electron = meet, we obtain
4= e
2
=1.4Xl0-
1
I) m
8ne
omec
2
As is shown in 72.5, the term "classical electron radius" usually
applies to a quantity twice as large reI = 2a = 2.8 X 10-
16
m. Com-
paring this result with the result obtained in Problem 14.23 we see
that the latter was two orders of magnitude greater. This implies the
incorrectness of the solutions of the two problems.. Tn modem science
the problem of the dimensions of elementary particles, including the
electron, is far from being solved.
24.18. The mechanical stress is equal to the energy density of the
electric field (see 37.8). 'Ve have p = Wo = B
o
E
2
/ 2. But we have
already found the field on the surface of a sphere (see Problem 24.14):
E = q/(4ne
o
R2). Hence the result sought.
24. t 9. The electrical forces extending the film must exceed the surface
tension forces:
q2 40
32n
2
oR4 R
24.21. Before the connection is made there is a charge ql = CCPl on
.the first capacitor, and q2 = C<P2 on the second. Aft.er the upper
plates of the capacitors are connected, the charze q = q] + q2 is equi-
partitioned between them. The potential of the unearthed plates is
cp
q ql +q2 <1'1 +<1'2
SYS=-C-= 2C 2
sys
The energy of the system before the connection is
W=W
1+Wa
= C:i + Ci
l
while the energy of the battery is
W _ _ 2C + _ C(<pt+Cl'2)2
sys - 2 - 2 X 4 - 4
This is less than before the connection. The lost energy is transformed
into other types of energy (heating the conductors, forming a spark,
electromagnetic radiation, etc.).
25. Dielectrics
25.1. If the plates are connected to the power supply all the time, the
charge on the plates, and consequently the electric field strength,
remain constant. The density of the polarization charge is equal to
the magnitude of the polarization vector of the dielectric:
Opol= p ==Xe80
E
=(e-i) Bo<r>/tl
218 Solutions
25.2. If the capacitor is disconnected from the source, the quantity
remaining constant is the charge on the plates; the potential and the
field strength change by a factor of 1/8 when the dielectric is inserted
between the plates. Therefore
(J (e -1) 8(,>
pol = ed
25.3. The aluminium foil sheets are connected alternately to form
a system of capacitors connected in parallel (Fig. 25.3). The number
of capacitors is equal to the
number of dielectric layers.
The breakdown voltage
Is Ubn = EMd, so the work-
ing voltage is chosen to be
1/3 to 1/2 of this.
25.4. The capacitance of a
capacitor will remain un-
changed, if we place a very
thin piece of foil on top of Fig. 25.3.
the insulating layer. For this
reason the capacitance sought should be regarded as made up of
the capacitances of two capacitors connected in series, with capaci-
tances
and
CE 8 ~ S
C
t
= _(;;,_0_ and C
2
= _Ci_O_
d do-d
25.5. Here we have two capacitors connect-
ed in parallel, with capacitances
C _ B
0
8
1
_ BoS (lo-I)
1 ~ . . . ; . . .
d a,
C _ eeOS
2
_ E8
0
Sl
2--
d
- - ar;-
25.6. The droplet is polarized in the electric
field and acquires a dipole moment Pe = PV,
where P = 'Xep.oE is the magnitude of the
polarization vector and V is the volume of
the droplet. In a non-uniform field a force
dE
F = P
e
dr acts on the droplet. We have
~
"'\\\
""\ \
I I ~ \ \\
I It\ \
I / ~ \ \
I \ \
I I \
, I \ \
I , \ \
I I + + \ ~
, , ~ ,
mg
Fig. 25.6.
This force is directed upwards (Fig. 25.6). The force of gravity mg =
= pVg acts downwards. The problem requires that the magnitude
Electrodynamics 219
of the electric force exceed the magnitude of the force of gravity:
(8--1) VQ2 V from which r V(8-1) Q2
8 2 0 p g, 8 2
n eor n 8
0
pg
25.7. When the liquid flows through a strongly non-uniform electric
field close to the edges of the plates, it is polarized and drawn into
the space between the plates. Since the charge on the plates remains
unchanged in the process, and the capacitance of the capacitor incre-
ases, this is accompanied by a decrease in the energy of the field. This
decrease is compensated by the increase in the potential energy of
the column of liquid held between the plates. From the law of con-
servation of energy
q2 q2 meh.
2C
o
=W-+-2-
Boab e b
Here Co = T' c = -t [a + (- - 1) h] (see Problem 25.5), m =
= pbhd. Substituting these values into the first formula we obtain,
after some simplifications,
(B-1) q2=e
o
p ghab
2
[a+(e-1) h]
Express the charge on the plates in terms of the potential:
Q=<P6
CO
= 8
o
<Poab/d
After simplifications we obtain a quadratic equation
h
2
+_
a
_ h- = 0
e-1 pgd
2
Solving it we obtain the height the liquid rises.
25.8. In this case, too, the capacitance of the capacitor increases as
the liquid rises, but the energy of the electric field is not conserved
but increases. Besides, the potential energy of the rising water incre-
ases as well. Is this not in contradiction with the law of conservation
of energy? Of course not. The power supply performs work to raise
the liquid, the increase in the energy of the system being equal to
the work of the power supply in displacing the charges to the capac-
itor plates:
W p
s
= +
But
wp.s == p=
AW Ccp2 Cijcp2 AW 1 h 1 bdl 2
L.1 el=T--2-' .:.l pot=T
mg
=T
Pg
L
Substituting, we obtain the result sought after some simple trans-
formations.
25.9. To plot the graph it is convenient to use new variables: z =
;: 10
3
/ T, where T is the absolute temperature, and y = 10
3X
e
where
220 Solutions
'Xl' is the electric susceptibility. The respective values are presented
in Fig. 25.9, from which it may be seen that the Debye law is well
satisfied in the experimental range (see 38.6). Because of this, let
us use Iormula (38.26) to calculate the dipole moment of the water
vapour molecule, and the gas equation to calculate the molecular
concentration.
It is advisable to carry out the computations for all four experi-
mental points and average the results.
25.10. A characteristic feature of the inert gases is their deformation
polarizabiltty. From the formulas of 38.5 we have 'Xe = na where
J = 1 ~ e
4.2
4.0
3.8
3.6
3.2 x= IOJ1r
2.0 2.1 2.2 2.3 2.4 2.5 2.6 I
Fig. 25.9.
n = plk T is the atomic concentration and a is the polarizability of
the atom. The dipole moment of an atom in an electric field is
1
Pe= a8
0
E = - XeEo
E
n
The atomic concentration at standard conditions is n = NL (the
Loschmidt number). Hence
x()eOE (8-1) 8
0
E
Pe= -;;r;:- = N L
Calculations show that even in such strong fields the dipole moment
of an argon atom is six prders of magnitude smaller than that of a
water molecule.
26. Direct Current
26.2. From the symmetry of the circuit it is evident that the poten-
tials of points 2 and 4 are equal, so no current flows through the con-
ductor 2-4 and it may be removed.. This gives the circuit of Fig. 26.!b,
whose resistance is easily found. I ~
Electrodynamics 221
Fig. 26.2b.
2 r - - - - - - . ~
1,.....----4
26.3. The resistances of the conductors are proportional to their
lengths:
x = r (1 - cos 36)
26.4. The problem may be solved- in- stages. First we replace the star
connection by an equivalent resistance shown in Fig. 26.4a, and then
by an equivalent resistance shown in Fig.
26.4b. It is evident from considerations of
symmetry that the potentials of points H
and K are equal and so the connection HK
may be removed. We obtain a circuit whose
resistance is easily found to be a half of the
resistance of either of the two parallel links
of three conductors.
26.5. It is clear from considerations ofsym-
metry that the potentials at points 2, 4 and
5 coincide, I.e, CP2 = <p, = CP5 = cp'. Hence
the potentials at points 6, 8 and 3 coincide
as well: CP6 = CPs = <1'3 = cp". If we shortcir-
cuit the points at equal potential, Le, con-
nect the points by means of conductors of
negligible resistance, -the resistance of the
circuit remains unchanged. The circuit obtained in this way is shown
in Fig. 26.5b. Its resistance is equal to the sum of the resistances of
Fig. 26.4a.
three series-connected sections of three, six, and three parallel-
connected branches -each.
2 6 ~ ~ . In a balanced bridge the current flowing through the galvanom-
eter is zero, so eve = q> D. According to Ohm's law for a homogeneous
222 Solutions
section, we have CPA - Pc = IR; fPc - CPB = IR
x
; CPA - CPD =
= iR
l
= ip1l/S; CPD - <J>B = iR
2
= ip12/ S Hence we find the un-
known resistance.
26.7. The relative error is B= i:. +-}+L h l where h
R
is the
error in the value of the calibrating resistance, h is the error in
the position of the slide wire. We have
() ._ hR + hL
-7[" l(L-l)
Hence the error is at its minimum when the expression y = 1 (L - l)
is at its maximum. But
y=lL-l2= ~ 2 _ ~ +21 ; _l2= ~ 2 - (l- ; r
is at its maximum when 1 = L12, i.e. when the slide is in the middle
of the scale. This will be the case, if the calibrating resistance is
4
Fig. 26.5b.
chosen as close to the resistance being measured as possible.
26.8. The diameter of an arbitrary cross section of the conductor
a distance x from the minimum cross section is y == a + x (D - a)/l.
The current density and the field strength in an arbitrary cross section
are
= i = = ~ E=p!= 4pi
S n
y
2 , n
y
2
26.9. The resistance of the conductor is
l l
R= r pdx =i. r !!=..
J S n J y!
o 0
Let us change the variables, noting that y = a for z = 0 and y = D
fOT . X = l. Dirfferentia.ting, we obtain
'dy ==dx D-a so ds =.!:..!!:lL-
l D-a
Electrodynamics
Substituting into the expression for the resistance, we'. obtain
D
R= 4pl f.2JL= 4pl
n (D-a) J y2 n (D-a) Ya - sial)
a
223
26.10. (a) When the cells are connected in series, their e.m.f. and
internal resistances are added.
(b) When the cells are connected in parallel, the e.m.I, remains
unchanged, but the internal conductances are added.
(c.) In a mixed connection we calculate first the e.m.I. and the
internal resistance of a group, and then the same parameters for the
battery as a whole. .
26.1.1 ...,The accumulators should best be connected in series in two
groups 'of 100 cells each, bothgroups being connected in parallel and
the entire battery.connected through
a nheosta t to the terminals-of J a dy-
namo (Fig. 26.11). The e.m.f, of the
battery is:.<t!> = ne/2 = 100 X 1.4:c::
= \,40. V, the internal' '. resistance is
R
i
= nr/4 = 0.5 ohm. The current
in the circuit is I = 2i = 60 A.
Ohm's law can be written in the fol-
lowing form: L\cp - = I (R + R
i
) ,
from which
L\m Fig. 26.1.1.
R- "t'- n,
- I
26.13. The power Ptot == + r) is at its maximum in conditions
of short-circuit (R = 0). Tha-shert-circuit power is Psh.C =
The power in the external circuit is at its maximum when R = r,
To check this consider the extremum of the expression
Rr
P
ex
= -r- (R+r)2
We have
y = (R+r)2
Rr
R 2 r R 2 r
-+ +-=-- +-.+4=
r R r R
=(V -v )2+
4
Obviously, 1/= 4 for R = r Isminimum value. In this case the power
in the external circuit is maximum. The corresponding graphs are
shown in Fig. 26.13. . .
26.14. Connect all the .tubes in series (Fig. ,26.14) together with' a se-
ries resistor:' (U - n L\cp)/I, =',5-73 ohm. The power dissipated
is P = t U = 66 W. The ratio of the power dissipated in the tubes
224 Solutions
to that in the series resistor is
nL\q>
T8 T2 'TI
R
26.15. One possible circuit is that utilizing a five-position switch
(Fig. 26.15). In positions 5 and 4 the instrument is an ammeter with
terminals u+" and A, in the other
three it is a voltmeter with termi-
nals u+" and V.
26.t6. The voltage drop of 40 V
takes place in a two-wire line of
Fig. 26.13. Fig. 26.14.
R
t
V
known parameters, and this makes it possible to find the current
in the circuit. The number of lamps connected in parallel is equal
to the ratio of the current flowing in the
circuit to the current in one lamp.
26.17. The circuit shown in the diagram
of Fig. 26.17 enables anyone of the
three powers desired to be obtained.
26.18. The length of the wire is l =
0".6:rtt"U
2d2
-
26.19. (a) Numerical calculation. Compile
Table 26.19 from the available data. The
quantity of heat dissipated in time = Fig 26 15
= 1 s is nQ
n
= ifiR The quantity of ..
heat dissipated during the whole time
is equal to the sum of .the individual quantities .of heat:
+AQ,=(ilav+
iiav+
... At=
= 2580 X 40 X 1= 1.03 X 10
6
J
(b) Integration. The current changes according to the law u =
= 5 + 2t. The quantity of heat is
10 10 10
Q= ) i
2
R dt=40 I (5+2t)2dt=20 J(5+2t)2d(5+2t)=
o 0 0
t , s
Electrodynamics
Table 26.19
i. A
225
0 5
6
36
1 7
2 ~
8 fj4
~ 11
10 100
4 13
12 144
5 15
14 196
6 17
16 256
7 19
18 324
8 21
20 400
9 23
22 484
10 25
24 57G
Total 2580
10
== 20(5-+ 2t)3 I = 20 (25
3
- 53) == 20X 775X 20= 1.03 X 10
5
J
333
o
We see that the result of the numerical calculation was accurate.
26.20. For this circuit the parameters present in Ohm's law n(j) == iR
"""' /""0--,
1 ,...... Off
Fig. 26.17.
take the form
A q, __ I' ( f1 f/ ) dq
L.l{P =- l- 1m -- =:::--
C' ~ t ...O !1t . dt
The minus sign is due to the decrease in the capacitor's charge in the
process of its discharge. Substituting into the expression for Ohm's
law, we obtain
dt dq
RC if
15-0360
220
Integrating, we obtain
Solutions
_t_ == _ In q +In A
RC
where A is a constant. Noting that for t =:" 0, q qo, we obtain
o= -In qo + In A, from which A === qo. Hence
t
RC == -In q+ln qo
Denoting the time constant (the relaxation time) T = RC, we obtain
In ..!L. == _!...- q ==: qoe-t/'t
go 't '
For the current we have
. dq qo -tiT: _. -tIT:
L == - - == - e - foe ,
dt 't
where i == == == .!!.JL
o 17 RC R
26.21. (a) When the e.m.f. of the battery is less than the breakdown
potential of the voltage stabilizer, its resistance is infinite and the
current in the circuit is zero.
(b) When the e.rn.I. of the battery exceeds the firing potential
of the voltage stabilizer, its "resistance" drops to zero and the current
in the circuit is determined by the resistance of the resistor and the
difference between the. e.rn.I. and the firing potential of the voltage
stabili zer.
26.22. The current in a circuit containing a barretter is determined
by the transconductance of the device and is independent of the
circuit's resistance. However, current flows only if - toR> 0,
i.e, if > ioR. Otherwise the voltage across 'the harretter drops to
zero, and the current through it ceases to flow.
27. Magnetic Field in a Vacuum
27.1. The rigidi ty of the spring is due to the existence of electrical
forces of interaction between the particles of the material. In a moving
reference frame the lateral force decreases and hence the rigidity of
the lateral spring decreases as well: -"- ..
k..1 :=: kO-l Y1-v
2jc2
But if the acting force changes according to the same law as the rigi-
dity (the spring constant) of the spring, the lateral dimension of the
spring remains unchanged in full agreement with the theory of rela-
tivity.
Consider the following imaginary experiment. Suppose a current
flows in a conductor from the left to the right (Fig. 27.3a) so that
the electrons move to the left with a certain "velocity v. Let a free
electron move in the same direction wi th the same veloci ty. I n the
reference frame xyz of the" conductor there are three forces acting
on the electron: the force of repulsion from the electron cloud, F_,
Electrorlynnmics 227
the force of attraction to the ionic lattice, F+, and the Lorentz force
F
m
acting in the same direction. I t is known from experiment that
the resultant or those three forces acts in the direction of the conductor.
Let us consider the reference frame Xo!loZo of the electrons (Fig.
27.::\h). i lieltl here is cx act ly the s.une, but. it does not
Fig.
;:(0)
Fig.
act on a stationary electron. The forces acting- on an electron arc the
force of repulsion from the electron gas, F'?', and the force of attrac-
tion F+O) to the moving ionic lattice. But if in the former reference
frame the electron was attracted to the conductor, in accordance with
the principle of relativity it wi ll be attracted to it in any other refer-
ence frame as well. Therefore > and consequently, >
> E(_O). \Ve see that the lateral field intensity of moving charges is
greater than that of stationary charges, i.e. greater than the Coulomb
field.
27.4. Neglecting the thickness of the wire as compared with the
radius of the coil and the other dimensions, we obtain an expression
for the field induction on the axis:
l-l oiwa.
2
2 (a
2
-f- h
2
):i / 2
where to is the number of turns, a is the radius of a turn, and h is the
distance from the centre to the point on the axis of the coil where
the field is to be determined.
27.5. The induction of the magnetic field in the centre of a long sole-
noid is B == I-loin == If the wire is closely wound (see Fig.
the diameter of the insulated wire is d == ll ur. Hence B == I-loi/d.
At the end of the solenoid the field is twice as weak.
27.6. Making use of the result of Problem 27.4, we obtain for the
field at the centre of the ring
1 ):.=J.!oiWa2(1_t__
2 a
3
2a
3
Y125 a
15*
228 Solutions
The field induction at the midpoint is
B
m
== 2f.to
iwa2
_ 0.9131-toiw
2 (a
2
+a
2
/ 16)3/2 a
I t may be seen that with a small error <S the field is almost uniform:
~ Bm-Be =0.913-0.858=6%
Em 0.913
27.8. Putting the speed of rotation in the formula B = f.toqv/4nr
2
to be equal to v == ror, we obtain the charge sought.
27.9. Consider the disk to be divided into thin concentric rings. The
area of a typical ring is 118 = 2rtr I1r, where i ~ r is the thickness of the
ring. The charge of the ring is I1q == 0 ~ == 2Jtorl1r. When rotating,
the charge creates a magnetic field at the centre of the ring with
strength
fj,!l = aq v = a w ~ r
4nr
2
2
The total intensity of the field at the centre is
1 1
H = I1H
1
+IiH
2
+...== 2 <TW(L\
r
1 +I1r
2
+...)=2 o(j)R
where R is the external radius of the disk.
The magnetic moment of the ring is
I1q 1
I1Pm = inr
2
::= nr
2
T = 2' aq wr
2
== nOffir
3lir
To find the total magnetic moment of the rotating disk we must add
up all these values. We have
R
Pm= j rn:Jwr
3
dr = -} nlJwR4
o
1 1
The moment of momentum L = fro == 2: mR2ffi == 2 :rtffiphR4,
where p is the density of the material. We have
Pm a
Y=2ph
27. to. The magnetic field induction at the point of interest is
3
B==2f-LoPm= f-LoPm (a2+x2) - Y
4rtr
S
2n
The gradient is
Electrodynamics
dB = l-toPm 2+ 2)-3/'2_
dx 2n dx a x --
229
28. Charges and Currents in a Magnetic Field
28.1. The momentum of the particle may be found from the condition
mu
2
R = 2eBu, since an alpha-particle is a doubly ionized helium atom.
