1. (10 points) SHORT ANSWER
(a) (5 points) Light incident on aluminum causes electrons to be ejected. How can
an experimentalist increase the kinetic energy of the electrons ejected from the
aluminum?
Answer: Einstein’s photoelectric effect says that the kinetic energy increases
with the frequency of the light. The kinetic energy K of an ejected electron
goes as
K = hf − φ
(1)
where φ is the work function of the metal. Note that if the amount or intensity
of light incident on the surface of a metal is increased, the kinetic energy of the
ejected electrons does not change, but more electrons are ejected because there
are more photons to eject more electrons. Classical physics would predict an
increase in the kinetic energy since a greater intensity means a higher electric
field and the energy density goes as the square of the electric field. The kinetic
energy increases with the frequency of the light, not with the intensity.
(b) (5 points) Consider the time-independent Schroedinger equation with a potential energy U (~r) that depends on position ~r. Can the energy eigenvalues
depend on position?
Answer: No, the energy eigenvalues E are independent of position.
1
2. (10 points) SHORT ANSWER
(a) (5 points) Can a wave function ψ(x), that is a solution of the one dimensional
Schroedinger equation, be infinitely large? In other words, is it possible to
have ψ(x) = ∞ for some values of x? If so, give an example. If not, explain
why.
Answer: No, a wave function ψ(x) that is a solution of the Schroedinger
equation cannot be infinite at any value of x because it has to satify the
normalization condition:
Z ∞
|ψ(x)|2 dx = 1
(2)
−∞
(b) (5 points) The wavelength of the maximum intensity of the sun’s radiation
is observed to be approximately 500 nm. Assume the sun to be a blackbody
and calculate the sun’s surface temperature. Give your answer in Kelvin to 2
significant figures.
Solution: Wien’s displacement law gives
λmax T = 2.898 × 10−3 m K
2.898 × 10−3 m K
Tsun =
λmax
2.898 × 10−3 m K
=
500 × 10−9 m
= 5796 K
T = 5796 K ≈ 5800 K
2
3. (10 points) SHORT ANSWER
(a) (5 points) What is quantum mechanical tunneling? Give an example.
Answer: In quantum mechanical tunneling a wave function representing a
particle with energy E can penetrate a barrier with potential energy U (x) > E.
An example is scanning tunneling microscopy (STM) in which electrons from
a tip jump across a vacuum gap to an metallic surface. Another example is an
electron jumping across a gap from one wire to another wire. Planar tunneling
of electrons between two metals separated by a thin insulating layer is another
example.
(b) (5 points) An oboe player tunes the orchestra with a note that has a frequency
of 440 Hz. If the note is played for 2.5 seconds, what minimum range of
frequencies is heard during this time?
Answer: From the uncertainty principle for energy and time:
∆t∆E ≥
E =
∆t∆ω ≥
∆ω ≥
2π∆f =
∆f ≥
∆f ≥
h̄
2
h̄ω
1
2
1
2∆t
∆ω
1
4π∆t
1
= 0.032 sec−1
4π(2.5 sec)
frequency range: 439.97 Hz to 440.03 Hz
3
4. (10 points) SHORT ANSWER
(a) (5 points) An electron in a simple harmonic oscillator potential makes a transition from the second excited state to the ground state and emits a photon
with an energy of 4.4 eV. What is the zero point energy E0 of the electron?
Solution: The energy levels of the harmonic oscillator are given by
1
h̄ω
(3)
En = n +
2
where n = 0, 1, 2, 3... and ω is the oscillator frequency. So the energy of the
photon equals
∆E = E2 − E0
5 1
=
h̄ω
−
2 2
= 2h̄ω
= 4.4 eV
h̄ω = 2.2 eV
The zero point energy is the ground state energy:
1
h̄ω
Eo =
2
1
=
(2.2 eV)
2
= 1.1 eV
E0 = 1.1 eV
(b) (5 points) A single crystal of table salt (NaCl) is irradiated with a beam of
x-rays of wavelength 0.250 nm, and the first Bragg reflection is observed at
an angle of 26.3o . What is the spacing between neighboring atoms in NaCl?
Give your answer in nm to 3 significant figures.
Solution: Use Bragg’s law for X-ray diffraction with n = 1, λ = 0.250 nm,
and θ = 26.3o .
2d sin θ = nλ
nλ
d =
2 sin θ
0.250 nm
=
2 sin(26.3o )
= 0.282 nm
spacing = 0.282 nm
4
5. (10 points)
In frame S, clock 1 is located at x = 0 and clock 2 is located at x = L. A flash
lamp is located at x = L/2 and releases a flash of light at time t = 0. The two
clocks start to run when they receive the flash of light from the flash lamp. An
observer O in reference frame S notes that the clocks start at precisely the same
time t1 = t2 = L/(2c) where t1 is the time when clock 1 receives the flash of light,
t2 is the time when clock 2 receives the flash of light, and c is the speed of light.
Now consider an observer O′ in reference frame S ′ that is moving in the x direction
with velocity v. Will observer O′ see the clocks start at the same time? If not,
what is the difference in time ∆t′ = t′1 − t′2 that O′ observes? According to O′ , t′1
is the time when clock 1 receives the flash of light, and t′2 is the time when clock 2
receives the flash of light.
