UNIT VII DUAL NATURE OF MATTER AND RADIATIONS

(4marks)

VERY SHORT ANSWER TYPE QUESTIONS:1. An electron and photon have same wavelength. Which one of the two has more
energy?
Relativistic energy of a particle, E= (m0c4+p2c2)
Hence the electron has more energy than photon.
3. If wavelength of electromagnetic waves are doubled what will happen to energy of
photon?
E = h
=

4. Alkali metals are most suitable for photoelectric emission. Why?

Alkali metals have too low work functions. Even visible light can eject electrons
from them.
5. Out of microwaves, UV, IR which radiation will be most effecting for emission of
electrons from a metallic surface?
UV are most effective since they have highest frequency hence more energetic.
7. If the intensity of incident radiation on a metal is doubled what happens to the K.E of
electrons emitted?
K.E of photons remains unaffected since they do not depend
8. What is the value of stopping potential between the cathode and anode of photocell?
If the max K.E of electrons emitted is 5eV?
Stopping potential V0 = Kmax/e = 5ev/e =5 V
9. It is easier to remove an electron from sodium than from copper, which has a higher
value of threshold wavelength?

w0 = h0 = hc/0
0 1/ w0
Since sodium has lower work functions than copper it is easier for electron
ejection. As it is lower work
function, higher wavelength.

11. An electron and proton possessing same K.E. Which one will have greater
wavelength?
1/2 mv2 = (m2 v2)/2m = p2/2m

e > p
electrons have greater De broglie wavelength than proton .
12. In Davisson Germer experiment if the angle of diffraction is 520 find Glancing
angle?
= 90 - /2
= 90 52/2 =64

13. What is the effect on the velocity photo electrons, if the wavelength of incident light is
decreased?
KE of photoelectrons is given by Einsteins photoelectric equation.
Ek =1/2 mv2
= h - w0
As wavelength decreases velocity increases.

14. The stopping potential for some material is 1.2 V. What is the energy range of the
emitted photoelectrons?
The range of energies of the emitted photoelectrons will be from 0 to 1.2 eV.
15.The intensity of incident radiations in a photoelectric experiment is doubled. How the
stopping is potential affected?
The stopping potential will remain unaffected because it does not depend on the
intensity of the incident light.
16. If the intensity of the incident radiation on a metal is doubled, what happens to the
kinetic energy of the emitted photoelectrons?
There is no change in the kinetic energy of the emitted electrons. This K.E. is
independent of the intensity of the incident radiation as long as its frequency remains
the same.

17. The frequency () of incident radiation is greater than threshold frequency (q) in a
photocell. How will the stopping potential vary if frequency () is increased, keeping
other factors constant?
When, ( >o) stopping potential will be increased.
18.What is the energy and wavelength of a thermal neutron ?

Kinetic energy of a thermal neutron = f/2 kBT.

Since degree of freedom of a thermal neutron is three.
K.E = 3/2 kBT = 6.06 x 10-21 J
1/2
1/2
Wave length
e = h/(2mK.E)
e = h/(3m kBT) =0.147nm.
19. An X ray tube produces a continuous spectrum of radiation with its shots
wavelength end 0.45 A.What is the maximum energy of a photon in the radiation in electron
volt?(ii) From your answer to (i) what order of accelerating voltage (fo0r electron) is required
in such a tube?.
3
eV
(ii)In this case energy of X-rays photon is 27.6 KeV, the striking electron must be of
energy higher than 27.6 KeV. Therefore an accelerating voltage of the order of 30 KV is
required.
20. It is difficult to remove free electrons from metal X as compare to metal Y . What
you infer?
Work function of metal X is higher than metal Y.
21. A particle behaves like a wave. What determine the wavelength of the wave?
Momentum of the moving particle.
22. Draw a graph giving variation of maximum kinetic energy of photoelectrons with
frequency of incident radiations. What is the slope of this
graph?
The equation of this straight line is
K.E. = 1/2mv2max= h ( - o)
23. The wavelength of radiations incident on a material is
decreased. Does the maximum velocity of photoelectrons increase or decrease?
A decrease in wavelength implies an increase in frequency. Since an increase in
frequency increases the maximum K.E. of the emitted photoelectrons, the maximum
velocity would increase.
24. For photoelectric effect in sodium, the figure shows the plot of cut off voltage versus
frequency of incident radiation. Calculate the threshold frequency (0) the work function
for sodium
(ii)The work function (W) is related to the threshold frequency (0) by the relation.
W o = ho = 6.6 x 10-34 x 4.5 x 1014 J
= 29.7 xl0-20J = 2.97 xl0-i9J