Having established the fact that this is a nonrelativistic particle,
\ve find its velocity and energy from the equations
u == 2eBR/m and K == 2e
2
B2R2/ m
28.2. The momentum of a muon is half that of the alpha-particle
of the previous problem. It may easily be established after calculating
the quantity .1 L = 1.66, that the muon in this problem
y moe
is a relativistic particle. We find = 0.856, from which the velocity
of the particle may be found.
The rest energy of a muon is 207 times that of an electron, i.e.
== 207 X 0.511 == 106 MeV (see Problem 8.1). The total energy is
= ,1 ,and the kinetic energy is K = - <t'o.
y
28.3. Since the charge of the particle and the magnetic field induction
remain unchanged, the radii of the tracks of the particles are propor-
tional to their momenta: R
1/R2
= Pl/P2. The relation to the kinetic
energy depends on the nature of motion.
(a) Nonrelativistic particles. The momenta of the particle are
proportional to the square roots of their kinetic energies, and there-
fore also proportional to the radii of their tracks.
(b) Relativistic particles. In this case the dependence of the momen-
tum on the kinetic energy of the particle is more complex:
1 --"--
p == - VK +K)
e
Hence we obtain the desired ratio of the radii of the tracks.
28.4. This is a relativistic electron, since its total energy == (go +
+ K = 0.511 + 1.5 = 2.0 MeV is much greater than its rest energy.
We obtain from mu
2
/R = euB
T _ 2nR _ 211m _
--u--es-e/k2
In the field the electron acquires a kinetic energy of 20 keV,
which is much less than its rest energy (511 ke V). Therefore in this
problem it is a nonrelativistic particle. Resolve the electron velocity
230 Solutions
into t\VO components: one along the line of induction, v II == v cos a,
and the other perpendicular to it, v
J
_ == v sin ex (Fig. 2H.5). No forces
act on the electron in the longitudinal direction, therefore the electron
will move at a constant speed along the z-axis according to the equa-
tion
Z :::-: Zo -{-- v II t = Zo + vt cos a
In the lateral direction (i.e. in the or!! plane) the electron is acted
upon by the Lorentz force, which makes the projection of its motion
in this plane a circle with radius
Ul,V.l Y2meq> sin a
R :.:.:: -e-B- =
and with period T = 2Jtm/eB. III space
the electron moves along a helix which
winds around the lines of induction. The
radius of the circle, R, was gi ven above.
The pitch of the helix is
2JtnzlJ cos ex
eB u-:2nRcota
28.6. An ion entering a uniform magnetic
field at an angle to the lines of induction
will move along a helix to one of the poles
of the magnet, and after some time will Fig. 28.5.
strike the dees and leave the bunch. To
prevent losses, the magnetic field should be made slightly nonuni-
form (in the shape of a barrel) (Fig. 28.6). It is easily seen that such
a field focusses the ions, c.oncentrating them in the middle plane.
28.7. The electrons in this problern are nonrelativistic, they enter
the magnetic field at a speed u == V2ecp/m = 1.33 X 10
7
m/s. In the
magnetic field they move along an arc of a circle of radius R ==
== umleB (Fig. 28. 7b). The electrons are deflected through an angle
L GCE == a. But the angle L GeE is congruent to the angle L COM
(as angles with mutually perpendicular sides). Therefore
. Me I eEL eBl
SIn a=--.: --=-==. --= ----
OC R III e
U
y 2m'eecp
As is evident from the figure, GD == GE + ED, or
i-cos ex a
d= L tan a +R (i-cos a):=: L tan a-r-l . ....:.= L tan a+1tan
SID a 2
Knowing the parameters of the device one may easily calculate
the displ accmont sought.
28.8. Making usc of the resul t of Problem 4.14 we obt.ain
eEL (. 1) El ( I )
d::==- 2ecp lJ+T == 2<p +2
whence
Electrodynamics
E == 2(pd
I (L ... j. 1/2)
28.9. First we find the charg-e of tho ion using the condition lnu
2/
R =
= quB. \Ve have q = ntu/BR. Taking aCC-OUIlt. of the fact that heavy
ions in the energy range of hundreds of megaelectron-volts move at
nonrelativistic speeds, ''"0 have m == A X 1.66 X 10-
27
, where
G
Fig. 28.6.
o
f\,
I
te;'R
I \
I "
I 'c
I ,
Mr---
L
Fig. 28.7b.
A ::-:: 20.18 amu is tho atomic mass of neon. The momentum of the
ion is found from its kinetic energy: mu Hence the charge
of the ion
V2 X 20.18 X 1.66 X 10-
27
X 100 X 1.6 X 10-
13
_
s> BR 1.55 X 1.1 -
=6.6 X 10-
19
C
Dividing by the electron charge we find the neon ions to be ionized
quadruply.
The total number of revolutions an ion makes is equal to its kinetic
energy divided by the energy acquired in the process of passing twice
through the accelerating gap:
lV= 100 X 10
6
=42
2qcp 2 X 4 X 300 X 10
3
The frequency of the change in polarity of the accelerating field is
equal to the circular frequency of the ion: .
n ::-= _u_ =-!l.!!-
2nR 231m
28.10. The magnetic induction is found from the condition
B==L=
ell ecR
232 Solutions
and the frequency from the condition
c V1-(e.t;o/'6:)2
11 === 2nR
(see 41.6). Note that towards the end of the acceleration cycle the
proton's energy is so great that n == c/2rrR approximately.
28. t t. The condition for the balance of forces is eE = euB from which
we get u === R/B. Since the change in the sign of the particle's charge
is accompanied by a simultaneous reversal in the direction of both
the electric and the magnetic forces, the sign of the charge cannot be
established.
The distance of interest is equal to the difference in the diarne-
tel's of the ionic orbits 2 (R 2 - R1)' The speeds of both ions
are identical and are determined by the conditions of their passage
through the velocity filter: u == E/ B. The radius of the orbit is R ==
mu/qB
o
= mE/qBB
o
. Therefore the difference fought is
2E
(m
2-,nt)
qBB
o
We obtain the difference sought by assuming the ion to be singly
ionized, i.e. by putting q == e.
28.t3. Making use of the result or the previous problem and noting
that R tz:': l12, we obtain the expression lor the ionic mass
el f/2 elB2 4.82 X 10
4
m t:t: 2E- (kg) == 2 X 1.66 X 10-27 E (amu) = E (amu)
28.14. The magnetic moment of the moving coil is Pm = un S; while
the torque is it! = PmB, since in our problem a == :rr/2.
28.15. A torque M == iol SB acts on the moving coil balanced by
a torque M =- fa due to the elasticity of the twisted thread, where f
is the torsion modulus, Equating, \\'C obtain fa tot SB: from which
we see that, other conditions being equal, the coil's angle of rotation
is proportional to the current.
28.16. It is easily seen that the moving coils hanging freely wi ll
arrange themselves so that their planes wi ll be perpendicular to their
. Thf' f F dB N' h
common aXIS. e oree 0 interaction IS = Pmdx. oting t at
dB 6f.1oPm.r
a:;-= - 4n (a2 -1- .r2 ) fi / 2
(see Problem 27.11) and that according to the conditions of the prob-
lem x r Q, we have
dB 61-loPm
-;[i = - 4nr4
The force of in teraction is
F=Pm dB = _
ax 4nr
4
Elcctrodvnarnics 233
I t is quite similar to the force of interaction het ween electric dipoles
(see 10.4 and 40.6).
28.17. Making use of the resul t of the previous problem, we obtain
F ...- _ Gf.low2i2S2 3nf1.ow2i2a4
-- 4nr4 - 2r4
28.18. The positive charge circulates as shown in Fig. 28.18a. The
direction of the magnetic moment is established with the aid of the
screw driver rule (sec 40.5, 40.6). The direction of circulation of
-... u
Fig. 28.18a. Fig. 28.18b.
a negative charge is the opposite (Fig. 28.18b), but the magnetic
moment is, just the same, directed against the field.
28.19. As was shown above (see Problem 28.5), the particle will move
along a helix winding around the lines of induction. Let us resolve
.Fig. 28.1UL.
the velocity vector into t\VO components: the lateral v-l (the orbital
velocity) and the longitudinal v 1/ (the drift velocity). Since the orbital
moment of the circulating charge is directed against the field, the
magnetic forces tend to push the charge out into the region of the
weak field (compare with 37.4 and 41.10, where the dipole and
magnetic moments are arranged in the direction of the field). vVe see
that as the charge approaches the magnetic mirror, its drift velocity
234 Solutions
decreases, becoming zero in the case of a suffieiently high field gra-
dient (Fig. 28.19b). From this moment it begins to drift in the opposite
direction towards the weaker field (Fig. 28.19c).
28.20. A charged particle entering a magnetic field perpendicularly
to the lines of induction wi ll move along the arc of a circle of radius
R = p/eB, where p is the momentum of the particle. Mirror reflection
I
Fig. 28.19c.
takes place if the entire path of the particle happens to lie inside
the field.
Since, according to the conditions of the problem, the electrons
move perpendicularly to the "magnetic mirror", they will be reflected
backwards, provided that the radius of the semicircle is less than
the thickness of the "mirror", Hence e; < d.
The total energy of the electron is =--= -t- p2c2 < 1/
the kinetic energy is K == -
Finally, we get
29. Magnetic Materials
29.1. According to the definition of the magnetization vector M,
the magnetic moment of a body in a magnetic field is Pm = M V =-=
== Xm
I1V.
29.3. The saturation magnetization is Msat == noPm' where 11
0
== 4/a'J
is the concentration of the atoms, Prn === 7.95f.lB is the magnet if
moment 0 f a single atom. Hence M sat == 4 X 7. a
3
23()
Solutions
29.9. From the graph 29. Ua we compile the table
JI,
A/m
50 75 100 200 I 500 1000 1500
15.0 X1031!UX10318.0 X10315 .OX10312.2X103\1.2X10310.8X lO
a
The graph for f.1 is given in Fig. 29.9b.
29.10. The right-hand half of the hysteresis loop is plutted using the
data given ill the table; the left-hand half is plotted symmetrically
8,T
1.4
1.2
1.0
0.6
0.4
0.2
............._-_....I....-__
-0.2
Fig. 29.10.
(Fig. 29.10). The coercive force is determined at the point of inter-
section of the graph with the H-axis; saturation induction is the
point where the upper and lower branches of the loop intersect. The
saturation magnetization is
M
- Usat II r<oJ B
sat
sat--- - sat ro../--
,to flo
for Bsnt The residual magnetization is M; = Br/flOt where
B; is determined at the point of intersection of the graph with the
B-axis.
Electrodynamics 237
30. Electromagnetic Induction
30.1. The bulb will not glow because the magnetic flux flowing through
the circuit made up of the wings, the wires and the bulb remains
unchanged.
30.3. The forces acting on a falling conductor are the force of gravity
mg directed downwards and the Ampere force F = iBl directed up-
wards. By Ohm's law i == and the induced e.m.I. is == Blv.
*
Fig. 30.4a. Fig. 30.4b.
Hence the Ampere force is F = R. The equation of motion
of the conductor is
B
2
l 2V
a= m
After a certain moment of time the acceleration becomes zero
and the conductor continues to fall uniformly at a constant speed
. (compare with Problem 5.13).
The rod moves perpendicularly to the lines of induction of the
kmagnetic field, and in a small section of it an elementary e.m.I,
r = Bv I1x is established, 11.1: being the length of the section and v
t. it s velocity (see Fig. 30.4a). The voltage across the rod is the sum of
elementary e.m.I. 'so Since v = wx, = Bwx Sx, The result
f may be found by t\VO methods:
(a) By integration. We have
r l
( h<p= JBwx dx = Bw [ .;2 =i- Bwl
2
o
(b) Graphically. Plot a graph of the strength of the induced field
:E* = = Biax, Since this is a linear function, its graph is
"'of the form shown in Fig. 30.4b. The voltage is numerically equal
to the area under the graph:
lBwl == Rwl
2
2 2
238 Solutions
30.5. The force applied to a conductor moving at a constant speed
is equal in magnitude to the magnetic force. But the current is i =
Blv
=.-=- . Hence
R R
F :.:.::..: iBl :::::: B'.!.l2V/R
30.6. Since the ring is small, the field inside it may be assumed to be
uniform and equal to the field on the axis. Therefore the magnetic
flux is
<D -.= BS
4n (a
2
'1- :r
2
) :{ /2
The variable in this case is the coordinate .1: == Xo - pi, where r is
the velocity of fall. The magnitude of the induced e.m. f. is
I I =-=-= dcD -= d (a2 x2) - 3/2
dt 2 dt
__2- 2 ( 2 -L 2)--5/ 2 .
2
..!!:=- _ 3""oPm
r2x v
- 4 f.toPm
r
a I x .r d --- 12
t
30 7 Th . I d f' cD . .. 11 th
. e mr nee e.m .. is C!> = - Lit= At ' since initi a y e
magnetic flux through the ring is zero. The charg-e 0 flowing through
the circui t is
. (})
q::o: Lf1t=jf
It is equal to the galvanometer constant multiplied by the number
of scale divisions: q :::::: C'N,
30.9. A change in magnetic flux is accompanied by an induced electric
field of strength E* = where r is the radius of the ring. This
field causes the polarization of the dielectric, i.e. a preferential orien-
tation of its dipoles in the direction of the induced electric field.
30.11. The current in the circuit is I == - Ind)1R, where cg
is the e.m.f. of the accumulator battery, R is the resistance of the
circuit (including the internal resistance of the battery) and (glnd
is the e.m.f. induced in the armature in the course of its rotation.
When the armature is stopped, = 0 and the current is 1
0
=
= '1>/ R, from which we find the resistance of the circuit. The power
of the motor is -
p'n (1- :0 )
30.14. To find the number of turns divide the length of the torus'
internal circumference by the wire diameter. We obtain
w =-= ota = 80rr/0.6 = 420
The magnetic field strength is
1 nD
m
dD
m
Electrodynamics 239
where D
m
== 120 mm is twice the distance from the axis of revolution
of the torus to the centre of the genera ting circle,
Knowing the magnetic field strength and making use of the graph
of Fig. 29.9a, we find the induction of the Held. Applying the formula
W
rrt
:-: BH/2 we find the energy density, and multiplying this by the
volume of the core we find the total energy of the magnetic field.
The inductance is found using the formula
Li
2j2
30.15. The energy of' the magnetic field in the absence of the ferro-
magnetic core is W
1n
1/2flo1J2V. It is significantly less than the
energy of the magnetic field in the presence of a ferromagnetic core,
despite the fact that the current Ilowing through the winding is the
same in both cases. The explanation is that in the absence of a ferro-
magnetic core the current in the coil attains its stationary value very
quickly and the work performed by the power supply in inducing
the magnetic field is much less than in the presence of such a core.
30.16. In the case of a good contact of the armature with the core,
the force of attraction at one pole is F =-:: zcmS. Therefore the armature
as a whole is attracted with the force F = 2w
mS
= where
S ::--: 60 X 60 X 10-
6
m
2
= 36 X 10-
4
n1
2
is the core cross section
(see Fig. 30.16). Since we assumed the gap between the armature and
the core to be negligible, we may assume the induction of the field
to be the same in the gap as in the core. The magnetic permeability
in the gap, on the other hand, is unity (u == 1).
We have Fsat = in case of saturation magnetization and
Fr=:..: eist, in the case of residual magnetization. The values of
Bsat and B, ale presented in the table following Problem 29.10 (see
also Fig. 29.10).
30.18. Ohm's law should be wri tten in the form + == iR,
where is the e.m.I. of the power supply and li'L = -L is the
e.m.I. oJ self-induction. Hence
di .
iR
dt
Divide by R and introduce the notation R =-= 11';1' the stationary
current, and L/ R ==- T, the relaxation time. \Ve obtain the equation
di . di .
I I\1-- 't i ; or -'t dt = Z - 1
M
Multiplying by dt, "'C obtain
-TrJi=--:(i-lM)dt,
Integrating, we obtain
or
di dt
i - J"AT = -- --=r
r di 1 r
J i-l1.1 = --:r J dt ,
t
yielding in (i-1"Ad= --+In C
1:'
240 Solutions
where C is the integration constant. Taking antilogarithms, we obtain
i-1M -til
--C-===e
When t = 0, the current is i (0) =-= 0, so C ::::- -1M. After some simple
transformations we obtain
i = 1M (1_e-
t
/'t)
The graph of this function is shown in 43.12, Fig. 43.6.
30.19. According to the result of the previous problem, we have for
t === 0.9/
IV1
an exponential equation
O.g[!vI = 11\1(1- e-
f
/
t
) , or e" tiT: = 0.1
Taking the logarithms, \\'C obtain
- _t_ log e= log 0.1, _..!..- 0.4347.": -1
Hence t == 2. 3-r.
30.20. The formula obtained in Problem 30.18 is no longer valid when
the difference between the stationary and the instantaneous currents
becomes equal to the thermal current fluctuations. After that a gradual
increase in the current is replaced by the usual thermal fluctuations
of the current around its equilibrium value.
r
Fig. 31.2b.
7'0
L\<p = = \ E* dr
()
The induced e.rn.f. may be found
either graphically (Fig. 31.2b), or by
integration:
31. Classical Electron Theory
31.2. Here the centrifugal force of inertia, I cf === lnw
2r,
where r is the
distance from the appropriate point of the conductor to the centre
of the disk is the non-electrical force
acting on the di sk. The strength of Eft
this force field is
E* =!.!:l.. = nH.u
2r
e e
TO
mw
2
r mw2r5
e J r dr= 2e
o
31.3. The electron concentration may be computed from the expres-
sion for the Hall voltage (see 44.2) and the electric conductivity
of copper found from the data on the dimensions of the plate, the
Electrodynamics 241
current and the longitudinal voltage. Knowing these quantities,
we may easily find the electron mobility from formula (44.8).
31.4. Since silver is a monovalent element, the electron concentra-
tion in it is equal to the concentration of atoms. The density is p =
== mon, where mo is the mass of the atom. But mo = A/NA, where A
is the atomic mass. Hence n == pNA/A and the Hall constant is
R
H
== A/epNA
31.9. A thermocouple may be regarded as a heat engine working
in the temperature range from T} to T2' where the indices 1 and 2 re-
fer to the hot and the cold junctions. When a charge passes through
a thermocouple, work is performed equal to W ::-= 'Y)Ql' The transported
charge is
W YJQ1
u: ~ = a. (T
1-T2
)
31.13. First find the volumetric heat (molar heat capacity). It may
easily be established (for instance, by checking the dimensionality)
that C == pc. From 45.3 we obtain
A= 3K == 3[(
aC a.pe
32. Electrical Conductivity of Electrolytes
32.1. The electrical conductivity of an electrolyte is l' == qn (b+ +
+ b_) =:-= o..qno tb; + b_), where a is the dissociation coefficient
sought. The charge of a monovalent ion is q == e. The concentration
of the solution is C == mono == Mno/NA, where M is the molar mass
of the dissolved substance. Substituting these values, we obtain
a= Vkl
eCN
A
(b++b_)
32.2. The mass of material deposited on the cathode is pSd= ~ ~ ;
the valence of nickel is Z == 2. The remaining information is con-
tained in the statement of the problem and in the tables.
