Name _________________
Solutions to Test 3
November 14, 2012
This test consists of three parts. Please note that in parts II and III, you can skip one
question of those offered. Some possibly useful formulas can be found below.
h  6.626  1034 J  s  4.136 1015 eV  s
  1.055 1034 J  s  6.582 1016 eV  s
1D square well:
 2 2 n2
En 
2mL2
2
  nx 
sin 
 n  x 

L
 L 
n  1, 2,3,
Harmonic Osc.
En    n  12 
n  0,1, 2,
Barrier penetration:
E
E
T  16 1   e 2 L ,
V0  V0 

2m V0  E 

Hydrogen
13.6 eV  Z 2
En  
n2
Reflection off a step:
 E  E  V
0


R   E  E  V0


1

2

 if E  V0


if E  V0
Part I: Multiple Choice [20
points]
For each question, choose the best answer (2 points each)
1. Suppose an electron has n = 4, l = 2, and m = 0. Such an electron would be a ___
electron.
A) 4p
B) 4s
C) 4d
D) 2s
E) 2d
2. Suppose a particle is in a spherically symmetric potential. What is the total angular
momentum squared L2 if l = 4?
B) 12 2
C) 16 2
D) 17 2
E) 20 2
A) 4
3. The operator whose expectation value is the average of the total energy is called the
A) Schrodinger B) Lagrangian C) Hamiltonian D) Momentum E) Potential
4. Suppose after a lot of work, I found a wave function   x  which satisfies
Schrodinger’s equation, but the normalization is wrong, so that    x  dx  4 .
2
What should I multiply the wave function by to fix this problem?
D) 12
E)
A) 2
B) 4
C) 16
1
16
5. What distinguishes a bound state from an unbound state?
A) Bound states have less energy than the potential at infinity
B) Bound states have more energy than the potential at infinity
C) Bound states have less energy than any barriers they might encounter
D) Bound states have more energy than any barriers they might encounter
E) Bound states are wave functions that cannot be normalized
6. In a complicated atom like Uranium, with 92 electrons, the reason you only get two
electrons in the 1s state is because
A) The Pauli Exclusion Principle says you can only have one particle per state,
and there are only two states (spin up and spin down) in the 1s orbital
B) The Pauli Exclusion Principle says you can only have two particles per state
C) Electrical repulsion makes it unfavorable for more than two electrons to be in this
state
D) Putting multiple electrons into this small space means they have enormous
momentum, according to the uncertainty principle
E) The US constitution states that just as you can only have two Senators from each
state, you can also only put two electrons in each quantum state. It wasn’t true
before 1791
7. If I send photons one at a time through a half-silvered mirror, and then put ideal
perfect detectors to see which way the photon went, what will the detectors see?
A) Each detector will register half a photon each time
B) Half the time one detector will see it, half the time the other will see it, but
never both and never neither, and the results will be random
C) Photons will alternate, one going one way and the next always going the other
way
D) Each detector sees the photon 50% of the time, so 25% of the time they will both
see it and 25% of the time neither sees it.
E) No photons will go either way, since quantum mechanical effects only happen if
you do many photons at once
8. Which of the quantum numbers l, m, and ms for an electron in a hydrogen atom can
sometimes be negative?
A) All of them can be negative
B) l and ms can be negative, but not m
C) l and m can be negative, but not ms
D) m and ms can be negative, but not l
E) None of them can be negative
9. When we calculate the expectation value of the position x , what does it tell us?
A)
B)
C)
D)
E)
Where a particle will be found
The average value of the position if you perform many measurements
The most likely place to find the particle
The uncertainty in the position
The root-mean-square value of the position of the particle
10. The radial wave function R(r) for an electron in hydrogen depends on which quantum
numbers?
A) n only
B) l only
C) m only
D) n and m
E) n and l
Part II: Short answer [20 points]
Choose two of the following three
questions and give a short answer (2-4
sentences) (10 points each).
11. When your professor was in high
school, his chemistry teacher taught
him that Schrodinger’s equation was
E  H , and he wondered why
you couldn’t just simplify this to
E  H . Explain why this doesn’t
make any sense.
In the equation E  H , the H is
an operator, not a number. In particular it
contains derivatives, coming from the term
2
pop
2m . Hence you can’t cancel it, since
it needs a wave function to act on to mean
anything.
12. A scanning tunneling microscope, to function, requires that electrons cross a gap
between the tip of the microscope and some object that they are studying. How
can the electrons cross that gap?
Although, classically, particles can only cross a boundary if they have enough
energy, quantum mechanically they can “tunnel” through the barrier, though the
probability is low. But for a small enough barrier (microscopic) and a light enough
particle (the electron), the resulting current can be measurable.
13. What is the electronic configuration for Cobalt,
with Z = 27 electrons? Your answer should look
like 1s2 2s2 …
You draw the little chart in the usual way, and
count out electrons. We have
Co: 1s2 2s2 2p6 3s2 3p6 4s2 3d7 .
1s2
2s2
3s2
4s2
5s2
6s2
7s2
2p6
3p6
4p6
5p6
6p6
7p6
3d10
4d10
5d10
6d10
7d10
4f14
5f14
6f14
7f14
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each).
14. A group of quantum mechanical congressmen with kinetic energy E = 400. J are
approaching a fiscal cliff with potential V0 = –9,600 J.
(a) Find the probability that a congressman is reflected from the fiscal cliff.
We use the formula for reflection, which gives
 E  E  V0
R
 E  E V
0

