CHEM 342. Spring 2002. Problem Set #1. Answers.
For the following three problems, please circle the correct answer.
Compton Effect
1. The Compton Effect was first observed by studying the scattering of X-rays by a
graphite target. Although the incident X-rays were monochromatic, the scattered X-rays
contained an additional component of longer wavelength.
Photoelectric Effect
2. If the frequency of incident light is above the threshold frequency, then as the intensity
of light increases, the kinetic energy of ejected electrons remains constant and the number of
electrons increases.
3. If the frequency of incident light is above the threshold frequency, then as the
frequency of light increases, the kinetic energy of ejected electrons increases and the number of
electrons remains constant.
4. What is the effect of incident light striking a metal surface if the frequency of the light
is below the threshold frequency for this metal?
No effect. Each individual photon lacks enough energy to remove an electron from the
metal.
5. Define in a maximum of 20 words for each
(a) Correspondence Principle: Quantum mechanics approaches classical mechanics in the
limit of large quantum numbers.
(b) Compton Effect: The elastic scattering of light by electrons. In the collision of an
electron with a photon, the electron gains energy, while the photon loses energy and emerges
with a lower frequency. Furthermore, the angles of scattering must be explained using the law of
h
conservation of momentum, with the momentum of the photon given by
.
c
Electron
Electron
Photon h/c
h'/c
Photon
(c) Photoelectric Effect: Involves shining light (of intensity I and frequency  ) on a
metal surface to remove electrons with kinetic energy (Ek) and flux (F, number of electrons
emitted). The Ek increases linearly with  , while F is unaffected by  . Also F increases linearly
with I, while Ek is unaffected by I.
6. The work function for metallic cesium is 3.43  1019 J. Calculate the kinetic energy
and the speed of the electrons ejected by light of 300 nm wavelength. [Hint: Planck's constant,
h,is 6.626  1034 J s; the speed of light in a vacuum, c, is 3.00  108 m/s; the mass of an
electron, me, is 9.109  1031 kg]
1
CHEM 342. Spring 2002. Problem Set #1. Answers.
KE electron  E photon  W
KE electron 
KE electron 
hc
W

6.626  10 34 J s 3.00  10 8 m s 1


300  10
19
KE electron  3.20  10 J
KE electron 
v
v
9
m
  3.43  10
19
J
me v 2
2
2 KE electron
me

2 3.20 10 19 J
9.109 10
31
  8.38 10
kg
5
m s 1
Note: W = the work function of the metal
Blackbody Radiation
7. The peak of the Sun's emission occurs at about 480 nm. Estimate the temperature on its
c
surface. Hint: The Wien displacement law reduces to T max  2 , where the constant
5
2
c 2  1.44 10 m
c
T max  2
5
c2
T
5 max
T
1.44  10  2 m

5 4.80  10  7 m

 6000 K
Bohr Atom: Spectral Lines
8. Calculate the wavelength of light emitted when an electron falls from the n = 100 to
the n = 99 orbit of the hydrogen atom. [Hint: Rydberg's constant, R, is 1.097  107 m1]
2
CHEM 342. Spring 2002. Problem Set #1. Answers.
 1
1
1 
 R 2  2 
n


 1 n2 
1
 1.097  10 7 m 1

1
 22.27 m 1



1

1
 99
2

1 

100 2 
 0.0449 m
22.27 m 1
DeBroglie Waves
9. Electrons are accelerated by a 1000 V potential drop. Calculate the de Broglie
wavelength. Also calculate the wavelength of the X-rays that would be produced when these
electrons strike a solid. [Hint:The electron charge is 1.602  1019 C; the mass of an electron, me,
is 9.109  1031 kg; Planck's constant, h,is 6.626  1034 J s.]
E electron  1.602  10 19 C 1000 V   1.602  10 16 J

E electron 

1
1 p2
mv 2 
2
2 m
p  2mE
h

p


h

2mE
6.626  10 34 J s



2 9.109 10 31 kg 1.602 10 19 C 1000 V 
  3.88 10
E photon 
 photon 
11
m
hc
 photon
hc
E photon