32.3. First find the charge passing through the solution, making use
of the graph in Fig. 32.3. Since the current changes linearly, the
eventual value being i
ev
== 6 - 0.03 X 180 == 2.4 A, the charge is
easily obtained: q = (io -1- lev) t/2. Incidentally, the same result may
be obtained by integrating the expression
180
q = \ i dt
0
(cheek this). Then the mass of copper may he found from Faraday's
law.
32.5. The e.rn.f. of the power supply should exceed the e.m.I. of
polarization, which may be found from the specific energy of the
16-0360
242
chemical reaction:
Solutions
'AA
eNAZ
32.6. The capacitor's charge is q = C the accumulated energy
1
is W
cap
= "2C = 1.8 J. The amount of hydrogen produced
is m = ;the quantity of heat liberated in the process of com-
bustion is Q = rnA = 9.1 X 10-
3
J, which is much less than the
energy of the capacitor. Obviously, in the course of the discharge
i,A
6
30 60 90 120 150 180 t.s
Fig. 32.3.
of the capacitor through the electrolyte, part of the energy stored
in it will be liberated in the form of heat, with only a small fraction
being spent on the chemical reaction.
32.7. The lift F =-= (Po - PH) Vg, where Po is the density of air and
PH the density of hydrogen. The mass of hydrogen is
m = pRV = PHF/g (Po - PH).
Applying Faraday's law, we find the charge q passing through the
electrolyte solution. The energy required to produce the hydrogen is
where Q represents the Joule heat loss. The polarization e.m.f. has
already been found in Problem 32.5.
33. Electric Current in a Vacuum and in Gases
33.1. The calculation is done using the Richardson-Dushmann formu-
la. For the sake of convenience one should first find the logarithm
of the current:
log isat=log n+log d+log l+log B+2log T-O.434 log Ao/kT
Electrodynamics 243
33.2. According to the Richardson-Dushmann formula we have
x== (Il..)2 exp [ A
o
(T
2
-T
1
) J
II T
1
kT
1T2
For the computation one should find log x,
33.3. The collision of the electrons with the auode is an inelastic
one, and this makes it possible to compute the force using formula
(17.19) from 17.5. Noting that the current is i == enSv, we obtain
r imo i '/-2-
1 =-e-=---e y ecpm
33.4. The ratio of the saturation currents of the two cathodes is
(!..L)2 exp
i
2
T
2
kT
2
kT
I
33.5. Calculations should be done for the linear section of the charac-
teristic curve. In. this case a change in the anode voltage a ==
= 150 V, with a constant grid voltage (for instance, with Ug = 0),
causes a change in the anode current of == 75 rnA. The tube's
internal resistance is R
i
= a The tube's amplification factor is
V- = a t i.e. V- is the ratio of the change in the anode voltage to the
change the grid voltage which causes a given change in the anode
current. It follows from Fig. 33.5 that a change of anode current
equal to 75 rnA may be obtained either by changing the anode voltage
by 150 V, or by changing the grid voltage by g == 7.5 v.
Therefore the grid is 20 times more effective in controlling the current
flowing through the tube (on the linear section of its characteristic
curve), i.e. amplification factor I.t == 20.
33.6. Away from the saturation region the current density is a func-
tion of the ion mobility ( 48.2), from which we get for the ion con-
centration
The concentration of air molecules in normal conditions is equal
to the Loschmidt number ( 26.9), which gives the ionization coef-
ficient ex. == n]NL.
33.8. Away from the saturation region the current will increase, since
the strength of the electric field increases as the plates are brought
closer together. In conditions of saturation the current will decrease,
since the effective volume of the ionization chamber is smaller. The
current-voltage characteristics are shown in Fig. 33.8.
33.9. The saturation current is isat = == veV, where 'V is the
number of ions produced per second per unit volume of the chamber.
33.10. Usually the computations are carried out using the relation
3/2kT = This, however, yields too large a value for the tem-
16*
244 Solutions
perature:
o
a,
Fig. 33.8.
T= :::::: 106 K
In fact, ionization takes place at lower temperatures. The reason
is the Maxwellian molecular speed distribution, according to which
in equilibrium conditions there is always a noticeable percentage of
molecules with speeds exceeding
the average. For example, from
Table 25.1 (see 25.2) it may be i
9
seen that 368 2.5% of the mole-
cules have a speed more than three
times the average. This means
that their kinetic energy is more
than nine times greater than the
average kinetic energy of the mol-
ecules.
33.11. Thermal motion of ions and
of electrons in a magnetic field
takes the form of motion along arcs
of circles, whose radii may be
computed with the aid of the formula used to compute the radius
of the ionic cyclotron orbit (see 41.2). We can express the momentum
of the particle in terms of the gas temperature: p == -V 3,nk T which
gives the cyclotron radius R == -V 3mkT .
eB
33.12. First find the Reynolds number:
13.6 X 10
3
X 5 X 10-
2
X 0.2 8.8 X 10' 2300.
11 1.55 X 10-
3
Hence the flow is turbulent and it should be assessed with the aid of
the Stewart number:
N == yB
2l
= 1.05 X 10
6
X 0.36 X 5 X 10-
2
=7
pv 13.6 X 10
3
X 0.2
But the Stewart number represents the ratio of the magnetic force
to the resistance of pressure. Hence, in this case the magnetic force
will appreciably affect the coefficient of hydraulic friction.
33.t3.
Re = 1,(.2 X 10
3
X 5 X 10-
2
X 0.2 ==4200 > 2300
2.44 X 10-
3
N = 6.5 X10-
3
1.
33.14. Since the lines of induction are frozen into the plasma, the
magnetic flux will remain unchanged as the star contracts: <I> = <1>0'
Vibrations and Waves
245
or BR2 = B
o
R5. Hence we find the magnetic induction of the pulsar.
For data on star radii (prior to contraction), and pulsars, see Prob-
lem 14.21.
33.15. The pressure of the magnetic field is equal to the energy density:
Pm ==W
m
= B2/2,...,0
The pressure of the gravitational forces will be assessed in the same
way as in Problem 16.7:
-- 3M yM R 3yM2
Pgrev == pgh=-:-. 4nR3' 2R2 . 2= 16nR4
Substituting the values of the pulsar's mass and radius (see Prob-
lem 14.21) we obtain the value of the pressure exerted by the gravita-
tional forces.
34. Harmonic Vibrations
34.4. The expression for the kinetic energy is transformed as follows:
K = 2.50 cos
2
( 20m+ 3
4
n ) ,= 1.25 [ 1+cos ( 40nt + 3; ) ] =
::= 1.25 (1 +sin 40nt)
The frequency of energy oscillations is VK == 40n/(2n) == 20 Hz,
the period is TK == 0.05 s.
34.5. The law of oscillations is of the form s == A cos (wt + rp).
Since So == 0, it follows that 0 == A cos tp, giving cp == (2k + 1) ~ .
The initial phase is always less than the period of the oscillations
cp <2n. Hence cp == n/2, or cp = 3n/2.
The particle's velocity is v == -Aw sin (wt + rp). The initial
velocity V
o
:.= -Aw sin (j) == 0.20 is, according to the statement of
the problem, a positive quantity, and this is possible only if cp =
== 3n/2. Hence, A w == 0.2. But w = 2Jtv = rr rad/s. Hence the
amplitude is A = 0.20/n = 0.064 m. Knowing the amplitude, the
frequency and the initial phase we may easily write down the equation
of the oscillations.
34.6. The equation of the oscillations is of the form s = A cos (wt -I-
+ cp). The total energy is W === mw
2
A2/2, the velocity is v =
== -A(J) sin (wt + cp). At -the initial point of time
So == A cos cp, lJ
u
:-:::- - Amsin cp, W::-- m(i)2A2J2
. P
n
1/ In. S' . 0 I > 0
gJVI n~ SI n cp -= - AU) == - "0 2W . ~ HI ce SI n 'P > an( cos 'P ,
it follows tha t the init.ial phase rp lies in the in terva I 0 < cp < n/2.
The circular frequency is ro == - V
o
cot cp, the ampl itude A ==
So
= l./ s& -I- ~ 2. The period of oscillations is 50 lOS, so the time of
246 Solutions
0.4 s spans 8 periods. During this time the particle will cover a distance
equal to 32 amplitudes.
34.8. Apply the vector diagram (Fig. 34.8). Here
-./( A)2 (A A)2 . (A/4)-(A/8) 1
B= V A-
T
+ ---8 ; A-(A/4) 2
34.10 Carry out the following transformations:
Fig. 34.8.
8=4 cos> y. sin 1000t =2 (1+cos t) sin 1000t=
=2 sin 1000t+2 sin 1000tcos t-=2 sin 1000t+sin 1001t-sin 999t
The spectrum is shown in Fig. 34.10.
A,m
2
999 1000 100/ w, radts
A,m
30/;6
1/16
496 500 504 fJJ,rad/s
Ing. 34.10. Fig. 34.11.
34.1t. \Ve have
1+cos
2
t+sin
4
t = 1++(l +cos 21)+i- (1- cos 2t)2=
-1
= T (4+ 2 2 Cos 2t +1-2 cos 2t + 2t) .;...:
Vibrations and Waves 247
1 r 1 J 1
="4 .7+T(1+cos4t) =S(15+cos4t)
Hence it follows
s = (15+cos4t)sin500t=
15 . 5.0 1. .
() l+gSlI1500t,cOS4.t==
= sin 500t + 11
6
sin 504t +11
G
sin 4961
The spectrum is shown in Fig. 34.11.
A,
30/16
8/16
1/16
496 498 500 502 504 w, rad/s
Fig. 34.12.
34.12. We have
1 1
1+ cos
2
t +cos- t == 1+2" (1-+-cos 2t) +T ('1-+cos 2t)2 =
1
= T (4+ 2+2 cos 2t+ 1+2eos 2t +cos
2
2t) =
1 r 1 -1 1
=T 7-1-
4
cos 2t +2(1 +- cos 4t) J= 8 (15-1-8 cos 2t +cos
Hence
s (15 -r- Rcos 2/-1- r os 4t) sin 500t --,- sin SOOt + cos 2t sin 5')Ot +
+--} cos 4tsin500t = sin 500t+ ; sin 502t+ sin498t+
_L sin 504t + t
1
(j sin 496t
The spectrum of the vibrations considered above is shown in Fig. 34.12.
248 Solutions
35. Free Vibrations
35.2. Q = klhw
o,
where h =:: Ffrl v == 6:rrMl.
35.3". The friction force Ffr == is independent of the velocity,
and this makes it possible to discuss the problem from the standpoint
of energy. Suppose the weight goes over from the initial state of maxi-
mum deflection from the position of equilibrium, characterized by
the amplitude A 0, to another, similar state with amplitude A1. Then,
according to the law of conservation of energy,
-} kAij-V-mgAo= kAi+f.lmgAI
from which i t follows that
A -- A _ 2f.lmf!
1- 0 k
The sanle will he true for all the subsequent oscillations, i.e. the
amplitudes will form an arithmetical progression:
A -A _
n -- 0 k
The pendulum will stop when its amplitude becomes zero. Substitut-
ing An === 0 gives n == 2
k Ao
. Since all the amplitudes except the
}J.mg
initial one are passed through twice the number of swings is
N = 2n - 1== kA
o
- t
Il
m
g
As may be seen, due to friction the mechanic-a} energy rather quickly
transforms into internal energy, and the oscillations cease.
35.4. When the volume of a gas is changed at constant temperature
we have, according to the Boyle-Mal iotte law
PI (d - x) S == P2 (d + x) S == p dS
The force acting on the piston is
2pxSd
F==(PI-P2) d
2
.- x
2
where V is the volume of one half of the vessel. As can be seen, the
force does not conform to Hooke's law, and the oscillations are not
harmonic. But for small deflections of the piston (when x d), the
force will be quasi-elastic: F == 2pVxld
2
, and the oscillations of the
piston will be harmonic. The rigidity of the system is k == Fix ==
== 2pV/d
2
35.5. For an adiabatic process we must make use of the Poisson equa-
tion, and of the approximations (for x <t: d)
(d::)" = (1+ i 1- I =(1-; r" 1-1-
Vibrations and Waves
For the force we obtain
u
Fig. 35.10.
Ck \
\
\
\
\
\
\
i.e. Fad == yF1SQt.
35.6. As may be seen from Fig. 35.6 the force which restores the body
to its equilibrium positions is F == pg AV == 2pgSx. Since this force
is proportional to the displacement, the natural frequency of the
vibrations can be found from the formula W
o
== 1/kl m. Here k ==
== FIx == 2pgS, m == pSI.
Hence W
o
== 1
/
2g/ l.
35.7. The restoring force is F = PogSx, where S == 20 X 20 cm
2
==
== 4 X 10-
2
m", Po is the density of water, and x is the increase in the
depth of immersion. Since the force is
quasi-elastic, the frequency can be obtained
using the familiar formula.
35.8. The period of oscillations of the pen-
dulum at the surface of the Earth is To ==
=2n liZ/go' At an altitude h above the sur-
face of the Earth it is T = 2n l
f
l/g. We
have L.\TT = 2.., where To = 8.64 X 10
4
s is
o To
the duration of a complete day and 't is" the
lag of the clock. Hence
't
o
L1 T=.-:: 'to =-. 'To (ygo/g--1)
To To
The acceleration due to gravity at the Earth's surface is go = -v;:
while at an altitude h it is g = (R where R is the radius of the
Earth. After some simple transformations we obtain
35.9. One should find the total acceleration with respect to the Earth:
W = II a
2
+ g2. Then the period of the pendulum wrll be T ==
== 2n VZ!w.
The equilibrium position will be deflected from the vertical by
an angle rp = arc tall .!!:...
35.10. The part icles velocity at. an arbitrary point on the circle is
(Fig. 35.10)
n:: V2g1(eosa-cos (Xo)=x y2gl
250 Solutions
where x = l/ cos a - cos 0..0' The body travels along the element
of arc I1s = I Ao.. in time
Put /),,0.. = 3 = 1/60 radian. Then
/
T 112
l -. -1.--
g 2q(),(itv
The period of oscillations is
rT Y2 (1 1 )
--60 -.-+-+...
g llav X2av
Table 35.H)
x
a I cos a Icos a - cos ao I
0 45 0.7071 0.0000 0.0000
0.005 10.50
1
42
0.7431 0.0360 0.1
0.228 4.38
2 39 0.7771 0.0700 0.265
0.292 3.43
3 36 0.8090 0.10in 0.318
0.:340 2.94
4 33 0.8387 0.1316 0.362
0.380 2.63
5
0.8660 0.15SU O.
0.414 2.42
6 27 0.8910 0.18:39 0.428
0.441 2.27
7 24 0.9135 0.2064 0.454
0.456 2.15
8 21 o. 0.2265 0.476
0.485 2.06
9
-tHo
0.nS11 0.2440 a.1t84
0.501 1.nU
10 15 0.2538 0.508
0.510 1.90
11 12 0.9781 0.2710 0.520
0.525 1.91
12
go
0.9877 0.280G 0.530
0.532 1.88
1:{
()O
0.9945 0.2874 0.535
0.538 1.89
14
')0
o 0.540
.J
0.540 1.85
15 0 1.0000 0.292') 0.541
Total I 44.23
Vibrations and Waves 251
Since the formula To == 23t 1
1
ZIg is used to compute the period of
small oscillations, it follows that T == kT
o
, where
k= V2 (_1+_1+...)
60 Xlav X
2
rl V
is the correction factor. Let us compile a table (see Table 35.10) Irom
the computed data. We have
k = V2" X 44. 23. ~ . = 1.042
60
Hence, the period T == 2nk V' l/g == 2.08 s, the value of To being
To = 23t1/ ZIg == 2.00 s. In this case the relative error due to the
use of the formula for small oscillations will be 4%.
35.13. The moment of inertia about the axis is
1 1
1=3 m
l
l
2
+
2
,n
2r
2
+'fl
2
(l+r)2
The distance between the centre of mass and the axis is
lienee we find the period.
36. Forced Vibrations. Alternating Current
36.1. Away from resonance, A == I ~ M 2) I' at resonance Ares ==
m wo-w
= QA s ta t === QFMlk.
36.2. Since V1\1 = wA, for co = 0 and for w - - ~ 00 the ampli tude
of the velocity becomes VM == o. The resonance curve is shown in
Fig. 36.2.
36.3. First \ve find the natural frequency (see Problem 35.1). We have
(J)o = 1/ g/A stat. The Q factor is found from the time of damping:
Q ~ wo't ::= 't ~ rI g/Astat. lienee we find the amplitude at resonance:
Ares = QAstat ~ l' VgAstat
Clearly, at resonance the system wi ll break down.
36.". To prevent the merging of t\VO successive impulses the Interval
between them should be appreciably longer than the time of damping
of the first impulse and the build-up time of the second. i.e.
'f
s p
> 2,; ~ 4Q/w
o
To find tho maximum volume of information transmit ted per
second, one should take into account that such information should
252
Solutions
Fig. 36.2.
be made up of N1 dots and N 2 dashes: N == N1 + N 2' where N is the
amount of information. But usually on the average N
1
N2 N12.
We have
1
Hence we find the amount of information: N R116L. .
36.5. Let the voltage be u == UM cos wt! . and the current l ==
:::::: ['!ott cos (wt + rp). According to the definltion of the back e.m.I.,
self-induced e.m.I. = -L , the voltage VI'U
M
drop across the inductive reactance is uL ==
- L = L (see 54.2). Substituting
di -I.Mw sin (wt-1-Cp) we obtain U
AI
cos wt=
dt
sin (wt + cp). Hence
UM n
XL= 1M =----=Lw,
36.6. Let the vol tage be U = U1\1 cos wi
and the current i = 1M cos (wi + rp). The
voltage drop across the capacitor is Uc =
:::::: qlC, and the current is
dq du .
i:.:= dt=== C dt= -UMCw SIn rot
Hence 1M cos (wt + cp) = - UI!t!Cw sin wt from which it follows that
UM 1 1t
",Yc=--= 1
M
= Cw' CP=T
36.7. \\Then a coil and a resistor are connected in series, the current
flowing through them is the same, and there is a phase shift between
the voltages. Therefore the vector diagram is plotted with the vector
representing the effective or the amplitude value of the current (see
Fig. 36.7).
36.8. The vector diagram is shown in Fig. 36.8.
36.9. When a capacitor and a resistor are connected in parallel, the
same voltage is applied to them, and there is a phase shift between
the currents. Therefore the vector diagram is plotted with the vector
representing the effective or the amplitude value of the voltage (see
Fig. 36.9).