2.97 x 10 -19
eV 1.856 eV
1.6 10 -19

25. Suppose the photoelectric effect occurs in a gaseous target rather than a solid. Will
photoelectrons be produced at all frequencies of the incident photons?

No; we are likely to get photoemission only for those frequencies whose photons
have an energy equal to or more than the (minimum) ionization energy for the gas
concerned.
26. Yellow light does not eject photoelectrons from a given photosensitive surface,
whereas green light does. What shall be situation in case of red and violet light?
We will not get any photoemission with red light since its frequency is less than
that of yellow light. We will, however, get photoemission with violet light since its
frequency is more than that of green light.
27. By what factor does the maximum velocity of the emitted photoelectrons change
when the wavelength "the incident radiation is increased four times? (Given that the
initial frequency used is five times the threshold frequency)
When the wavelength is increased four times, the frequency goes down by a
1
1
mv12 = h ( 1 0 ) and
mv 22 = h ( 2 0 )
2
2
v12 1 0
=
2
factor of four. v 2 2 0
Now,

4 0
v12 5 0 0
=
=
=16
2
v 2 5 0 0 1 0
4
4

v1
=4
v2

Thus, the maximum velocity goes down by a factor of 4.

28. A cesium photocell, with a steady potential difference of 90 volt across it, is
illuminated by a small bright light placed one metre away. The number of electrons that
cross the photocell is n. What will be the number of electrons crossing the photocell
when the same light is placed half metre away?
When the light is brought to a distance of 0.5 m, the intensity of the light falling
on the photocell goes up four times. Since the number of photoelectrons emitted is

directly proportional to the intensity of incident light, the new number of photoelectrons
emitted would become 4n.
29. The sun rays are focused on a metal surface , and it produces a current. The lens
forming the image is then replaced by another lens of the same diameter but only half
in focal length. What will be the effect on the photoelectric current?
Hence, the intensity of light falling on the metal surface gets reduced in case (b)
and, we, therefore, get a reduced photoelectric current.

30. To work functions 2ev and 5ev for two metals x and y respectively. Which metal will emit
electrons, when it is irradiated with light and wave length 400nm and why?
= 400 x 10-9m=4 x 10-7m
E= h c/ = (6.6 x 10-34 x 3 x 108)/ (4 x 10-7) = 4.98 x 10-19 j
E= (4.98 x 10-19)/ (1.6 x 10-19) =3ev
Hence, metal x will emit electrons.
31. A photon and an electron have same de-broglie wavelength. Which has greater total energy.
Explain ?
For a photon E1=hc/
For an electron =h/mv or m=h/ v
E2=mc2
= (h/ v) x c2
E2/E1=c/v>1
Therefore,E2>E1.thus, electron has total energy greater then that of photon.

32. The de-broglie wave length of a photon is same as the wave length of electron. Show that K.E.
/h times K.E. of electron. Where m is mass of electron, c is velocity of light.
ph= e= =h/mv
K.E. of photon Eph=h=hc/
K.E. of electrons E=1/2mv2=1/2 m [h/m ]2
=h2/2m 2

Eph/Ee= (hc/ ) x 2m 2/h2

=2mc /h
Eph=Ee(2mc /h)

33. How may photons are required for emission of one photo electron if frequency
of incident radiation is
(i)

less than threshold frequency.

(ii)

More than threshold frequency.

(i)No photo electron will be emitted and photons are absorbed by electrons.
ii)One photon will emit one photo electron.