 94  44.4% .
2
  400  400  9, 600 J   20  100  2  80 2  2 2
  
   20  100    120    3 

400
400
9,
600
J





2
All the units cancelled out, so we don’t have to actually keep track of them. We note that
55.6% of congressmen would end up jumping off a cliff, which might or might not be
good for our country.
(b) The potential V0 is now raised until it takes on a positive value, but less than
E. It is nonetheless found that the same fraction of congressmen are
reflected from the fiscal step. Find the new potential V0.
The probability is still the same, so we have
 E  E  V0
4
 R
 E  E V
9
0


2

3
E  E  V0
E  E  V0
2

 ,


.
Since the potential is now positive, we can be sure that
E  E  V0 , so this quantity
must be positive. Hence we have
E  E  V0
2

3
E  E  V0
,
2 E  2 E  V0  3 E  3 E  V0 ,
5 E  V0  E ,
25E  25V0  E ,
24 E  25V0 ,
V0 
24
25
E
24
25
 400 J   384 J .
15. A proton of mass m  1.6726 1027 kg is trapped in a one-dimensional infinite
square well of unknown size L . A physicist studying this discovers that he can
make the electron jump from the ground state n = 1 to the state n = 4 when he
bathes the system with microwaves with
frequency f  6.13 1010 Hz  6.13 1010 s 1 .
(a) What is the energy of one photon with this frequency?
Since the problem is all in SI units, we might as well keep it that way.
Remembering that E = hf, we have
E  hf   6.13  1010 s 1  6.626 1034 J  s   4.062 1023 J .
(b) What is the size L of the infinite square well?
The energy for the infinite square well is given by En   2  2 n 2  2mL2  . The
difference between n = 1 and n = 4 is then
E  E4  E1 
 2  2 42
2mL2

 2  212
2mL2

15 2  2
.
2mL2
Solving for the length and then substituting numbers, we have
2mL2 E  15 2  2 ,
2
15 2 1.055  1034 J  s 
15 2  2
17 J  s
L 


1.213

10
2mE 2 1.6726  1027 kg  4.0617  1023 J 
kg
2
2
 1.213 1017 m 2 ,
L  3.48  109 m  3.48 nm .
(c) The atom then transitions from n = 4 to n = 3. What is the energy of the
resulting photon that is emitted?
In a manner very similar to before, we compute
7 2 1.055  1034 J  s 
7 2  2
E  E4  E3 



2mL2
2mL2
2mL2 2 1.6726 1027 kg 1.213  1017 m 2 
 2  2 42
 1.896  1023 J
 2  2 32
2
16. A particle has wave function given by
 30  ax  x 2  a 5/ 2
  x  
0

 a
for 0  x  a ,
otherwise .
This wave function has already been
properly normalized. It has
expectation values x  12 a and
x a
x 2  72 a 2 .
(a) Find the expectation values p and
p 2 for this wave function.
Since the wave function is real, we automatically know that p  0 . To find
p 2 , we use
p
2
2
a
 d 
2
* d 
dx
   p  dx    
  dx    0 

0
dx
 i dx 
2
30 2 a
60 2 a
60 2 1 2 1 3 a
2 d
2
2
  5   ax  x   ax  x  dx  5   ax  x  dx  5  2 ax  3 x 
0
0
0
a
dx
a
a
60 2
10 2
 5  12 a 3  13 a 3   2 .
a
a

*
a
2
op
2
*
(b) What are the uncertainties x and p for this particle?
In general, the uncertainty is given by  
x 
x2  x
p 
p2  p
2
2


2
7
2  
a 2   12 a   a
2
2
7
2
, so we have
 14  a
1
28
,
10 2
  10 a .
a2
(c) Does this wave function satisfy the uncertainty principle?
We then have
 x  p   a
1
28


10 a  
10
28

5
14
 0.5976 .
The uncertainty principle says this must satisfy  x  p   12  , which it does.
17. The wave function for the lowest energy (ground state) of the harmonic
oscillator is
 m x 2 
m
 0  x 
exp  


2 

4
(a) What is the probability density of the particle being at an arbitrary position
x?
The probability density is the wave function times its complex conjugate, which
works out to
 0  x
2
 m x 2 
m
exp  

.

 

(b) What is the probability density (in m-1 or nm-1) of an electron with mass
m  1.6726  1027 kg being at the origin x = 0 if it is in a harmonic oscillator
potential with angular frequency   1.50 1013 s 1 ?
We now just substitute everything in, which is pretty straightforward:
 0  0 
2
m


1.6726 10 kg 1.50 10
 1.055 10 J  s 
27
13
34
s 1 
 7.57 1019 m 2
 8.70 109 m 1  8.70 nm 1 .
(c) If the wave function of a particle at t = 0 is given by   x, t  0    0  x  , what
is   x, t  at other times?
In general, if we have a solution to the time-independent equation with energy E,
we simply multiply by e iEt  to get the solution at a general time. The energy for this
state is just E    0  12   12  . Therefore, the wave function at arbitrary time is just
  x, t    0  x  eiEt    0  x  e it 2 
4
 m x 2  it 2
m
exp  
.
e
2 