36.10. III this circuit Z - 1/R2 -+ (Loo - ;00 r.Let us take the
resistance from under the square sign and the capacitative reactance
Xc = 1/Cw out of the brackets, we obtain
Vibrations and Waves 253
The natural frequency is 000 = Vlc I Q = V = I giving
LC = R2C2 == 1/Q2wa. Substituting into the expression for the
impedance, we obtain
Z=R 1/-1+ (:;_1)2 =R l/1+ (y2_1)2
(J)
where y = -
(a)o
36.11. The vector diagram is shown in Fig. 36.11b (see Problem 36.9),
where
Zo== II R2+L
2
w
2
, cos CPo:=7- R/Z
o,
sin CPo = LwjZo
u
Fig. 3f>.7. Fig. 36.8. Fig. 36.9.
It j s clear from the diagram that
1
2
=
but since a == (n/2) - CPo, it follows that
/2 = c sin <Po
Substituting the values of the currents and sin CPo, we obtain after
some simple transformations the required effective value of the current
in the unbranched section of the circuit.
. I cos fP UR
From the vector diagram cos cp == 0 I 0 1Z2
u
36.12. To find the power we apply the formula P == I U cos cpo Knowing
the amplitude to be UM == 312 V, we must find the effective voltage
and substitute the data into the formulae of the previous problem.
To simplify computations assume n
2
== 10.
36.14. An electrodynamic wattmeter has two coils (Fig. 36.14). The
stationary coil wound with thick wire is connected in series with the
load. It sets up a magnetic field whose induction is proportional to
the current:
B<Xi,orB=k1IMCOS(wt+cp)
254 Solutions
The moving coil is connected in parallel with the load, so the magnetic
moment of this coil is proportional to the voltage:
Pm ex: u, or Pm = k 2 U ~ cos wt
Here k
1
and k
2
are proportionality factors. The instantaneous value
of the torque acting on the moving coil is
!tl:.= PmB== k
1k2I MUM
cos wtcos (wt+ q
== k
1k21
U [cos cp +cos (2ffit -1- rp) 1
The average value of the torque is
M av == k
1k21
U [cos cp -f- cos (2ffit ~ q [av ==k
tk2lU
cos cp
since the average value of the term cos (2wt + cp) is zero. We see that
the average value of the torque acting on the moving part of the
instrument is proportional to the aver-
age power consumed in the circui t, so
the electrodynamic wattmeter measures
the active power. ~
36.16. The glow appears due to the in- e
stantaneous value of the voltage, not to "
the effective value of the voltage mea- ......
sured by voltmeter. Since the amplitude
of the voltage is UM = U y2= 85 V,
the lamp will glow for a certain part of U
each period (see Fig.36.16).
36.17. The amplitude of the voltage ex-
ceeds the breakdown voltage.
36.19. If we neglect the phase shift we
may easily find the current in the sec- Fig. 36.11 b.
ondary 1
2
= kl
1
, and the number of
turns W
2
= wllk.
Assuming the maximum current density in the wires to be the
same, we find the cross section of the wires to be proportional to the
currents, therefore 8
2
= k8
1
-
To find the resistance of the secondary we must know the length
of the wire. Note that according to the statement of the problem,
the winding is wound in a single layer, so the lengths of the wires
are proportional to the number of turns:
t; w
2
1
7;= WI =k
Since the wires are of the same material, the ratio of their resistances is
R
2
l281 1
1f;= l1
8
2 =/i,2
This allows us to find the resistance of the secondary.
The copper losses in the windings are Pcopper == IfR
l
+ IiR
2
,
and the efficiency is
P2 PI-Pcopper
11==1\"= PI
Vibrations and Waves 255
36.20. In the no-load condition, the secondary does not consume power.
Hence PIlo-load = I;1o-load
r
l + P
i ron
where r
1
is the resistance of
-the primary. But the current / no-load of correctly designed: transform-
ers is very small due to the enor-
niOUS inductive reactance, and also
because the resistance of the wind-
ing is small. For this reason the
first term may be neglected in the
power balance and, Pno-I and =:
= Plron.
36.21. Consider the high frequency
case when the resistance of a solid
copper or aluminium ring is neg-
ligible compared with its inductive
reactance. Suppose a sinusoidal
current flows in the primary. The
magnetic flux permeating the ring
will also be sinusoidal. The e.m.I,
induced in the ring is proportional Fig. 36.14.
to the time derivative "of the mag-
netic flux with a minus sign. Hence the phase lag of the induced
e.m.f. behind the current in the primary is = -n/2. The phase
i,s
-70
-80
-85
u,v
85
80
Fig. 36.16.
lag of the current oscillations in the ring, as in any other inductive
reactance, wi ll be the same: q>i = -rr/2. Therefore the phase shift
between the currents in the ring and in the primary will be q> =
= + <Pi -n, i.e. these currents are in opposite phaseS; which
means they flow in opposite directions. Such currents, as is well
known, repel each other. The force of repulsion balances the force
of gravity, with the result that the ring "floats" in the air.
36.22. The secondary feeds a resistive load, so CP2 == o. The power
factor is
cos <PI = PI//lU!
The active power P
l
is obtained from the efficiency.
256 Solutions
37. Elastic Waves
log 2
Therefore
37.5. First we must find the density of air and the velocity of the
wave u == Vyplp at a temperature of 27C, and then apply the for-
mulas of 55.3 to calculate the energy and the amplitude of the wave.
37.6. Find the intensity of the wave 1 == PI41tr
2
, assuming the source
to be a point source. Then, as in the previous problem, find the ampli-
tude of the wave.
37.7. The intensity levels are connected with the intensities by the
relation
Zl - Z2 = 10 log (1
1/12
)
But for a point source, by the results of the previous problem, 1
1/12
==
= So
:1 - 2
2
== 20 log (r
2/rl)
37.8. For a small (i.e. a point) source, the wave intensity is inversely
proportional to the square of the distance from the source. I f in addi-
tion we take into account the attenuation, one would obtain for the
wave intensities at distances r1 and r2 from the source the expressions
I - I 2-1'I!L I __ lorg 2--T 2/ L
I-T' , 2-rr-
where L is the half-thickness. Hence we ohtain
T
2
- r
t
I
2--
_I L
1
2
rf
The difference in intensity levels is
s: Z -101 11 -
20 1
r2+10(r2-r1 )
1 - 2 - og 7; -_. og . L
37.9. The wave intensity is 1 == lo2-
x/ L
==
2-
x
/
L
= en J.lX
Taking the logarithms, we obtain
;r 1
L n2=l.l
x
which gives the required relation.
37.11. To solve the problem we should make use of the expression for
the frequency of the signal from a moving source as measured by
a stationary observer:
V
o
I v -
VI =:::: 1+x anr 2 -- 1- x
where z =: v/u is the ratio of the velocity of the source to the wave
velocity. The beat frequency is equal to the difference in frequencies:
2.x'v
n
2vv
o
V=V1-
V2:-"";:
1-x2 = u(1-v2ju2 )
Vibrations and Waves 251
We obtain a quadratic equation vx
2
+ 2v
ox
- v = 0, which gives
- "0 -f- vVf+V2 ~ ,,2 - "n " v
x = - ~ v " ("0-1- V ~ +,,2) Vo -1- Vv ~ -t- v
2
~ 2"0
since the beat frequency v is much lower than the natural frequency
of the tuning fork, Vo. Hence
~ ~ ~ from which v ~ ~
u lV
o
' 2"0
37.12. \Vhen the source approaches the receiver, the relation
1+V/u
v="o 1-v/u
holds, where V and v are respectively the speeds of the receiver and
the source relative to the transmitting medium. In the case of the
source moving away from the receiver, the signs in the numerator
and the denominator should be reversed.
37.18. In the presence of an oil film the ultrasonic vibrations enter
the part. In the presence of an air gap between the transducer and
the part, the wave is completely reflected by a layer of air back to the
transducer, and does not enter the part.
37.19. The reflected and the direct pulses can be seen separately only
if there is an interval of more than half the duration of the pulse
between them. This interval wi ll be equal to 't == 30T == 30/". During
this time the wave should twice travel through the thickness of the
metal in the direct and in the reverse directions, i.e. 't == 21/u. Hence
1== 30 u/2v == 15A.
38. Interference and Diffraction
38.1. Let the wave equation in a medium with wave resistance Zl =
= P1U
1
be
$1 == Al cos (wt - kx)
The wave is reflected by the medium with wave resistance Z2 =
= P2U2' and the equation of the reflected wave at the boundary is
8ref :=: A
ref
cos (wt +kx)
But
Zl-Z2
Ar el == Al --+-
Zl Z2
Therefore for Z2 > Zl the amplitude of the reflected wave is negative,
i.e, the phase of the wave is reversed. To simplify the computation
put Z2 ~ Zl. We obtain Aref = -At and the equation of the wave
in the first medium in the form
8 == 8t +8ret == Al cos (wt - kx) - Al cos (wt +kx) = 2A
1
sin kx sin rot
At the boundary Xo == 0 so So == 0 for any t, Thus, the node of the
standing wave appears here.
17-0360
258 Solutions
Similarly, for Z2 Zl we have A ref = AI' and the equation of
the standing wa ve is
S ::.:.::: 1 COS kx -cos wt
The amplitude at the boundary between the two media (x
o
= 0)
is 2A
1
, i.c. an antinode appears.
38.2 The wave resistance of quartz is greater than the wave resistance
of air (the lower surface) or of water (the upper surface). Therefore
there will be antinodes at both boundaries (although their amplitudes
wil! be different). lienee it follows that an integral number of half-
waves fits into the plate's thickness: 1 = n'A/2, and this enables the
fundamental frequency and the harmonics to be found.
If the upper surface of the quartz plate is covered by oil, the fre-
quency wi ll remain unchanged, although the power radiated by the
lower and the upper surfaces will be redistri bu ted.
38.5. The beat frequency is equal to the difference in frequencies:
'21
or
Hence
T
1-T2
lv
YT
1+YT2
10
But T
1
T
2
== T, therefore sr = 0.2lv YT.
38.6. Resonance sets in when an odd number of quarter waves fits
along the air column: 1 == (2n + 1) ').../4. In our case n = 0; 1; 2.
38.7. The first-order interference minimum will be observed for an
auxiliary angle al :.:= rr. In this case the angular width of the prin-
cipal maximum will clearly be
e 2 . Aa1 . U
'V =-= 2 1 = arcsin nD = 2 arcsin vD
39. Electromagnetic Waves
39.2. A standing wave is established along the aerial, as shown in
Fig. 39.2. In the middle of the aerial the current is at its maximum
(current anti node) and at the ends it is at its minimum (nodes), and
since the magnetic field strength is proportional to the current, distri-
bution of the standing wave of the magnetic field vector will be simi-
lar. As to the electric field strength, it forms antinodes at the ends
of the aerial (whyr), and a node in the middle.
Thus 1 = "-./2, whence 'A == 2l.
39.3. According to the theory of forced oscillations, the frequency
of a ",'ave does not change when it crosses a boundary between two
media. The parameters subject to change are the velocity and the
wavelength. \Ve have v = == i: where 'A = 2l is the wavelength
o
Vibrations and Waves 259
Fig. 39.2.
IE
,
,
I
I
I
,
I
I
I
I
I
I
,
HI
I
I I
\,
r
1\
,\
, \
I \
I
EMHM .. /' Eo
WM:::=: U C V P'o
The average value of the energy is equal to one half
of its amplitude value:
W= eE&: "1/ Eo
2c J.lo
In the case of an ideal absorbing surface the pressure is equal to the
average energy density.
The energy absorbed by an area S in time t is
- .I-u
'fb==- wuSt = -2-' JI
The of the is fAt == e.nSVM where VM == ooA
IS the amplitude of the velocity of the oscIllations of the charge.
The magnitude of the oscillating charge is q == enSl hence 1M ==
== qA ro/ l, The radiation power is '
in the liquid, "-0 is the wavelength in air. Noting that u ==
we obtain "'0:= 2l V-e.
39.4. First find the velocity of the wave and the medium in which
it propagates. We have u == oolk == elYe, and the dielec-
tric constant of the material e == e
2
k
2
/ 00
2
The magnetic permeability of the material is J.l =
= 1.0. Find the amplitude of the magnetic field
strength:
HM==E
M
Yeo/llJ.lo
The amplitude of the energy density of the electro-
magnetic wave is
But for a half-wave antenna wl = rrc (see Problem 39.2), so
p = ;2 !!oclL:
The power is P == f2R == so the antenna is equivalent
to a resistance
fi 1t J/ flo
R=-flo
c
= - -
6 6 eo
39.6. The power of the synchrotron radiation may be found from the
formula
If q
2
a
2
p=_t""_o__
12nc
t7*
26c)
Solutions
V
f
y )/2 V
Fig. 39.7.
where vaud is the maximum frequency of
the audio signal which must pass through
the circuit without substantial distortion.
Noting that 'Vaud 2 X 10
3
Hz, we obtain
RjL?;:; 4n X 10
3
(ohm-Hr")
The capacitance of the capacitor is C == __1_ . The natural
4n
2v2L
frequency of the resonance circuit is V
o
== cIA== 1.2 Ox 10
7
Hz. Hence
Q 'VoJvaud
Expressing the Q-factor and the natural
frequency of the circui t in terms of i ts pa-
rameters, we obtain
RjL >2nv
au
d
where a is the acceleration and 9 the charge of the bunch. Since a ==
== w
2
r and the current is I ql T == qw/2n, where T is the period of
- circular motion it follows that q == 2nI/w. Substituting into the
expression for the power, we obtain
p =
But roT == v is the velocity of the electron, therefore
P == 3tJ.1
0
12v2/ 3c
39.7. The spectrum of a modulated signal is shown in Fig. 39.7. For
there to be no substantial distortion of the signal the half-width of the
resonance curve should not be less than the
half-width of the spectrum of the modulated A.
signal: f!v > VI - V, or 'Vo/Q Vaud' From
this the Q-factor of the resonant circuit is
determined:
1
C =5.7 X 101DL (F)
If we choose an ohmic resistance of R == 0.1 ohm we obtain
reasonable values for the inductance and the
39.8. Equation (59.22) follows from the phase invariance (see 59.8).
Applying the Lorentz transformations, we obtain
cos 6 xo+vt
o
C Y1-V2Jc2
X(l cos 8
0
c
Vibrations and Waves
Removing brackets and regrouping the terms, we have
cot
o
(1- v cos 8)+ ( _ cos 8)
Y1-V
2/C2
C cY1-V2/C2 C
coz
o
sin 8 COOX
o
cos 8
0
c == cooto- c
261
Noting that Xo, zo and to are independent variables we see that
equality obtained is possible only if the factors preceding these vari-
abIes are equal. Hence putting == vlc we obtain
(1) 8) _ 00 (cos _ e
y-- -00
0
, f-- -ooocos 0' oosin8=coosin80
1
The first equality is the expression for the Doppler effect. Dividing
the second aquation by the first, we obtain the relation for the cosines:
cos
1 8 ==cos eo
- cos
39.9. Let the source and the observer approach each other at a speed
v == According to the classical Doppler effect, the frequency of
the approaching source is 00' = 00
0
/ (1 - and of the approaching
observer (0" = 000/(1 + Considering the periods, we obtain
But we must also take into account the time dilatation. In the
first case (that of the moving source) , To in the formula should be
replaced by the quantity "'I To where "'I is the relativistic factor. In the
second cise (that of the moving observer), the quantity T" should
be replaced by yT". We obtain
Since, according to the principle of relativity, T' == Til, we have
y2T
o
= T
o/(1--t-
from which it follows that
1 r1+l3
1'= T=T
o
V w=w
o
V 1-P
39. t O. The Doppler broadening is
L\vDnp .. /' 3RT
v - c - V Mc
2
where M is the molar mass. The gra vi ta tional shift of the spectral
line is
L\vg r nv !.pgrav ym
--v- = -c
2
- ="7if
262 Solutions
where m and r are the mass and the radius of the star, respectively.
On a "white dwarf" exceeds L\vDop by about an order of mag-
nitude.
39.11. Since the ions move towards the observer, the wavelength is
A= A f -
o
The kinetic energy of an ion is 40.0 MeV, the rest energy is four times
the rest energy of a proton (Problem 8.1). We have
V
222 V 22 CI!
1_,nuc= 1_E....:-=_f'l-_
O
m2c4
= <tJ.
o
=3.7284==0.989
3.7684
The reciprocal quantity is
1 K
y= ../ 1+ce-==1.01
y 00
Hence we obtain the ratio of the ion velocity to the velocity of light:
u/c= y 1-0.989
2
YO.01l X 1. 0.148
I-fence the observed wavelength is
A= 410 (1 - 0.148 cos 8) X 1.01 = 361 nm
39.12. The velocity of rotation of the Sun is much less than the veloc-
ity of light, so' we may use the classical expression for the Doppler
effect. From a surface element moving towards us, Al == Ao (1 -
from the element on the other side of the Sun, A
2
== AD (1 +
Therefore
L\A== =-.= . -
cT
ev
where Rev is half the Sun's equatorial diameter. The period of rota-
tion is
41tA
ORG)
T
0
=
39.15. The shift of the spectral lines is a maximum when one star
moves in its orbit towards us and the other away from us. Since the
orbital velocities of the stars are much less than the velocity of light,
the broadening of the spectral lines may be found from the nonrela-
tivistic Doppler formula:
/)''A == AD (t - == == 2A
o
v/ c
where v is the projection of the orbital velocity on the line of sight.
39.16. The period of revolution of the stars about their cornmon
centre of mass is twice the period o-f the spectral line broadening.
Vibrations and Waves 263
Knowing the orbital velocity and period, we may find the radius
of the orbit:
R = riz
Then, applying the law of gravitation, we have
1'M2 Mv
2
Jlf __ 4v2R 2v
3T
----- giving .H'
(2R)2 - R l' 111'
40. Interference and Diffraction of Light
40. f. Obviously, the principal (the zero-order) maximum will be
observed in the centre of the interference pattern (Fig. 40.1). Let us
find the coordinate of the m-th maximum, which we shall denote
Z 5
4
3
2
I
o
Fig. 40.1.
by Zm. This maximum will be observed 'when the propagation differ-
ence is r2 - rl = 2m'A/2.
But )2, ri=L2+{Zm+ r. Subtract-
ing, we obtain (r
2-rt)
(r
2-rt)
= 2z
md,
Since d <t L and since in
practice only the interference maxima of low order can be observed
(i.e. Zm L), we may put r
2
+r
1
== 2L. Fence
2 (r
2
- rl) == 2z
m,d,
or mAL =
Hence we obtain the coordinates of the maximum: Zm == mlcLld,
The separation between successive maxima (or minima) is
Az= ZTrl+l-Z1n == AL/d
40.2. The interference pattern will be blurred if the red maxi-
mum of order m will coincide with the violet maximum of order
(m + 1) : == Substituting the values, we obtain
" 1) 'l m -- A
v
i 0 I . 1 6 2
mAred= m-r I\.vlol gIvIng - , l.e. nt= <
Ared - 'Av 101
This means that the zero- and the first-order maxima will be clearly
seen together wi th the first- and the second-order minima (black
bands). The second-order maximum will be blurred, the third-order
and the subsequent maxima will not be visible at all. The zero-order
264
Solutions
maximum will be white, while the first-order maximum will be
spectrally coloured, with the red outside and the violet inside and
the other parts of the spectrum in between.