34. State the dependence of work function on the kinetic energy of electrons emitted in
a photocell. If the intensity of incident radiation is doubled, what changes occur in the
stopping potential and the photoelectric current?
According to Einstein

theory of photoelectric effect, kinetic energy of emitted

electron is

1
2
Ek = mVmax
= h W0
2
Greater the work function of the metal, lesser the kinetic energy of the
photoelectron. on doubling the intensity of the incident radiation stopping potential
remains the same, whereas photoelectric current is doubled.

35. Using Davisson and Germer Experiment to establish the existence of de Broglie
waves.
The experimental set up consists of an electron gun connected to a low tension
battery. The electrons emitted by the gun are accelerated to a desired velocity by
applying a suitable potential difference V using a high tension battery. Value of V can be
varied.
Using Bragg's law of diffraction of light through crystal lattices, the wavelength of the
wave showing diffraction peak at 50 was calculated as 1.65 A.

The predicted de Broglie wavelength of electrons accelerated through 54 V is

h
h
h
12.27
12.27
= =
=
=
A =
A =1.66 A
p
2mE
2meV
V
54
The excellent agreement between theoretical and experimental values of de Broglie
wavelength confirms the wave nature of electrons and hence the existence of the de
Broglie waves.

36. Through what potential difference an electron be accelerated so that it may have de
broglie wavelength 0.5
Let v = potential through which electron is accelerated.
Therefore, Energy = mv2 = eV
1/2
= h/(2mev)1/2
2
=> v = h2
here, h = 6.625 10-34Js
m = 9.1 10-31kg
e = 1.6 10-19 C
10-10 m
hence v = 6.03 104 V
37. An e- and photon have the same energy of 100eV. Which has greater associated
wavelength?
Here, e = h/(2mEe)1/2
2
=> Ee = h2
..(1)
e
Now,
2 2
2
Ep
p => Ep = h c ..(2)
2
p

as Ee = Ep = E = 100eV ..(3)
Dividing equation (1) by (2) and using equation (3) we get
E = h2c2 p2
2
h2
e

E = 2mc2

=>

2
p

2
e

= (E/2mc2)1/2
as E = 100eV
=> 2mc2
therefore E << 2mc2
e

Therefore the wavelength associated with a photon is greater than an electron

for the same energy.
38. the frequency of the light falling on the metal is doubled, what will be the effect on
photocurrent & the maximum Kinetic Energy?
The photo current does not depend on the frequency of incident radiation as:
Ek = h W
If the frequency is doubled
Ek/ = 2h! - W
=> Ek//Ek = (2h! W)/(h W) = (2h! 2W + W)/(h! W)
= 2 + W/(h! W) > 2
i.e. maximum KE will increase slightly more than double.
39.e work function of Li and Cu are 2.3eV & 4eV. Which of these metals will be useful
for the photoelectric cells working with the visible light?
The threshold wavelength of a metal is:
12475
/(W in eV)
o = hc/W =
12375
=> Li o =
/2.3
and Cu 0 = 12375/4
The wavelength 5380
photoelectric cell.
39. An e- has a speed of 5 106 m/s in a magnetic field of 10-4 T. What is the acceleration
of e- if e/m = 1.76 1011 C/kg.
BeV = mv2/r as the force due to magnetic field provides the
necessary centripetal force.
=> v2/r = Bev/m
as acceleration = v2/r
=> a = B(e/m)
=> a = v2/r = 10-4 1.76 1011 5 106
=> a = 8.8 1013 ms-2.
40. Assume that the potential difference between cathode and anode is the same as that
between two deflecting plates. If this potential difference is doubled, calculate by what
factor the magnetic field should changed to keep the electron beam undeflected?
e
/m = E2/2vB2 But E = v/d
e
/m = v2/2vB2d2 = v/2B2d2
or, B2 =
v
.
2(e/m)d2
v
or
B2 v => B
if v is doubled, then
B//B = (2)1/2 or B/ = (2)1/2B
Thus, the magnetic field should be increased by a factor of (2)1/2