6.626 10
34

J s 3.00 10 8 m s 1
1.602 10
16

J
9
 photon  1.24 10 m
Well-Behaved Wave Functions
10. List the characteristics of a well-behaved wave function.
(a) The function must be single-valued; i.e. at any point in space, the function  2 must
have only one numerical value.
3
CHEM 342. Spring 2002. Problem Set #1. Answers.
(b) The function must be finite and continuous at all points in space. The first and second
derivatives of the function must be finite and continuous.
(c) The function  2 must have a finite integral over all space.
all space 
*
d  a finite number
11. Which of the following functions are well-behaved? For those functions that are not
well-behaved, explain why not.
(a) f(x) = x for x  0 , and f(x) = 0 for x  0
(b) f x   x 2
(c) f x   e
x
(d) f  x   e  x
(e) f x   cos x
(f) f x   sin x
(g) f  x   1  x 2 for 1  x  1, and f(x) = 0 for x  1 and for x  1
(a) Not well-behaved because f(x) becomes infinite as x   , and the first derivative at
x = 0 is not continuous
(b) Not well-behaved because f(x) becomes infinite as x  
(c) Not well-behaved because the first derivative is not continuous at x = 0
(d) Not well-behaved because f(x) becomes infinite as x  
(e) Well-behaved
(f) Not well-behaved because the first derivative is not continuous at x = 0
(g) Not well-behaved because the first derivative is discontinuous at x  1
12. A particle can move only along the x-axis and has a wave function   e ikx . Give
the expression for
(a)  * , the complex conjugate of  (express answer in terms of k and x)
(b) P, the probability density (express answer in terms of  and  * )
(c) the probability that the particle is between x1 and x2 (in terms of  and  * ). Hint:
You only need to set up the appropriate integral.
For any normalized  , what is the probability that the particle is between   and   ?
[Hint: this question has a numerical answer.]
(a)  *  e ikx
(b) P   *  or P   2
(c)

x2
 * dx
x1
4
CHEM 342. Spring 2002. Problem Set #1. Answers.
For any normalized  , the probability that the particle is between   and   is

  dx  1.
*

Harmonic Oscillators: Hermite Polynomials
13. Calculate the zero-point energy of a harmonic oscillator consisting of a particle of
mass 2.33  1026 kg and force constant 155 N/m. [Hint: Planck's constant, h,is 6.626  1034 J
s.]
1 h k

E  v  
2  2 m

At the zero-point energy, v = 0. Therefore, we obtain the following:

34
 1  6.626 10 J s
E  
2
2

155 N m 1
2.33 10  26 kg
E  4.30 10  21 J
14. For the harmonic oscillator, substitute the wave function for the ground state into the
 2 d 2 1 2
Schrodinger equation Hˆ   
 kz   E and derive the expression for the ground
2m dz 2 2
state energy (also known as the zero-point energy). Hint: The wave function for the ground state
1/ 4
2
km
a
is  0    e  az / 2 where a 
. The expression for the ground state energy is


1
E   hν where ν is the frequency.
2
5
CHEM 342. Spring 2002. Problem Set #1. Answers.
 2 d 2 1 2
Hˆ   
 kz   E
2m dz 2 2
1/ 4
 1 2  a 1 / 4  az2 / 2
 2 d 2  a 
 az 2 / 2
ˆ
H  
  e
  kz   e
2m dz 2   

 2
2  a 
Hˆ   
 
2m   
1/ 4
2  a 
Hˆ  
 
2m   
1/ 4
2  a 
Hˆ  
 
2m   
1/ 4



2
d
 az e az / 2  1 kz 2  a 
dz
2



2
a  d z e az / 2  1 kz 2  a 
dz
2



a  az 2 e  az
2
/2
 e  az
2
/2
1/ 4
1/ 4
e  az
e  az
2
/2
/2
 12 kz  a 
2
2
1/ 4
e  az
2
/2
 
1/ 4
 2 a
2
1 2  a 
2
ˆ
H  
 az  1  kz   e  az / 2
2
 2m
  
 2a2 z 2 2a 1 2 
Hˆ   

 kz  
2m
2m 2


a
km

  2 z 2 km  2 km 1 2 
Hˆ   

 kz  
2
2m 
2
 2m 

 kz 2  km 1 2 
Hˆ   

 kz  
2m
2
 2

  km 
Hˆ   

2
m


 h 1 k 
Hˆ     
   E
2

2
m




E
h 1 k
 
2  2  m
1
1 h k
This can be rewritten in the more familiar form E   
or E   hv .
2
 2  2 m
15. In the vibrational motion of HI, the iodine atom essentially remains stationary
because of its large mass. Assuming that the hydrogen atom undergoes harmonic motion and that
the force constant k is 317 N/m, what is the fundamental vibration frequency  0 . [Hint: The
mass of a proton is 1.67  1027 kg.]
6
CHEM 342. Spring 2002. Problem Set #1. Answers.
1 k
1
317 N m 1


 6.93 1013 s 1
 27
2 m 2 1.67 10 kg
7