The separation on the screen between the red and the violet bands
is
"03 A b r?-r
1
h
CA s may e seen from Fig. 40.3b, -d- ==T. Since we con-
sider here two successive maxima (or minima) and since the light
Fig. 40.3b.
twice covers the distances fl and f
2,
the following relations hold:
2
from which
Iv h
Thus 2d-== T' from which the wavelength of the light can be
found.
40.4. When the mirror is displaced by a half-wave, the pattern shifts
by one band.
40.5. The optical propagation difference is J1::= n
2
l
2
- 11}ll =
= (n - 1) 1. This propagation difference accomodatcs N == 47.5 half-
waves. Hence (n - 1) I == NA/2 and n = 1 + N'A/21.
40.7. The interference pattern will disappear if the maxima or 'one
wavelength coincide with the minima of the other.
40.9. It follows from the condition d sin e == mt, that mkld 1.
Therefore the highest order of the visible maximum is m dlA,
m being the maximum integer. When computing the total number
of visible maxima one should take into account the presence of the
zero-order (the principal) maximum and the symmetry of the inter-
ference pattern about the principal maximum.
40.10. The first-order maximum is visible at an angle 8
1
, which is
determined from the condition d sin 8
1
= A. The second-order maxi-
mum is visible at an angle 8
2
= 8
1
+ 15, determined from the
condition d sin 8
2
== 2A. Hence it follows that sin 15) =
Vibrations and Waves
== 2 sin 8
1
, and this reduces to the equation
sin 15
tan 8
1
== 2 150 0.2503
-cos
265
Knowing the deflection angle 8
t
of the first-order maximum, we may
easily compute the wavelength,
40.11. Find the total number of slits: N = = = 8000.
Since Nd == 1 is much greater than the wavelength, y == 2Alz, which
gives the angular width of the principal maximum.
To find the resolving power of the grating we should find the
number of maxima that can be obtained with it. We have (see Prob-
lem 40.9)
d 1.50
m r=530 X 10-
3
2.8, so m=2
i.e. with this grating only the first- and the second-order spectra may
be observed. The resolving power is A == 'A/6'A == mN.
40.12. The spectral interval that can be resolved is 6.A == A
2
- At,
the resolving power is A == 'A
1/6'A
== mN, from which we may find
the total number of slits. The length of the grating is 1 == Nd ==
== A
1dlm
6.A. The highest order of the spectrum is found from the
condition m alA (see Problem 40.9). In our case this is 5.
40. f3. To see these spectral lines separately, we should have a grating
with 3 resolving power
The resolving power of our grating for the first-order spectrum is
A == mN == 990, so the spectral lines wil'l be resolved, but the reso-
lution will be poor. Measurements can be done better from the spect-
rum of higher orders.
The angular distance between the maxima of the second-order
spectrum is found from the conditions d sin 8
1
== 2A
1
and d sin 8
2
==
== 2A
2
The computations should be carried out using four- or five-
place sine tables; slide rule accuracy is not enough to solve this prob-
lem.
40.14. Let a parallel heam , that is, a plane wave, fall on the grating
at a glancing angle of o: (Fig. 40.14). The direction of the zero-order
intcrference maximum will obviously be the same, As to the maximum
of order m, its direction will be at a glancing angle so that the
propagation difference is 6. =--: a - b = mk, Noting that a ==
== do cos a, b == do cos where do is the grating constant, we obtain
do (cos = mA
This is just the condition for the interference maxima when the
light rays are incident at an angle to the diffraction grating. Now
let us express the interference condition for the angle 8 equal to the
266 Solutions
deflection angle of the diffraction maximum from the original direc-
tion of the beam. Since = a + a, .it follows that
cos a-cos cos a-cos (a-t-a) == cos a-cos a cos e+sin a sin e
Since the angle 8 is usually very small, cos e 1, and cos a -
Fig. 40.14.
- cos sin a sin 8. The condition for the maximum assumes the
form
do sin ex. sin 8= ml;
I.e. the instrument behaves wi th respect to the inclined beam in the
way it would have done if a grating with constant d == do sin ex were
placed perpendicular to the rays.
41. Dispersion and Absorption of Light
41.1. The condition for Cerenkov radiation which takes account of
dispersion is cos e == cl nu, Since the protons are relativistic, it follows
that
Y
- - v-
2
y- In
2v2c 2
y----P '$0
1--= 1---= 1--=-
c
2
m2c4
Hence
c 1
- = ---;:-===:;::::;:-
V V1-
11tc -+ K)
where (0 == 0.939 GeV is the rest energy of the proton (see Prob-
lem 8.1). The refractive indices for these parts of the spectrum are
nl == 1.48 and n
2
== 1.51.
v 1
41.2. The ratio
cos B
1.3428 cos 4110'
The kinetic
energy is
Vibrations and Waves
267 -
U==
( 1 -1)
-V
41.4. According to the definition, the group velocity is
. 1\0> do> 1
U=== Iirn -=-=.:--
ak-+O Jik dk dk
dO)
B t th b . k 0> nffi Diff .. b .
u e wave num er IS == u= -;- . I erentiating, we 0 tain
do> cedro c dro
Hence
u= c
dn
n-t-w dO)
41.5. In the region of normal dispersion
ex. e
2n
o
n
2
= 1+ ,where (X=--
W5-
OO 2
eom
e
The derivative of the refractive index with respect to the frequency
is found by differentiating this equation. We have
dn ex 2ao>
2n 7iW == - (W5-
w2
r>. (-2w) = (oo5-
W2
)2
Therefore
dn (J) (X > 0
am= n (W5-0>2)2
i.e. in the region of normal dispersion the derivative of the refractive
index with respect to frequency is everywhere positive.
Now let 000 > 0>. Then n > 1, and it immediately follows from
the formula U == C dn (see the previous problem), that U < c.
n+ro dw
If 000 < 00, then n < 1. Substituting the value of the derivative
into the expression for the group velocity, we obtain
c en en
(X,00
2
(X,W
2
(X,00
2
n+ n
2
+ 1+ 0
n (oo5-
W2
) 2 (005-
002
) 2 (Wij-W
2)2
We see that the number in, the denominator of the fraction exceeds
unity. And since in this case the refractive index is less than unity,
it follows that here too U < c.
41.6. Since the free electron concentration in the plasma is small,
the second term in formula (63.15) at high frequencies is much less
.268 Solutions
than unity, and the dielectric constant is close to unity. Applying
the approximate equality 1/1 + x == 1 + x/2 (for x ~ 1), we obtain
the expression for the refractive index.
The group velocity is found by computing the derivative of the
refractive index with respect to frequency:
dn e
2
n
o
aw= f
o
rn
e
(i) 3
and by substituting this value into the formula for the group velocity.
Incidentally, it is easier to apply the last formula of the solution
of the previous problem, substituting ro == 0 into it.
41.8. There can be no Cerenkov radiation in a plasma because in
a plasma the phase velocity exceeds the velocity of light in a vacuum,
and because the particles responsible for the Cerenkov radiation must
move with a velocity exceeding the phase velocity of light in the
medium.
41.10. The concentration of the valence electrons in aluminium is
shown in Table 44.1 ( 44.2). For the refractive index make use of
the result of Problem 41.6.
41.13. The transit time of the light pulse from the toothed wheel
to the mirror and back is 't == 21/c. During this time at n. ::::= 283 r.p.s.
the toothed wheel will turn through k teeth, k == ZT! T1 == Z't1l1'
where z == 720 is the number of teeth on the wheel. In this time the
toothed wheel rotating at n
2
::::= 313 r.p.s. will turn through k + 1
teeth, so k + 1 == z'tn
2
Subtracting, we obtain 1 = Z't (n
2
- nt),
from which
c === 2lz (n
2
- n
1
)
41.15. Make use of the data in the table of refractive indexes for
various wavelengths, choosing the spectral interval between the
yellow (5461 A) and the blue (4861 ~ parts of the spectrum. Substi-
tuting <0 = 2nc/n"A and k == 2rr/"A into the expression for the group
velocity
we obtain
u=== c (nIA] -n
2"A2)
s (At - "A
2
)
The phase velocity is u == cl n,
4t.16. Consider the energy balance in the case when light passes
through a plate. Suppose a beam of intensity I
o
falls normally on the
plate. Part of it will be reflected, the rest, of intensity]' === TI 0'
where T is the transmittance (Fig. 41.16), will enter the plate. Because
of absorption, the intensity of the light which reaches the other face
will be I" == ]'e-tJ,d. Finally, the intensity of light passing out into
Vibrations and Waves
2eQ
the air will be I == TI II = T2I oe-lJ,d. Hence
I T2/ -Jld 1
1 oe == eJ.L(d2 - dd
T; T2I
o
e - ,... dll
The half-thickness is L == (In 2 ~ (see Problem 37.9).
47.17. The transmittance of a material layer is the ratio of the inten-
sity of a beam of light passing through the layer to the intensity
~
~ ~ %
1
0
I' ~ ~ /"
I
%
~
~
%
Fig. 41.16. Fig. 41.17.
of the beam entering the layer: k = 111
0
(Fig. 41.17). In the figure
r
2
4n
I I = T 1
0
, I" = I I . r ~ e-/L(r
2
- r.) I = TI", T = (n +1)2
Substituting the appropriate data we obtain the result sought.
41.18. Suppose white light of intensity lin == A2 falls on the light
filter. At the resonance wavelength A
o
the intensity of the transmitted
light will be 1
0
= A 2e-J.Lod. The intensity of transmitted light of
other wavelengths will be
1== A2
e-Jl.d
= A2
e
- J.L od
e
- a d(Ao- A)2== Ioe-adOoo-A)'l
Weare interested in the case when
I ~ 1
0/2,
i.e. when e-
a d
(A
o
- i.,)2 ~ 2-
1
Taking logarithms, we obtain ad (A
o
- A)2 ~ In 2, giving Ab =
-./ln2. . 0 .. /ln2
= x, V ado The WIdth of the spectral Interval IS I:1A ==2 V a:d.
The transmittance at the resonance wavelength is k
o
= I o I A ~ =
= e" .....
od
where V
o
is the volume of the counter, and the auxiliary angles ex
and are expressed in terms of energy of a photon in the following way:
sin e = sin e
cx,= he P he
45.23. The photon will either pass through the polaroid, or it will
be absorbed by it. The probability of a photon passing through the
polaroid is wpas = Cl, while the probability of it being absorbed
286 Solutions
which gives
is Wabs == sin
2
a, where a is the angle between the optical axis of the
polaroid and the direction of the electric field vector of the electro-
magnetic wave corresponding to the photon.
45.24. Prior to collision, the electron and the photon fly head-on
towards each other; after collision they will move in the initial direc-
tion of the electron's motion. The laws of conservation of energy
and the momentum will assume the form
hv , hv'
P--c-==P +-c-, +hv'
Hence it follows
+pc=<S' +p' c+2hv', <S - -p'c
Multiplying, we obtain
_ p2c2+2h" pc) = - p'2C2 +2hv' - p' c)
But - p2C2 = - = Hence
hv pc) == hv' (!' - p' c), or h (! +pc) ==hv' pc+2hv)
Multiplying both sides of the equation by the expression +pc,
we obtain
h" +pC)2 == hv' +2h" +pc)]
But in the ultra-relativistic case (g pc (see Problem 8.12). Hence
,
hv == +
46. Elementary Quantum Mechanics
46.1. The kinetic energy of a particle is K = +p2C2_
- from which we obtain for the momentum p = !. y K +IlK),
c
and for the de Broglie wave
A= !!:.=
P
For K <{::' we obtain the nonrelativistic approximation:
At _ he h
nonrel - y oK y 2nl,K
The error due to the substitution of the nonrelativistic formula
for the relativistic one is
6= Anonrel - A
A
Fundamentals of Quantum Physics 287
since fJ ~ 1. Hence, the error introduced by the substitution of the
nonrelativistic formula for the relativistic one will be less than 6,
if K ~ 4 6 ~ o
46.2. The kinetic energy of the particle is equal to its charge multi-
plied by the accelerating potential: K = erp, Substituting this value
for the kinetic energy into the formulae obtained in the previous
problem, we express the de Broglie wavelength in terms of the ac-
celerating potential.
46.3. We use the expression for the resolving power of a microscope
( 66.8), putting sin u == 0.02. We can find the wavelength from the
nonrelativistic formula, since the kinetic energy of the electron of
10 keY is much less than its rest energy which is 510 keY.
46.4. If one assumes the apertures of the electron microscope and
the ion projector to be approximately equal, the difference in the
resolving powers will be determined by the difference in the wave-
length. Assuming the accelerating potentials to be approximately
equal as well, we see that the difference in the wavelengths is due
mainly to the difference in the masses of the accelerated particles,
the mass of the electron being about three orders of magnitude less
than that of the ions.
46.5. The de Broglie wavelength for these electrons is A== hIoJ! 2mecp =
== 12.25/Y15== 3.2 A. The first-order diffraction minimum is observed
at an angle e such that sin e== AID, where D is the width of the
slit ( 57.9). Since the angle is very small the width of the principal
maximum is
: = 2l tan 6 == 21A/D
46.6. Using the Bragg law for the first-order maximum (see 62.7),
find the de Broglie wavelength for neutrons: A== 2d sin a, where a
is the glancing angle. From the wavelength we find the kinetic energy
of neutrons (see Problem 46. f), their velocity and the corresponding
ternpera ture:
h
v---
- mk:"
h
2
T= 3mk).,2
46.7. The root-mean-square momentum of the molecule may be found
from the condition ~ kT = pI/2m, from which Pr.m.s. = Y3mkT
- h h
and the de Broglie wavelength A= y'
Pr.m.s. 3mkT
46.8. (1) At the accelerating potential (J)1 == 10
2
V the electron is
a nonrelativistic particle. I ts momentum is PI == V2mecpl and the
velocity is VI == V2ecpt/m. But the group velocity of the de. Broglie
wave is equal to the particle's velocity: U
I
= VI. The phase velocity is
c
2
V---m
Ul==-=C
2
--
Vi 2 e ~
288 Solutions
(2) At the accelerating potential fP2 == 10
5
V the electron is a rel-
ativistic particle. Its momentum is found from the condition
V +p
2
C
2
== 0 +eq>2' from which it follows that P2==
= Veq>2 + e2) The mass is found from the condition
c
1
m2c2== whence m2==2" The group velocity is
c
U
2
=v
2
=J!L =c ye<P2
m
2
The phase velocity is
c eCV2)
Ve<p2 +erp2)
Note that in this case the energy is conveniently expressed not in
joules, but in kiloelectron-volts, since == 510 keV.
46.9. For a particle in the ground state one de Broglie half-wave fits
into the length of the potential well: L == "A/2. The momentum of the
particle is p == hl"A == h12L. The recoil of the particle from the wall
of the well is perfectly elastic, so the change in its momentum is
Isp == 2p == hIL. The average force of pressure is equal to the product
of the change in the momentum variation and the number of collisions
per unit time:
V p2 h
2
2p 2L = = 4mL3
p2 n2h2
46.10. The energy of a particle is = 2m = 8mL2' where the quan-
tum number n, according to the conditions of the problem, assumes
the values 1, 2, 3.
46.11. The zero-point energy is == 1/2hv. The energy of the first
excited state is = }h'V, so the excitation energy is = -
- <t'o == h, The vibrational degrees of freedom will no longer be
excited when the energy of thermal motion becomes less than the
excitation energy. The usual criterion is } kT <::;; h, which yields
the minimum temperature
Tvtb ==2hv/3k == 4 X 103 K
This does not agree with experiment, since the lines of the vibrational
spectrum of hydrogen molecules are observed at lower temperatures.
This is because of the Maxwellian molecular speed distribution
(see 25.2), which shows that a gas contains molecules whose speeds
are far in excess of the average. For instance, about 2% of the mole-
cules have speeds three times greater than the average. The energy
of such molecules is more than 9 times the average kinetic energy.
Fundamentals of Quantum Physics 289
L
46.12. The kinetic energy of an orbiting electron is K ==....!- m
ev
2
=
1 2
=="'2 mero
2rl
, the zero-point energy is = firo/2. The natural fre-
quency is <0= ]I == ]IF/rneA, where F is a quasi-elastic force
and A the amphtude of the oscillator. Assuming the quasi-elastic
U
o
---1----
'A
-1-
1
:-
&F
Fig. 46.14a. Fig. 46.14b.
force to be of the Coulomb type and the amplitude to be equal to
the radius, we obtain
ro==" / e
2
V 4ne
omer
3
Putting K = and substituting the circular frequency, we obtain
after some simple transformations
41tE
o
li
r= e2me
Despite certain arbitrary assumptions, we have obtained a correct
expression for the first Bohr radius.
46. t3. The probability of tunnelling through a potential barrier is
2L
w=DjDo==e-
a
, where a=T ]12m (Uo-e<p)
(see 70.6) .
46. t4. In the absence of an external electric field the electron in a me-
tal is shielded by an infinitely wide potential harrier of height U0
(Fig. 46.14a). In the presence of a strong electric field of intensity E
the potential barrier assumes a triangular shape with height U
o
and
width L == E == A aleE (Fig. 46.14b), where A 0 is the work function.
Neglect the shape of the barrier and assume it to be rectangular.
Since the energy of the electrons in the metal is and their work
function is A
o
= U
o
- (see 75.3), the parameter a which deter-
mines the probability of the electron tunnelling through the potential
barrier assumes the form
2L ..i 2A
o
..i-
a===T y 2m eEIi y 2mAo
tfl-0360
2Yu Solutions
47. Atomic and Molecular Structure
47. t , The approach distance of an alpha-particle to the nucleus will
be a minimum in the case of a central collision, when the entire kinet-
ic energy of the alpha-particle is transformed into potential energy:
K = U == Z4jZ2e2, where z, and Z2 are the respective atomic num-
nor
bers.
47.2. \"le obtain the first equation from the condition that the centrip-
etal acceleration is due to the Coulomb force:
e
2
-r-= 4ne
or
2
'
which gives
The second equation stems from the rule of orbit quantization: mer ==
= nh, Dividing the first equation by the second, we obtain
1 e
2
v=----
n 4ne
o
li
The maximum speed corresponds to the first (principal) energy
level. Its ratio to the speed of light in a vacuum is the fine structure
constant:
1
7.3 X 10-
3
= 137
47.4. From the formula = heR (A - ;2) = 13.6( 1 - , we
obtain the number of the excited level: n = 3. Direct transitions
from the third level to the first, or second, and from the second to
the first level are possible. We obtain three spectral lines.
