Tutorial Set 1
Tutorial Set 1
Tutorial Set 1
EXCERSICE 1. A He2+ ion is accelerated from rest through a voltage drop of 1.000kilovolts.
What is its final de Broglie wavelength? Would the wavelike properties be very apparent?
SOLUTION: Since a charge of two electronic units has passed through a voltage drop pf 1.000
× 103 volts, the final kinetic energy of the ion is 2.000 × 103 ev. To calculate λ, we first convert
from ev to joules: KE≡ p2/2m =(2.000 × 103ev)(1.60219 × 10-19 J/ev) =3.204 × 10-16J.mHe
=(4.003g/mol)(10-3kg/g)(1mol/6.022 × 1023atoms)=6.65 ×10-27kg; P=√(2mHe.KE)=[2(6.65 × 10-
27
kg)(3.204 ×10-16J)]1/2=2.1 ×10 -21kgm/s. λ=h/p=(6.626 × 10-34Js)/(2.1 ×10-21kgm/s)=3.2 ×10-
13
m=0.32pm. this wavelength is on the order of 1%of the radius of a hydrogen atom-too short to
produce observable interference results when interacting with atom-size scatters. For most
purposes, we can treat this ion as simply a high-speed particle.
EXCERSICE 2. A retarding potential if 2.38 volts just suffices to stop photoelectrons emitted
from potassium by light of frequency 1.13 × 1015s-1. What is the work function, W, of potassium?
SOLUTION: Elight =hv =W+KEelectron, W=hv –KEelectron =(4.136 ×10-15 ev s)(1.13 ×1015 s-1)-2.38ev
=4.67ev-2.38ev=2.29ev.[note convenience of using h in units of eVs for this problem].
EXERCISE 3. Spectroscopists often express ΔE for a transition between states in wavenumbers,
e.g. m-1 or cm-1, rather than in energy units like J or ev (usually cm-1 is favored, so will proceed
with that notice)
a. What is the physical meaning of the term wavenumber?
b. What is the connection between wavenumber and energy?
c. What wavenumber applies to an energy of 1.000J? of 1.000ev?
SOLUTION:
a) Wavenumber is the number of waves that fit into a unit of distance (usually of one
centimeter). It is sometimes symbolized ϖ=1/λ, where λ is the wavelength in centimeters.
b) Wavelength characterizes the light that has photons of the designated energy.
E=hν=hc/λ=hcϖ (where c is given in cm/s).
c) E=1.000 J= hcϖ, ϖ=1.000J/hc =1.000 J/[(6.626x10-34 Js)(2.998x1010 cm/s)] =5.034x1022
cm-1. Clearly this is light of an extremely short wavelength since more than 1022
wavelengths fit into 1 cm. For 1000eV, the above equation is repeated using h in eVs,
and this gives ϖ=8065 cm-1.
EXERCISE 4. The life time of an excited molecule is 2x10-9 s. What is the uncertainty in its
energy in J? in cm-1? How would this manifest itself experimentally?
SOLUTION: The Heisenberg uncertainty principle gives, for minimum uncertainty ΔE.Δt=h/4π.
ΔE=(6.626x10-34 Js)/(4π)(2x10-9 s)] = 0.001 cm-1. Larger uncertainty in E shows up as greater
line width in emission spectra.
EXERCISE 5.
a) Show that sin(3.63x) is not an eigenfunction of the operator d/dx.
b) Show that exp(-3.63ix) is an eigenfunction of the operator d/dx. What is its eigenvalue?
c) Show that (1/π)sin(3.63x) is an eigenfunction of the operator ((-h2/8π2m)d2/dx2). What is
its eigenvalue?
SOLUTION:
a) d/dx(sin(3.63x) =3.63cos(3.63x) which is different from Constantxsin(3.63x).
b) d/dx(exp(-3.63ix) =Constant.exp(-3.63ix). Eigenvalue=-3.63i
c) ((-h2/8πm)d2/dx2)(1/π)sin(3.63x) = (-h2/8πm)( 1/π)(3.63)d/dx(cos (3.63x))
=[(3.63)2h2/8π2m].(1/π)sin(3.63x)=Constantx(1/π)sin(3.63x). With Constant being the
Eigenvalue= (3.63)2h2/8π2m.
EXERCISE 6. (Question without solution for student to provide answers)
Give three quantum hypotheses as well as their authors of the early twentieth century that paved
the way to the development of quantum chemistry and its applications. State how each
hypothesis solved each given problem.
EXERCISE 7. Give seven key differences between classical and quantum mechanics.
EXERCISE 8. Give an account of the electromagnetic wave.