47.5. The energy of transition from the excited to the ground state
is shared by the photon and the atom: = h + D, where D ==
= p2/2M is the atom's recoil energy, and p is the momentum due
to the emission of a photon. In accordance with the law of conserva-
tion of momentum, P = Pph = hvlc, Hence D = h
2v2/2Mc2
and the
transition energy is
(1+ )
Solving this quadratic equation, we obtain the expression for the
energy of the photon:
hv == 2(g
1+V
Since the transition energy in a hydrogen atom is below 13.6 eV,
and its rest energy is 1 GeV, it follows that M c
2
10-
8
, and
so this term in the denominator may be left out without appreciable
loss in accuracy. Since, according to the statement of the problem,
Fundamentals of Quantum Physics 201
the transition is from the fifth to the first level, it follows that
(
1 1) 24
12--V = 25
24hcR . . 24
2
11, 2R2
Hence h = 25 ,the recoil energy IS D=-= 2 X 25
2
LV} ,the vo-
. 24hll
locity of the atom IS u= -.-
2J.ilf
_ Z2
hcR
_ ,
47.6. Noting that for a hydrogen-like ion (Q ,,2 we
obtain the generalized Balmer formula
...!.-=Z2R (_1__1 )
Iv m
2
nt.
For hellum (Z == 2) we obtain
.i- =--= 4R (_1__1)
Iv ln
2
n
2
The principal line of the Lyman series is the result of the transition
from the second. to the first level (m =: 1, n == 2) and of the Balmer
series (Ha) the result of the transition from the third to the second
level (m = 2, n == 3).
47.7. The diameter of the excited hydrogen atom is d == 2n2ao, where
n is the number of the energy level and a
o
the Bohr radius. The con-
centration of atoms is
1 1
no -d
3
' or no -86 3
n au
Since the Balmer series is exci ted as the result of the transition of an
electron to the second level, the maximum number of the level is two
units more than the maximum number of observed lines, Hence in
a gas-discharge tube n === 14, in a celestial body n == 35.
47.9. The spectral lines in question closely resemble the first three
lines of the Balmer series of the hydrogen spectrum: 6563 A, 4861 A
and 4340 A.. To make sure that they are in fact the lines, find the
ratio of the wavelengths of the galaxy spectrum to the wavelengths
radiated in the laboratory. We shall obtain identical ratios:
Iv 6877 4989 4548 r::
6563 = 4861 = 4340 =1.04a
The red shift is due, obviously, to the motion of the galaxy away
from us (the Doppler effect). \Ve have
whence
r
.. / 1--1- = 1.045
V
Solving this equation we find the speed at which the galaxy moves
away from us.
19*
292 Solutions
47.10. Since the muon mass is only one ninth the mass of the proton,
we should consider the proton and the muon as revolving about a com-
mon centre of mass (Fig. 47.10). To find the radii, we obtain a system
of equations:
mlVr = m2vi e
2
(1)
'1 r2 41t8
0a
2
a = Tl + '2 (2)
mlvirl + m2v2T2 = nil (3)
mlTI = m2
r2
(4)
It follows from the first and the fourth equations that vl/TI = v21r2'
nn nli
and from the third and the fourth that VI + V
2
= -- = -- .
mirt 1n2r2
r
2
muon
r,
02
i
t>
proton
0
Fig. 47.10.
Hence we obtain the orbital speeds: v\ = nhlanu; V
2
= nlilam2. Sub-
stituting this result into the first and the second equations, we obtain
the radii:
2 4nB
o
n
2
2 4nB
o
!i
2
rl=n r
2=n
mle
2
, m
2e
2
Hence the Bohr radii of a mesoatom may be obtained:
2 43tB
o
n
2
mp+ m"..
an=n 2 .----
e mpmJ.L
The energy of an electron occupying an arbitrary energy level 'is
2 2 2 2 2'
_ ml VI + m2v2 e e e ao
n - 2 2 4n8
oan
8nB
o
a
n
- 8nB
o
ao
.-;;;
-ncR..!:2..
an
where ao is the first Bohr radius of the hydrogen atom.
47.12. The solution is similar to that of Problem 47.10. One may make
use of the formula obtained in Problem 47.10, putting ml = m2 = mo.
We have an == n22ao, where ao is the first Bohr radius. The energy
in the ground state (n = 1) assumes the form = -hcR/2.
47.13. The first potential jump takes place when the electron goes
over from the first to the second level:
(_1__1)
1
2
2
2
Fundamentals of Quantum Physics 293
47.14. The orbital quantum number corresponding to the s-state
is l = o. Therefore the appropriate magnetic quantum number is also
m = O. Hence the electrons may differ only in their spin projections:
s = 1/2 and s = -1/2. Thus there are two possible sets of quantum
numbers: n, 0, 0, 1/2 and n, 0, 0, -1/2.
The orbital quantum number corresponding to the p-state is
1 = 1. Therefore the magnetic quantum number can assume three
values: m = 1, m = and m = -1. Since there are two possible
spin projections corresponding to each magnetic number, the pos-
sible sets of quantum numbers are six in all: n, 1, 1, 1/2; n, 1, 1,
-1/2; n, 1, 0, 1/2; n, 1, 0, -1/2; n, 1, -1, 1/2 and n, 1, -1,
-1/2.
47.16. The valency of an element is determined by the number of
electrons occupying the upper partially filled electron level. All
these elements have one electron on this level.
47.17. The Pauli exclusion principle does not hold for bosons. There-
fore in a system in a state of minimum energy, all three particles
will occupy the first energy level. Since in this case their momenta
are equal, the force of pressure turns out to be three times greater.
47.18. Of the three fermions, only two (with opposite spins) can occupy
the lowest energy levels. The third fermion must go over to the second
level, and this will be the state with a minimum energy of the system.
On the second level the length of a potential well holds t\VO half-
waves, i.e, L == A. The particle's momentum turns out to be p =
= hl): == h/ L, i.e, it is twice that of a particle occupying the first
level. The corresponding increase in "force is four times (see Prob-
lem 4u.9). For the resultant we obtain
2h
2
4h
2
3h
2
Fav = 4mLS + 4mL3 = 2mL3
47.19. The shortwave threshold is determined by the kinetic energy
of the electrons bombarding the anti-cathode: hcl]; ~ K. But the
kinetic energy of the electrons is itself determined by the accelerating
potential: K = ecp. Hence
'" >= hclecp
1
47.20. Apply the Moseley la \v in the form r== R (Z - 1)2 X
X ( 1
12
- ;2 ). lIenee it follows
Z=1+V ~
Knowing the atomic number, we can easily find the material
of the anti-cathode.
47.21. The wavelength of the Ka-line for vanadium (Z = 23) may
be determined from the Moseley law. This line can be resolved only
if it doesnot lie outside the continuous spectrum, i.e. if Iva. > hclei,
he 3(Z -1)2hcR
Hence <p > --;r-, or cp > ~ .
e/\,a, {e
294 Solutions
47.22. From the conditions of the problem, (A
a-A)/Aa
== O.1. Hence
he
it follows that A==O.9A
a,
or he/ecp=O.9Aa,. We have {p== O.geA
a
=
(Z -1)1 heR
1.2e
47.23. The angular momentum of a rotating quantum system, includ-
ing a molecule, is L == V1 (l + 1) Ii, where 1 == 0, 1, 2, ....
The kinetic energy will accordingly be
Krot ==.!!-= l (l +1) 1i2
2J 2J
where J is the moment of inertia. The kinetic energy in the first excited
state is K
1
== 1l
2
/J , the angular momentum is L
1
== Ii 1
1
2 and the
angular velocity is
L
1
21i V2
ffi
1
= 7 = md2
Here d == 0.74 Ais the distance between the centres 01 the atoms
in the molecule, and m == 1.67 X 10-
27
kg is the mass of hydrogen
atom.
47.24. The energy of the molecule on the first vibration-rotational
level is
<.evib-rot _ <'vib+ c.erot _ 11m +_!!:.
01 - 00 0 1 -- 2 J
The transition to the zero level results in the radiation of a photon
of energy (gPh == ==;"2/J. Expressing the photon energy in
terms of its wavelength, == 2nnc/A, we obtain
A== 2nJclIi
l ==. 1, 2, . ., '1:3 where
47.25. The diagram of vibration-rotational energy levels is shown
in Fig. 47.25. There are only t\VO purely vibrational levels: =
== nw/2 and == 3nw/2, and thirteen intermediate vibration-
rotational levels:
t/; == vib + rot == nw -l- I (l--t-1) 11,2
l 0 l 2' 21 '
It is clear from the diagrarn that the fourteenth vibration-rota-
tional level coincides with the first purely vibrational level:
'{g _ nw L.14x151l2 31lw whence 14x15h
2
-::-:noo
14 - 2"1 21 2 21
This makes it possible to determine the molecule's moment of inertia
about its centre of mass: J 105 til t, On the other hand, the moment
of inertia is J == ln
n
ri
I
-1- where r
H
r
F
== d is the distanco
sought between the centres of atoms and But m
F
= 19mn.
Therefore r
H
=--= 1nr
F
19d/20. Substituting into the expression for
Fundamentals of Quantum Physics
the moment of inertia, we obtain
For the distance between the centres we obtain
d= .. /2.1 X 10
3
/l
V 19wmn
Fig. 47.25.
------------Evib
o
____________Grib
1
---------- '2
----------c
l
47.27. The spectral lines of a helium atom are due exclusively to the
electronic transitions from one energy level to another. These transi-
tions result in a line spectrum.
In a hydrogen molecule there
is a set of vibrational and ro-
tational levels, in addition to E/
4
the electronic levels (see 74.4). t/
2
Because of this the spectrum of
a hydrogen molecule consists -------------
not of isolated lines, but of
bands.
47.28. The orientations of the
magnetic moments of the pro-
ton and the electron may be
ei thor parallel, or anti-parallel
to one another. Therefore the
total energy of interaction of the
electron and the proton is
e
2
211 uJ!pf.le
= --IJ-- 4 3
:incur rtr
(see 40.6,41.10 and Table 10 in the end of the book). The relative
error is
where a
o
is the Bohr radius.
Every energy level is' seen to split into t\VO sub-levels: the upper
== + I I, and the lower == tb - I mag I, where n
is the number of the level and I 'f!J mag I is the magnitude of the energy
of magnetic interaction. Dashed lines in Fig. 47.28 show the first
three energy levels stemming from Bohr's theory, while the solid
lines show the sub-levels due to magnetic interaction. The diagram
is, of course, not to scale.
47.2!l. The transition from the upper to the lower sub-level of the
ground state in hydrogen results in the emission of a photon with
an energy = - (gj = 2 I I. The corresponding wave-
296
Solutions
-----c;
-------c;'
6
1
----------
length is
'A =....!!:.:- =
t/;ph
4n X 6.62 X 10-
34
X 3.00 X 10
8
X 5.29
3
X 10-
33
- =056 m==56 cm
- 4 X 431 X 10-
7
X 1.41 X 10-
26
X 9.28 X 10 24
It follows that the classical calculation does not produce the
correct wavelength, but the order of magnitude is right.
47.30. The frequency of the nearest red companion in the combination
scattering spectrum is known to be = "0 - flW>vib/h, of the
violet companion "iiOl= "0 + C'
where "0 is the frequency of light. But --------- (J,
= h", where" is the natural fre-
quency of vibrations of a molecule. There- E ----------- 2
d 2 [2"
fore "re = V
o
- V, V
1
viol = Vo + v,
from which we find the natural frequen-
cy of the molecule to be
"riol_vied
v= 2
C (1 1) Fig. 47.28.
2 ",vior- ",red
1 1
47.31. To make the operation of a laser possible, a mechanism of stim-
ulated emission must be provided which produces absolutely iden-
tical photons. The photons must have identical frequencies (energies),
identical phases and identical spins (i.e. their polarization must be
identical). This is possible only because photons are bosons and there
may be an unlimited number of them in the same quantum state.
Fermions, on the other hand, obey the Pauli principle, which forbids
the presence in a system even of two particles with identical quantum
numbers. Consequently, there can be no induced radiation in a system
made up of fermions. Therefore a laser operating on fermions is not
feasible.
47.32. The lowest angular divergence can be found from the. condition
29 2'AID.
48. Quantum Properties of Metals and of Semiconductors
48.3. Find the total energy of the electron gas W = i FV, where
V is the volume of the metal. Imagine the gas compressed by a small
amount dV. This wfll require work against the forces of pressure equal
to = -P dV, where P is the pressure of the electron gas. This
work is equal to the change in the energy of the electron gas W =
= dW. To find the differential of energy let us express the electron
gas concentrationIn terms of its volume: n == N/V, where N is the
Fundamentals of Quantum Physics 297
total number of electrons. We have
w:=l- N (2-) 2/3 N2/3V-2/3
5 2m 8rt
Differentiating, we obtain
2 3 h
2
( 3 ) 2/3
dW== -3X"5 2m 8'1 N5/3V-
5
/
3
dV ==
2h
2
( 3 ) 2/3 2
= - 5m 8n n
fJ
/
3
dV = -5n'fgF dV
The pressure of the electron gas is P = } F
48.4. Making use of the result of the previous problem, we obtain
p == 2h
2
(2-) 2/3 N5/3V- 5/ 3
5m 8n
which yields
PV
5
j 3 :.::.-: const
The "adiabatic index" is '\' == 5/3.
48.5. The pressure in the "white dwarf" is due to gas consisting of free
electrons and of helium nuclei. These particles are in a degenerate
state, like free electrons in a metal. The mass of an electron is almost
1/8000 the mass of a helium nucleus. Therefore the Fermi energy
of the electron and the pressure of the electron gas are 8000 times grea-
ter than the corresponding quantities for helium. Hence the helium
pressure may be neglected. Making use of the result of the previous
problem for the pressure of the electron gas, we obtain
(2-)2/3 N
5
/
3
5m
e
8n e
There are two electrons to each helium nucleus, so N e == 2Na ::=
= 2M/rna, where AI is the mass of the star, and rna, == 4.002 X 1.66 X
X 10-
27
kg is the mass of a helium nucleus.
Hence PV
6
/
3
== AM6/3, or P = A
p
5/ 3 , where
i ( 3 ) 2/3
A= - =3.2x1U
6
Parn
Skg5
/
3
s, (ma,/2)6/3 8n
48.6. The number of electrons rising above the Fermi level is es-
timated with the aid of an approximate formula
(see 75.7).
48.7. The specific heat of a kilomole of electron gas is == Rk F
'(see 75.8), the lattice heat is = 3R (see 45.2). We have
==
48.8. The mean free path is found from the quantum expression for
the electric conductivity: l' = e
2n'A/PF
(see 75.9). The Fermi momen-
tum is found from the known concentration of conduction electrons.
298 Solutions
The interatomic distance is found from the concentration of atoms:
d = n"A
1
/
3
= 2.3 A (see 44.2).
48.9. Let the concentration of pairs in the superconductor be n' == n/2.
Then the current will be i = qn'Sv = 2e i- S ' where p is the mo-
mentum of the pair. According to Bohr's rule, the angular momentum
is quantized: pr = NIi, where N = 1, 2, 3, ." .. (the principal
quantum number). We have for the current
nS en
i==N-.--
r 2,n
e
where S is the conductor's cross section, and r is the radius of the
ring. Clearly the current is quantized: i == N if)' where i
o
is the mini-
nS en
mum current i
o
== - -
, r 2m2
Since the magnetic flux is proportional to the current, it too turns
out to be quantized:
<D = Li = N Li
o
= N<P
o
where L is the inductance of the ring, and <Po is the minimum magne-
tic flux. Rigorous theory yields <Do = h/2e = 2.07 X 10-
1 5
Wb.
48.10. The electric conductivity of electronic (N-type) semiconductors
is proportional to the number of electrons in the conduction hand.
In the assumption that the transition probability of the electron
from the valence to the conduction band can be computed with the
aid of a barometric distribution, we obtain n = where
is the forbidden band width (see 26.11, 34.3, 35.1). Hence we obtain
for the electric conductivity
1'= e
2
AA e-h'fJ/hT
PF
where B is a constant characteristic of the material (at a specified
temperature).
48.11. The mean free path of the electron is much less dependent
on the temperature than the exponential term, therefore the tempera-
ture dependence of the factor B can be neglected in the first approxi-
mation. We have
rJ/
k T
2
1!-=__
'\'1 e-tltJ/hT1
48.12. In the case of intrinsic conductivity, the electron and the hole
concentrations are equal. Therefore 'V = en ib ; + b_), where b; is
the mobility of holes ann. b: is the mobility of electrons. Hence n ==
'\'
e(b++ b_} .
To find the Hall coefficient, note that the Hall potential difference
of the electron and the hole components are of opposite signs {see
Nuclear and Elementary Particle Physics
44.2). \Ve have
299
RH=R(-> -R(+> = b_-b+
H H y.
48.13. Since indium is a trivalent element, the indium impurity acts
as an acceptor and produces hole-type conductivity. Knowing the
concentration of holes and their mobility (see Problem 48.12), we
obtain l' = en+b+. Antimony is a pentavalent element, so the anti-
mony impurity acts as a donor. The electric conductivity is l' = en.b.i.
49. Nuclear Structure
49.2. Solve the system of two equations:
10.013 x + 11.009 y == 10.811; x + y = 1
where x and yare the fractions of the light and the heavy isotopes,
respectively.
49.3. The radius of the nucleus can be estimated from the formula
R ~ R
o
VA (see 80.6), where R
o
== 1.4 X 10-
15
m and A is the
mass number. The height of the Coulomb potential barrier is U
o
==
Ze
2
h Z' h . b
== -4R' were IS t e atomic num era
nEo
49.5. The binding energy of the tritium nucleus 11-1
3
is ~ IH3 ==
== (1.00783 + 2 X 1.00867 - 3.01605) X 931.5 = 8.5 MeV. The bind-
ing energy of the helium nucleus 2He3 is ~ H e == (2 X 1.00783 +
+ 1.00867 - 3.01603) X 931.5 =7.7 MeV.
49.6. The number of alpha-disintegrations is obtained by dividing
the change in the mass number by 4, which is the mass number of the
alpha-particle. We have
226-206
4
5
After five alpha-disintegrations the decrease in the atomic number
will be 10, and ZR1 - Zph == 88 - 82 = 6. It Iol lows from this
that there are in addition four beta-disintegrations, each of which
results in a unit increase in the atomic number.
49.7. The transition energy is equal to the difference between the
energy of the original nucleus and the rest energy of the reaction
'Products:
~ = [209.93297 - (4.00260 + 205.97446)] X 931.5 = 5.5 MeV
This energy is equal to the sum of kinetic energies of the alpha-particle
and the recoil nucleus (see 17.2): 0 -== Kry" + Kph and Ka./KPh =
== Mpl/Ma.'
49.8. The sum of the masses in the final stage of the possible reaction
exceeds tho mass of the original nucleus. The reaction is impossible
because it contravenes the law of conservation of energy.
49.9. See the previous problem.
300 Solutions
49. to. The mass of a silicon nucleus exceeds the mass of a phosphorus
nucleus by Sm == 30.97535 - 30.97376 = 0.00159 amu, the corre-
sponding energy being == 0.00159 X 931.5 = 1.48 MeV. This
exceeds the electron rest energy (0.51 MeV), and therefore beta-decay
is possible:
14Si31 15
P 31
+-1eO+oV
The total energy of the beta-particle and the antineutrino is
1.48 MeV.
49.11. Let the kinetic energy of the extracted neutron be zero. Then
the work needed to extract a neutron will be equal to the difference
between the total energy of the reaction products and the rest energy
of the original nucleus:
!:Ja'J == (12.00000 + 1.00867 - 13.00335) X 931.5 = 4.96 MeV
49.12. To evaluate the probability, apply the formula
2R
w=e-
a
, where a=-/i- V2m(U
o
- K )
Here Uis the height of the Coulomb barrier, and K is the kinetic
energy of the alpha-particle. For data on the radius of the polonium
nucleus and on the barrier height see Problem 49.3.
49.13. The activity of the specimen is equal to the number of decays
per unit time: :
dN '
Q == - de 'AN (see 81.4). The ratio of the activities is
t2- t t
:c=.!l.!.-=.!!...!..= e-
A
( t 2-
t
1) ==2 - -
T
-
Q1 s.
Knowing the half-life, we can easily compute the decrease in the
activity.
49.15. Knowing the activity Q='AN and noting that 'A= =
= ,we find the half-life T = O.693N/Q. The number of nuclei
is N = mNAlA, where m is the mass of the specimen, A is the mass
number of the isotope and NA is the Avogadro number. Finally
T ==O.693mNAI AQ.
49.17. We might at first reason that the energy of gamma-photons is
= 5.30 - 4.50 == 0.80 MeV. However, this estimate is too rough,
since it takes no account of the recoil energy of the nucleus. Certain
problems in nuclear physics demand a much higher accuracy. Consider
the energy level diagram for the case of polonium decay (Fig. 49.17).
The letter Ph* with an asterisk denotes the excited lead nucleus
which emits gamma-photons. The total energy of the transition is
= KIa. + R
1
, where KIa. == 5.30 MeV, and R
I
is the recoil energy
of the nucleus. Since R
1
== 4K
1a./206
(see 17.2), it. follows' .that
=== 5.30 + (5.30 X 4/206) = 5.40 MeV
Nuclear and Elementary Particle Physics 301
II Pb* ____ L... _
P02JO
84
Similarly, the energy liberated in the course of the second transition is
02 = 4.50 + (4.50 X 4/206) == 4.58 MeV
The energy of the gamma-photon is
t5
v
= 1- 2 = 5.40-4.58=.:::0.82 MeV
Note that because the mass of the lead nucleus is large, we can neglect
the recoil energy resulting from the emission of a gamma-photon.
49.18. This problem is similar to
Problem but because the
energy of the gamma-photon is
much higher than that of an
ul traviolet photon, the recoil is
in this case much greater. Since
the energy of a gamma-photon
is == 14.4 keY, and the mo-
mentum ispv = the recoil
energy is 'iR = =
= c
2
Therefore the transi-
tion energy is
(1+ 2;'I'c
2
)
. 82 Pb
206
The relative change in energy is .
_ ss; FIg. 49.17.
B- Gv - 2Mc2
The natural relative line width is 6
na
t == where 't is the
lifetime of the nucleus in the excited state.
49. t 9. The absorption of a gamma-photon by a free nucleus results
in it acquiring a momentum equal to that 'of the photon and, conse-
quently, a kinetic energy = Therefore the
energy of the absorbed photon is = + where = +
+ is the energy of the transition, and is the energy of the
emitted photons. Hence
= (1+ ;:a )
This distorts the resonance absorption. If we start bringing the source
and the absorbing substance closer together, the gamma-photon' s
energy increases because of the Doppler effect: =
0V (1 + At a certain speed we obtain = and resonance
absorption sets in. We have
(1+ ;:2 )= giving ;:2 I and v =
"The decay probability is the ratio of the number of the nuclei
that experienced decay to the total number of nuclei: w = _ d: .
302 Solutions
dN
It is proportional to the time of observation, w = Adt, i.e. N
-A dt, \\'e have
r dNN ~ .. - _ ~ r dt,
J flwJ so InN+C:=::-'At
At the initial point of time t == 0, N == No, so In No + C == 0
and C = -In No. Substituting this into the above formula, we obtain
InN-lnNo=--At, or N=Noe-'At
49.2t. According to 16.6, the kinetic energy of the neutron is K ~
~ h
2
/ 2ma
2
~ 0.2 MeV. It may be seen at first glance that only very
fast neutrons can penetrate the nucleus, but this is in contradiction
with experiment. However, it should be taken into account that in the
nucleus the neutron is not a free particle, but interacts strongly with
the other nucleons (nuclear forces). The mean binding energy per
nucleon ( 80.4) is known to be several mega-electron-volts, which
greatly exceeds the localization energy calculated above. This enables
all the neutrons, including thermal ones, to penetrate the nucleus.
50. Nuclear Reactions
50. t. The number of the nuclei that took part in the reaction is N =
== mNAIA, where m is the mass of uranium reacted, N A is the Avo-
gadro number, and A the mass number. Multiplying the number of
the nuclei by fj,'e = 200 MeV, we obtain the energy of the explosion:
Dividing by the calorific value of TNT, q, we obtain the TNT
equivalent: mTNT == wt
50.2. The energy released is found from the masses equation:
~ == (6.01513 + 2.01410 - 2 X 4.00260) X 931.5 ~ 22.4 MeV
The energy release per nucleon is 22.4/8 = 2.8 MeV. This is three
times as much as is released in fission of uranium: 2001235 =:: 0.85 MeV
per nucleon.
50.3. To initiate the nuclear reaction, the deuterium nuclei must he
brought together to within the radius of action of nuclear forces.
To do this they must surmount the Coulomb potential harrier (see
Problem 49.3). Take account of the Maxwellian distribution (see
Problems 33.10 and 47.26).
50.4. Let the pi-meson be at rest in some reference frame. Then its
momentum is zero. If a pi-meson decays into two photons, they will
fly in opposite directions, so that in this reference frame the com-
bined momentum will be zero although each photon separately has
momentum. The transformation of a pion into a single photon is
impossible, since it contravenes the law of conservation of momentum.
Nuclear and Elementary Particle Physics 303
/
"
/
/
/
/
/
/
,
"
"
"
<,
<,
-,
<,
-,
L\Vb = (inn - X 931.5 =-=
:=140-106=34 l\IeV
The energy of each photon is = 135/2 = 67.5 l\1eV (see 83.7).
50.5. From the uncertainty relation for the energy we obtain t!:t.C6'
hl, where 't is the Iifctime of the particle. The accuracy with which
the mass (and the energy) can be measured is
8.= :1In == t1ctJ
m tJ
50.6. From the law of conservation of energy we obtain for the gamma-
photon e1' 2moc2, where mo is the rest ll18SS of an electron.
50.7. Suppose that the photon produces a pair of particles with iden-
tical momenta (Fig. 50.7). In this ease the laws of conservation of
energy and the momen tum will be wri t-
ten in the form
Hence 21nc
2
= 2mvc cos Cl, or c = v cos a;
which is impossible.
50.8. Find the total energy released in
the reaction:
The rest energy of a muon is Fig. 50.7.
= 106 MeV, the rest mass of a neut-
rino is zero. Assuming the dacaying pion to beat rest, we obtain
that the muon and the neutrino momenta are equal in magnitude
and opposite in direction. We have
Pv== <tb
v
,
C C
which yields
V :"= 1/ KJ! -]-- KJ!)
But, from the law of conservation of energy == il'fD' - KfJ,' so
- KJi = V (20oJ! + Hence it follows that the kinetic
energy of the muon is
K == 2 ( Vb Of.L -+ tb )
50.9. The decay of a-neutron results in the release of energy
= (m
n
- m,p) X 931.5 = 939.6 - 938.2 == 1.4 MeV
This energy is shared by the electron and the antineutrino: =
== + where the total energy of the electron is the sum of its
v
rest energy 0 and its kinetic energy K
r
, i .e. == -t- Ks, In
accordance with the law of conservation of momentum for a proton
at rest
or 'fb
v
=.i-
c c
304 Solutions
Substituting the value of the neutrino energy, we obtain after simple
transformations
Li
Fig. 50.11.
p
2Ka, = (NIH-1-l'rfLl-
-2Ma,) X 931.5-t-KH == 24.2 MeV
Since the kinetic energy of the
alpha-particle turns out to be much
lower than its rest energy, the alpha-
particles produced in the reaction
are nonrelativistic. The separation
angle is found from the law of
conservation of momentum
8
PH=:: 2Pa cos 2
But for a nonrelativistic particle
p == V2mK, so
50.10. Let us denote the total energy of the neutral pion by and
its rest energy by == 135 MeV. By the law of conservation of energy
0n == where is the pho-
ton energy. From the law of con-
servation of momentum (see Fig.
50.10) we obtain P == 2pv cos 45.
Noting that the photon momentum
is Pv == c and the pion momen- Ji.o
tum is Pn. == - we
c
obtain
cos
2
45, or, -
- cos- 45
It then follows that = -Vi Fig. 50.10.
The kinetic energy of the pion
is Kn. = (V2 and the photon energy
50.11. The nuclear reaction takes the form IHI + aLi? == 2
2He
4
2. Force
k
1k2
2.1. k= k
1+k2
2.3.
2.4. d = F
2
al (Fi - F,J.
PeDs a
2
2.6. TI == . ( + )'
SIll al a
2
mg sin ~
2.7. F
I
== sin a + ~ ,
2.8. . mg and 2mg.
F
t
X
t+
F
2
X
2 + F
S
x
3
+ ...
Xo= F
1+F2+Fs
+...
2.5. k = k
1
+ k
2
T = P cos a l
2 sin (a
l
+cx
2
)
F = mg sin a
2 sin (a+p)
3. Particle Dynamics
3.1. 8.4 X 10
7
Pa.
3 2
- m2-ml F- 2mlm2g
. a-g ,- ,
m
2+ml
m2+
ml
3.3. It is greater in the second case.
F
pres
= 2F.
Answers and Hints 307
3.4. (1)
(
2) m (g+b)
a2 m+M'
m (g-b)
(3) as= m+M '
3
5 - -m- M sin a F
a-g m+M '
mMg (1+sin a)
m+M
3.6. a = _g m2
1 m
2
tan a +ml cot a '
Q
M sin a cos a
3.7. ax==g M+msin2a '
(M+m) sin
2
a
ay=-g M+msin2 a
b = _ m sin a cos ex.
x g M +msin
2a'
mMg
Q== M+msin
2
cx '
.. /l cos cx lo
3.8. T=2n V-
g
- . 3.9. l== 1-mro
2jk.
3.10. 3912'.
3.11. In the lower point the overload is 8, in the upper it is 6.
r r 1
3.12. H == 2 c o t ~ ex.; T ~ 2 cot a.
r 2 3/2 2 2 A ( A
313
- (1+
X
) 315 b- voslnpCOSa-
p)
r-p -2 - 2
P g cos a,
3.16. Vo== VgH (1+ l
2
jH2) ; tancx=Hjl.
4. Gravitation. Electrical Forces
4.1. 5.96 X ~ ~ kg. 4.2. 6.03 X 10
24
kg. 4.3. 1.99 X 1()3o kg.
4.4. The Sun attracts the Moon 2.1 times more strongly than does
the Earth.
4.5. 0.63 astronomical units = 1.08 X 10
8
km.
4.6. 0.414 of the planet's radius.
4.7. 8.52 mjs2; 1.62 m/s"; 270 mjs2. 4.8. 1 h 25 min.
4.9. 4 X 10-
8
C. 4.10. E = 0; E = q V ~ .
nBoa
20
308 Answers and Hints
(3)
4 13
- .. /- eEL cos 4 14 4!) 10
4
N C
.vo-v ...tJX /.
4h mV5 sin" a < E < mV5 sin 2a
4.15. 0 < tan ex < -l- ; 2eh el
5. Friction
5.1. 2 s; 4 m.
5 2 (t)
== (m-!J.M) g F == mMg (1+J.1) .
at m+M' 1 m+M '
(
2) F _ mM[g(1+J.1)+b]
a
2
=== m+M ' 2 - m+M
m (g-b)-J.tMg F _ mM [g
a3= m+M ' 3- m+M
.
5.3. a==g m+M [Hl,nt. In solving the problem
one should take into account the initial velocity of the block.]
mlm
2
g cos (ex.-cp) t
5.4. Q= q> == arc an u,
ml cos- ex cos <p +m
2
sin ex. sin (a-cp) ,
[Hint. The force of sliding friction of the wedge against the
table should be included in the equation of motion of the wedge
(Problem 3.6).]
mMg cos a cos <p
5.5. Q== M cos <p+ 1nsin ex. sin (a-<p)
5.6. Q= mMg [Hint. See Problems 5.4 and
M cos cp+ m SIn a SIll ex.-cp)
5.5.]
5.7. g tan (a, - cp) a g tan (a + <p), where (p == arctan p.. [Hint.
Note that the direction of the force of static friction is not known.]
5.8. At a distance less than 8 cm from the centre of the disk.
5.9. Over 11 m/s.
5 10 .. /' g tan to .. /' g tan where m==arctan 11..
V R SIll ex. V R SIn ex. , -r r
[Htnt, Take into account the fact that the direction of the
force of static friction is not known.]
5.1t. 98m; 57. 5.12. 23 cmls; 5.1 m/s
2
; 0.09 5; 1 em.
5.f3. Hint. Making use of the results of the previous problem, express
the instantaneous acceleration in terms of the instantaneous
velocity.
5.4. 0.2 m/s; 15 s, 5.15. 11 m/s.
Answers and Hints 309
6. Theory of Relativity
6. f. II into Compare the dimensions of the rod and of the hole which
at rest are precisely the same.
6.2. B = uv/ c
2
6.3. 0.99995c.
6.4. Hint. Does the result obtained have the meaning of the law
of addition of velocities?
6.5. Hint. Consider the reference frame fixed to the medium which
moves at a speed u with respect to the source of light.
6.6. Hint. Make use of the results of the previous problem.
6.7. 55 m; 7.7 m. 6.8. 6.3 X 10-
5
s. 6.9. 2.83 X 10
8
m/s.
6.fO. p = pJ(1 - where = ole,
6.11. Hint. Make use of the Lorentz transformations.
6.12. 2.6 X 10
8
m/s; 5.3 X 10
8
m/s.
6. t 3. The particle moves in a circle. The expression for the force
will be formally the same as in Newtonian mechanics.
6.t5. l = lo Vi -
7. The Law of Conservation of Momentum. Centre of MIlSS
7.1. 670 m/s. 7.2. -3.25 mls; 9.3 m/s. 7.3. 87 em.
7.4. 20 m/s
2
7.5. 8.1 km/s; 5.3 km/s. [Hint. Make use of the Tsiolkovsky formula.]
7.6. Hint. Write equation (15.7) of 15.5 in differential form' and
integrate it, keeping in mind that the velocity of the exhaust
gases is a constant.
7.7. Hint. Take into account the change in the rocket's mass as the
fuel burns.
7.8. Hint. Make use of the properties of the centre of mass ( 15.8).
7.10. The centre of mass is at a distance of 4.3 units from the left-
hand edge of the plate.
r
2
7.11. At a distance of x == R +r ==1.13 em to the left of the centre
of the large circle.
7.16. T2/R3 = 4n/y (ml + m
2
) .
8. Total and Kinetic Energy
8.1. Electron: 8.19 X 10-
14
J == 0.511 MeV; proton: 1.504 X
X 10-
10
J = 938.3 MeV; neutron: 1.506 X 10-
10
J = 939.6 MeV.
8.2. 2.6 X 10
8
m/s.
8.3. 1.28 X 10-
13
J = 0.8 MeV; 6.4 X kg -ttsls.
8.4. 5.8 X 10-
18
kg .m/s; u = = O.996c.
8.5. u = = O.976c. 8.6. 0.8%; 69%; 92%.
8.7. 3.8 m/s. 8.8. ct== arctan -.!- . 8.9. 2.7 X 10
2
kW.
8.10. Hint. The work done by the electrical forces is equal to the
change in the kinetic energy.
8. f 1. Less than [Zpc,
310 Answers and Hints
9. Uncertainty Relation
9.1. Hint. Express the kinetic energy of the electron in terms of the
radius of its orbit; assume the characteristic dimension of the
region of localization to be equal to the radius.
9.2. 200 MeV. 9.3. 1 eVe 9.4. 9 MeV.
10. Elementary Theory of Collisions
to.1. Horizontally at a speed of 3.1 m/s. 10.2. 0.14 MeV.
10.4. The distance from the vertex of the parabolic mirror to the
focus is f = p/2. [Hint. Use the result of Problem 3.14.]
10.6. 83; 43.
10.7. V1 == vd/2r; V
2
== v V1-(d
2
/4r
2
) ; sin a
2
= d/2r; al +a
2
== '1t/2;
[Hint. Use the result of Problem 10.5.]
. d 21111V cos (X,2
10.8. slnct2 =- + ; V2= + ;
Tl T2 ml m
2
_ ... /"1 4m1m2 cos
2
cx,2
1-
V
V (ml +m
2
)2
[Hint. See solution of the previous problem.]
10.9. p = 2nmv2 cos
2
a. to.10. 5 m
2
dx 4nrf} ,
28. Charges and Currents in a Magnetic Field
28.1. 2.8 X 10
7
m/s; 16 MeV. 28.2. O.856c = 2.57 X 10
8
m/s;
99 MeV.
28.3. (a) Nonrelativistic particle: R
1/R2
=== y](JK;.
(b) R I
t i " ti ti I R /R .. ;- K
1
e a I v18 IC par 1C e: 1 2 = V K2 0 +K2)
318 Answers and Hints
28.4. DS. '. I
28.5. "A helix winding around the lines 'of induction. The radius of
the circle is 23 mm, the pitch is 39 mm,
28.6. Hint. Use the results of the previous problem.
28.7. 51 mm.
28.8. 1.38 X 10
4
VIm. [Hint. Use the result of Problem 4.14.]
28.9. 4; 42; 3.0 MHz. 28.10. 6.3 X 10-
3
T; 87 kHz; 1.1 T; 0.2 MHz.
28.11. 10
8
m/s; no. 28.12. 2.3 mm.
28.13. 4.02 amu (2He4) and 3.01 amu (2He3, or IH3).
28.14. 1.2 X 10-
D
N -rn. 28.15. 22.5; 10-
10
N -m/deg = 5.7 X
X 10-9 N -m/rad,
28.16. F= - 28.17. 2.9X10-. N.
28.18. Hint. Find the orientation of the velocity vector of the cir-
culating charge with respect to the induction vector.
28.19. Hint. Determine how the magnetic field acts on the circulating
charge, if its orbital magnetic moment is directed against
the field.
28.20. Less than 2.5 MeV.
29. Magnetic Materials
29.1. 1.5 X 10-
9
Am
2
; against the field. 29.2. 1.5 X10-9 Am
2
;
along the field. 29.3. 9.0 X 10-
9
AIm.
29.4. (a) 1.1 % greater; (b) 4.2% greater.
29.5. 33 AIm; 130 AIm.
29.6. Hint. Compare the energy of thermal motion with the energy
of interaction of two Bohr magnetons, separated by the inter-
atomic distance.
29.7. Above 9.0 X eVe [Hint. Estimate the energy of thermal
motion at the Curie temperature.]
29.8. 1.27 X m 0.01 mm, 29.9 Ilmax = 9.6 X 10
3
at H =
= 75 AIm.
29.10. 100 AIm; 1.26 T; 10
6
AIm; 1.8 X AIm.
29.11.
00 16.4 X10
2
! 1. 6X
102
1 8X 10
2
H,
A/m
f-L'
50 75 100 200
500 11000 \1500
4X10
2
Ito I 5
30. Electromagnetic Induction
30.1.0.2 V; it will not. 30.2. V2g(H-h).
30.3. It will move downwards at a stationary velocity
Vstat = lngR/ B
2
l 2.
30.4. dcp = Boo1,2/2. 30.5. 0.01 N.
Answers and Hints 319
I
q; 1- 3ttJl
o
i wa
2
r
2
v (xo-vt)
30.6. Ul - 5
2 [a
2
+(x
O
- vt )2] /2
30.7. 1.8 X ws. 30.8. 2.5 T.
30.9. The dielectric is polarized. 30.tO. 1.25 flC.
se.u. 25 W. 30.12. 5.9 mHo 30.13. 29.5 V.
30.14. 41 H; 0.82 J; 1.36 X 10
2
J/m
3
30.t5. 3.1 X J/m
3
; 1.8 X
X 10-
6
J.
30.16. 4.5 X 10
3
N; 1.5 X 10
2
N. 30.17. 1.5 H.
30.18. i = 1
M
(1 - e-t/'t), where 1
M
= 't = LIR.
30.19. t = 2.3LIR.
3t. Classical Electron Theory
31.1. = = 1. 8 X 10
11
C/kg.
31.2. 1.1 X 10-
8
V; will not change. 31.3. 1.1 X 10
29
m-
3
; 3.2 X
X 10-
3
m
2
/ (V -s),
31.4. 1.07 X 10-
10
mS/C.
31.5. 6.2 X 10
20
m -3; 4.0 m
2/(V.s).
31.6. 0.24 mm/s.
31.7. 4 X 10-
14
s: 400 A. 31.8. 131C. 3t.9. 65 C.
31.10. 3.6 X 10
3
m/s. 31.11. 38 kW/m
2
31.12. 90 MJ.
3t.13. 4 A. 3t.14. 4.1 X 10
2
W/(m.K); 68 W/(m.K). [Hint. Use the
Wiedemann-Franz law.]
32. Electrical Conductivity of Electrolytes
32.1. 78%. 32.2. 65 urn. 32.3. 200 mg. 32.4. 5 g.
32.5. 1.52 V. 32.6. 6.25 X 10-
11
kg; 9.1 X 10-
3
J.
32.7. Over 10
3
kW.h; more than 40 roubles.
33. Electric Current in a Vacuum and in Gases
33.t. 160 rnA. 33.2. Increases 21 times.
33.3. 4.1 X N.
33.4. A coated cathode is 3.3 times more efficient.
33.5. 2 kohm; 20. 33.6. 1.15 X 10
14
m-
3
; 4.3 X 10-
12
e
2n
A,2
41.10. n=1-
8
2
0
2=1-2.3x10-
7
n Bornec
(
e2no ) 2
41.11. R= 4 2
Bornew
41.12. 0.345; 0.361; Rvlo1lRred = 1.05. 41.13. 3.0 X 10
8
m/s.
41.14. Hint. Use the results of Problem 41.6.
4"1.15. U = 1.9 X 10
8
rols; u = 2.0 X 10
8
m/s.
41.16. 35 m-
1
; 2.0 em,
k
- ~ r ~ 16n
2
-J,t(Ta-TI)
41.17. - 1
0
- r ~ (n+"1)4 e
41.t8. 98%; 10 A. 4t.19. 6 layers. 41.20. 1.6 em.
42. Polarization of Light
42.1. It will be reduced to one fifth of its initial value.
1.2 2 I 1 I 2 2 45
.. = 2 0SIn. ex cos. ex,
42.3. There will be no interference pattern.
21*
324 Answers and Hints
42.4. (2m+ 1) x 0.856 flm; (2m+ 1) X O.78n rad == (2m+ 1) X
X 140.
42.5. dquartzldcalclte = 19. 42.6. 2.1 X 10
2
kg/rn",
42.7. (2m+ 1) X 5.19 mm; (2m + 1) X 2.75 mm.
43. Geometrical Optics
43.1. 2.6 mm. 43.2. 41. 43.4. i.a m. 43.5. 18.7; 7.4.
43.6. 2.2 diopters; -1.0 diopter.
43.7. In the posterior focus of the right-hand lens.
43.8. Hint. Bring the lenses in Fig. 43.7 together until they are
in contact.
43.9. Hint. Use the result of the previous problem.
43.10. -3.8 diopters; -1.1 diopters. 43.11. <I> == (n - 1)IR.
43.12. 2.4 em, 43.13. s: = /12; ~ == klH = 1/2. 43.14. 81
040'.
43.15. a = ~ (for :;c ~ a - f).
43.16. An elongated ellipsoid of rotation.
43.17. 2.4 cm; 6%. 43.18. 13.3 diopters.
43.19. Hint. See 65.5, Fig. 65.9.
43.21. A convex lens. 43.22. A concave lens.
43.23. Hint. See Problem 10.4. 43.24. No.
44. Optical Instruments
44.1. 1.76 W; 8.8 W; 2.9 W. 44.2. 11 Vim; 3.7 X 10-
8
T.
EWalI D 2/ 0.381 2J
44.3. -E--=-2h' 44.4. Eedge= ..I =--2-; Ecentre==-2
floor 3r
2
y 3 r r
44.5. E ==5/4E
o
44.6. 10 Ix; 2.8.lx.
R D2
44.7. 5.0 lx: 2.5 lx. 44.8. E ==T f2 .
44.9. It will decrease to 1/3.2.
44.10. -7 diopter glasses.
44.11. 57" ~ l' (see 66.4). 44.12. 370.
44.t3. 0.6 11m. 44.14. 24 diopters; 8.3 mm; 1.2"; 830 km. 44.15. '\'=
=1 ~ I
44.t6. 35 m; 5.2 km, 44.t7. Aradlo = 2.5 X 10
3
; Aopt = 5.4 X 10
6
48.12. n = 2.4 X 10
19
m-
3
; R
H
= 9.4 X m
3
/ C.
48.13. 5.8 X 10
2
ohm-
1
.m-
1
; 3.0 X 10
2
ohm-
1rn-I
48.14. 0.55 V; 8.6 V. [Hint. For the expression for the internal con-
tact potential difference see 78.1.]
49. Nuclear Structure
49.t. Helium 2He3-two protons, one neutron; tritium IR3-one
proton, two neutrons.
49.2. 20% of the light isotope and 80% of the heavy one.
49.3. Deuterium: R = 1.8 X m; u, = 0.8 MeV.
Polonium: R = 8.3 X 10-
15
m; U
o
= 14.6 MeV.
49.4. 2.23 MeV; 1.12 MeV per nucleon.
49.5. 2.84 MeV/nucleon; 2.56 MeV/nucleon.
49.6. Five alpha-disintegrations and four beta-disintegrations.
49.7. K
a
=.: 4.45 MeV; K
p b
= 1.05 MeV.
49.8. It cannot. 49.9. It cannot.
49.10. It can; an electron and an antineutrino; 1.48 MeV.
49.1t. 4.96 MeV.
49.12. w = e-
21
= 2.8 X 10-
1
.
49.13. Will decrease to 1/1.3 of its value.
49.14. 3.7 days.
49.15. 1.4 X 10
17
s = 4 X 109 years.
49.16. 8800 years.
49.17. 0.82 MeV. 49.18. Srec = = 1.4 X 10-
7
; 6
na
t =
= = 3.2 X 10-
13
49.19. v = l Mc = 81 tal.
49.20. N = Noe-'A.t. 49.2t. 0.2 MeV.
328 Answers and Hints
50. Nuclear Reactions
50.1. 3 X 10
7
kg = 30 000 tonnes.
50.2. 22.4 MeV; 2.8 MeV per nucleon.
50.3. They can.
50.4. It contradicts the law of conservation of momentum; 67.5 MeV.
50.5. s = = 6 X 10-
8
50.6. -s, 1.02 MeV.
50.8. KJJ. = 4.1 MeV; Wry = 29.9 MeV; = 7.3.
50.9. K
e
= 0.28 MeV; = 0.61 MeV.
50.10. s; = 5.6 MeV; = 95 MeV. 50.11. 12.1 MeV; 158.4.
50.12. _teO + Ipl -+ on
1
+ ov
o;
0.24 MeV.
1. Astronomical Data
Equatorial
Average
Period
distance from
half-diameter, Mass, kg
the Sun,
of revolution
10
6
m
10
9
m
about the Sun
The Sun 700 1.98X10
30
- -
Venus 6.2 4.9X10
24
108.11 227.70 days
The Earth 6.4 5.98X10
2
4: 149.46 365.26 days
Mars 3.4 6.5X10
23
227.7 686.98 days
The Moon 1.7 7.4X10
22
-
-
2. Mechanical Properties of Solids
Permls- Breaking
Bulk Young's
sfble stress for
Dens ltv o,
Material modulus modulus
stress extension
10
3
kg/rn
3
K, 10
1
0 Pa E, 10
1
0 Pa O'per' O"br'
107 Pa 108 Pa
Steel, soft 17 20 14 4-6 7.8
Steel, chrome-
nickel 17 22 30 10-15 7.8
Silver 10.4 8.0 3-8 0.9-1.5 10.5
Aluminium 7.6 7.0 3-8 0.9-1.5 2.7
Copper 14 13 3-t2 1.2-4.0
8.9
Nickel 16 20
8.9
Lead 4.1 1.7 0.2 11.3
Ice (-2C) 0.28
0.92
330 Tables
3. Thermal Properties of Solids
Specific heat Specific
I Linear
Coefficient
Melting
expansion
Material
C, kJ/(kg K) heat of
point
coefficient
of thermal
(from 0 to fusion A,
tro,OC a, x-i
conducttvitv
100
0C)
10
5
J/kg
10-6
K, W/(mK)
Steel 0.45 2.7 1440 10-11 50-70
Silver 0.23 0.88 980.6 20.5 4.2X10
2
Aluminium 0.84 4.0 660 27 2 .3X10
2
Copper 0.38 2.1 1080 20 3.8X10
2
Nickel 0.46 3.0 1453 18 0.9X10
2
Lead 0.126 0.45 327 30 35
Ice (-2 C) 2.1 3.4 0.00 - 2.2
4. Properties of Liquids
Material
Oil
Water
Water (8) C)
Mercury
Castor oil
H
2S04
, 30%
0.8
0.9999
0.9718
13.595
0.95
1.02
1-10 - - - - - -
1.00 72.8 2.1 0.61100 2.28 4.18
0.36 62.8 - - - - -
1. 55 475.0 1.8 10 356.6 0.25 0.14
986 - - - - - 2.13
2.44 - - - - - 1.4
Tables
5. Properties of Gases
331
Effective Viscosity Heat con-
Specific hea t
"l=
Density
cross 11, J.LPas ductivity K,
": KJ/(kg.K)
=2
o, kg/m
s
Material section (at stan- 10-
2
W/(m.K) (at stan-
G. dard con- (at standard
(from 0 to
C
v
dard con-
10-
2 0
m2
ditions) conditions)
100C)
ditions)
Helium 3.1 18.6 14.15 0.523 1.630 0.1785
Neon 7.0 29.8 4.65 1.05 1.642 0.8999
Nitrogen 10.8 16.6 2.43 1.04 1.401 1.2505
(N
2
)
Hydrogen 5.7 8.4 16.84 14.3 1.407 0.0899
(H
2
)
Hydrogen* 3.5 - - - 1.667 --
(H)
Oxygen (0
2
) 9.6 19.2 2.44 0.913 1.400 1.4289
Air 10 17.1 2.4 1.01 1.40 1.293
Steam
(100C) 6.0 12.8 2.6 1.951 1.334 0.598
Carbon dio-
xide (CO
2
) 16.2 13.8 1.38 0.91 1.300 1.9769
CO
2
, 100C 18.6 0.-90 1.30
CO
2
, 300-c 26.7 0.85 1.22
CO
2
, 500
0
e 0.83 1.20
* Computed with the aid of the formula (1 = 4 t a ~ where ao = 0.5 A is the Bohr
radius.
6. Electrical Properties of Materials (20 "C)
Material
Dielectric constant I Breakdown field
e strength EM' MV/m
Paraffin paper
Mica
Glass
Transformer oil
Ethyl alcohol
2
6-7
4-10
2.2
26
40-60
80-200
20-30
332 Tables
7. Velocity of Sound (Longitudinal Waves)
Material U, km/s Material U, km/s
Quartz, (X-cut)
Nickel (in a mag-
netic field of
0.3 T induction)
5.72
4.86
Water (17C)
Sea water
Castor oil
1.407
1.446
1.50
8. Refractive Indexes
Light
I
wave-I FIUO-I MOltenl Rock I S"l-I Water I Glass
length rite quartz salt vtte (')OC) J
A, nm Naci KCI" crown flin t
nfra-red 1256.0 1.4275
-
1.5297 1.4778 1.3210 1.5042 1.6268
{
670.8 1.4323 1.4561 1.5400 1.4866 1.3308 1.5140 1.6434
ed 656.3 1.4325 1.4564 1.5407 1.4872 1.3311 - -
643.8 1.4327 1.4568 1.5412 1.4877 1.3314 1.5149 1.645:1
range 589.3 1.4339 1.4585 1.5443 1.4904 1.3330 1.5170 1.6499
ellow 546.1 1.4350 1.4602 1.5475 1.4931 1.3345 - -
reen 508.6 1.4362 1.4619 1.5509 1.4961 1.3360 -
-
lue 486.1 1.4369 1.4632 1.5534 1.4983 1.3371 1.5230 1.6637
Dark blue 480.0 1.4371 1.4636 1.5541 1.4990 1.3374 - -
iolet 404.7 1.4415 1.4697 1.5665 1.5097 1.3428 1.5318 1.6852
{
303.4 1.4534 1.4869 1.6085 1.5440 1.3581 1.5552 -
UItraviolet 214.4 1.4846 1.5339 1.7322 1.6618 1.4032 -
-
185.2 1.5099 1.5743 1.8933 1.8270
- -
R
o
y
G
B
V
9. Masses of Some Neutral Atoms (amu)
Mass Mass
Hydrogen 1HI
Deuterium IH2
Tritium IH3
Helium 2He3
2
He4
Lithium sLi
6
sLi
1
Berillium 4Be7
4
Be9
1.00783
2.01410
3.01605
3.01603
4.00260
6.01513
7.01601
7.01693
9.01219
Carbon 6C10
6
C 12
6
C 13
6
C 14
Aluminium 13A130
Silicon 14Si31
Phosphorus 15P31
Lead 82Pb20G
Polonium 84P0210
10.00168
12.00000
13.00335
14.00324
29.99817
30.97535
30.97376
205.97446
209.98297
Tables
10. Fundamental Physical Constants
333
Constant
I Notation
Value*
'A
c
= Ii/mec 2.4263089X10-
12
m
IlB= eli/2me 9.274078 X10-
14
J. T-l
1. Magnetic constant
2. Electric constant
3. Velocity of light in a
vacuum
4. Elementary charge
5. Planck's constant
6 . Avogadro number
7. Atomic mass unit
8. Rest mass of:
electron
muon
proton
neutron
9. Specific charge of
electron
10. Faraday number
11. Bydberg's constant
12. Bohr radius
13. Compton wavelength
of electron
14. Bohr magneton
15. Magnetic moment:
electron
proton
16. Gas constant
17. Volume of 1 kilo-
mole of an ideal gas
18. Boltzmann constant
19. Stephan constant
20. Gravitational con- -
stant
21. Energy equivalent
of 1 amu
c
e
h
Ii= h/2n
NA
1 amu
f1e
R
'Y, G
431 .10-
7
G m-?
8.85418782X10-
12
Fm-
1
299792458 m S-1
1.6021892X10-
1 9
C
6.626176X10-
34
Js
1.0545887x10-
3
4. Js
6.0220943X10
26
kmole-
1
1.6605655 X 10-
27
kg
9.109534 X 10-
31
kg
5.4858026X10-
4
amu
1.883566X 10-
28
kg
0.11342920 amu
1.6726485X 10-
27
kg
1.007276470 aDlU
1.6749543X 10-
27
kg
1.008665012 amu
1.7588047 Ckg-
1
9.648456Xf0
7
Ckmole-
1
1.097373143X10' m-
1
0.52917706X10-
10
m
9.284832 X 10-
24
J. r-:
1.4106171 X10-
26
J .r-i
8.31441X10
3
J/(kmoleK)
22.41383
1.380662 X10-
23
J. K-l
5.67032X10-
8
W/(m
2
. K' )
6.6720X10-
11
Nm
2
/ kg
2
931.5016 MeV
* The above constants are trom the paper "Recoffirnended approved val-
ues or fundamental physical constants - 1973" (the periodical "Uspekhi
fizicheskihh nauk", v. 115, No.4, April 1975, pp. 623-633).
TO THE READER
Mir Publishers would be grateful for your
comments on the contents, translation
and design of this book. We would also be
pleased to receive any other suggestions
you may wish to make.
Our address is: Mir Publishers
2 Pervy Rizhsky Pereulok,
1-110, GSP, Moscow, 129820,
USSR
Printed in the Union of Soviet S ocialtst Republtcs
Basic Concepts of Quantum Mechanics
L. TARASOV, Cand.Sc.
This book gives a detailed and systematic exposition
of the fundamentals of non-relativistic quantum mecha-
nics for those who are not acquainted with the subject.
The character of the physics of microparticles and the
problems of the physics of microprocesses (interference
of amplitudes, the principle of superposition, the spe-
cific nature of measuring processes, casuality in quantum
mechanics) are considered on the basis of concepts about
probability amplitudes. Besides, the quantum mechani-
cal systems-microparticles with two basic states-are
analyzed in detail. The apparatus of quantum mechanics
is considered as a synthesis of concepts about physics
and the theory of linear operators. A number of specially
worked out problems and examples have been included
in order to demonstrate the working of the apparatus.
This book is meant for use by students of engineering
and teachers-training institutes. It may also be used
by engineers of different profiles.
Contents. Physics of Micro-particles. Physical Founda-
tions of Quantum Mechanics. Linear Operators in Quan-
tum Mechanics. Brief Historical Survey.
Handbook of Physics.
B. '..YAVORSKY, D. Sc. and A. DETLAF, Cando Sc.
A companion volume to Vygodsky's Handbook of Higher
Mathematics, designed for use by engineers, technicians,
research workers, students, and teachers of physics.
Includes definitions of basic physical concepts, brief
formulations of physical laws, concise descriptions of
phenomena, tables of physical quantities in various
syetems -o! units, universal physical constants, etc.
This is a third English :edition.
Contents. Physical Basis of Classical Mechanics. Funda-
mentals of Thermodynamics and Molecular Physics.
Fundamentals of Fluid Mechanics. Electricity and Magne-
tism. Wave Phenomena. Atomic and Nuclear Physics.