Solution: To find t′1 and t′2 , we use the Lorentz transformation.
t′ = γ t −
vx
c2
(4)
q
√
where γ = 1/ 1 − β 2 = 1/ 1 − v 2 /c2 . Note that in the S frame, x1 = 0 and
t1 = L/(2c), while x2 = L and t2 = L/(2c). So
t′1 = γ t1 −
= γ
L
2c
vx1
c2
vx2
c2
vL
L
− 2
γ
2c
c
t′1 − t′2
L
vL
L
γ
−
− 2
2c
2c
c
vL
γ
c2
vL/c2
t′2 = γ t2 −
=
∆t′ =
=
=
= q
1 − v 2 /c2
5
2
∆t′ = γvL/c2 = √vL/c2
1−v /c2
6
6. (10 points) Find the kinetic energy K of an electron whose de Broglie wavelength
is the same as that of a 2 eV photon.
Solution:
Strategy: Find the wavelength λ of the photon using
E = hf =
λ =
hc
λ
hc
E
Then find the momentum p of the electron using
p=
h
λ
(5)
Plug this into the (classical) kinetic energy expression:
K =
=
=
=
=
=
=
=
=
p2
2m
h2
c2
×
2mλ2 c2
h 2 c2
2mc2 λ2
h 2 c2
2mc2 (hc/E)2
E2
2mc2
(2 eV)2
2(0.511 × 106 ; eV)
2 eV
0.511 × 106 ; eV
3.91 × 10−6 ; eV
3.91 µeV
Note that K ≪ mc2 , i.e., the kinetic energy is much less than the rest energy of
the electron. This means that using the classical expression for the kinetic energy
is valid.
K = 3.91 × 10−6 eV= 3.91 µeV
7
7. (10 points) The negative pion, π − , is a subatomic particle with the same charge as
the electron but with mass mπ = 273me where me is the mass of an electron. A
π − can be captured into Bohr orbits around an atomic nucleus with a radius given
by the Bohr formula, except that me must be replaced by mπ .
(a) (5 points) Derive an expression for the speed v of a pion in the nth Bohr orbit.
Which orbit has the highest speed? Assume the negative pion is orbiting
around a nucleus with atomic number Z.
Solution: The quantization of the angular momenta is given by
L = mvr = nh̄
(6)
The radius r of an orbit is given by
r = n2 aBπ /Z
(7)
The formula for the Bohr radius aBπ with a π − is
h̄2
aBπ = 2
ke mπ
(8)
So
n2 h̄2
n2 aBπ
=
Z
Zke2 mπ
So the quantization condition becomes
r=
mπ vr = nh̄
n2 h̄2
mπ v
= nh̄
Zke2 mπ
vn2 h̄2
= nh̄
Zke2
Znh̄ke2
v =
n2 h̄2
Zke2
=
nh̄
Since the speed v ∼ 1/n, the highest speed is for n = 1.
v = Zke2 /nh̄
n(max speed)=1
8
(9)
(b) (5 points) Calculate the wavelength λ of the photon emitted (in nm) when the
π − makes a transition from the n =
2 state into the n = 1 state. Assume the
negative pion is orbiting around a 12 C nucleus.
Solution: Let aB be the Bohr radius for a hydrogen atom with an electron.
The formula for the Bohr radius aBπ with a π − is
aBπ =
h̄2
me
=
a
B
ke2 mπ
mπ
(10)
So the Rydberg energy for the pion orbiting around a proton (Z = 1) is
ER,π =
ke2 mπ
= (13.6 eV)(273) = 3.71 × 103 eV
2aB me
(11)
For Z = 6, the energy of the emitted photon is
2
Eγ = Z ER
1
1− 2
n
2
= Z ER,π
1
3
1−
= 36(3.71×103 eV)
= 1×105 eV
4
4
(12)
hc
λ
1240 eV
hc
= 1.24 × 10−2 nm
=
λ =
Eγ
105 eV
Eγ =
λ = 1.24 × 10−2 nm
9
8. (10 points) A particle of mass m is put into a rigid two-dimensional rectangular
box of length a along the x-axis and length b along the y-axis, where a > b. The
walls of the box are infinitely high. The wave function ψ(x, y) is shown in the figure
(ignore the actual values of a and b). The dark regions are dips where ψ(x, y) < 0.
(All questions below pertain to this wave function.)
(a) (2 points) How many nodal lines are there?
Solution: There are 3 nodal lines: one parallel to the x axis and two parallel
to the y-axis.
Number of nodal lines = 3
(b) (6 points) What are the wavelengths λx , λy , wave vectors kx and ky , quantum
numbers nx and ny , and the energy E, of the particle in the box? Some of
these quantities will depend on a and b. (This problem is continued on the
next page.)
10
Solutions:
Since the number of nodes is n − 1 and we have 2 nodes along the x-axis,
nx = 3. Similarly, ny = 2. The wave functions are of the form
ψ(x, y) = A sin(kx x) sin(ky y)
(13)
Requiring the wave function to vanish on the boundaries of the box means
that
sin(kx a) = 0
sin(ky b) = 0
kx a = nx π
ky b = ny π
nx π
3π
ny π
2π
kx =
=
ky =
=
a
a
b
b
2a
2π
2πb
2πa
2πb
2π
=
λy =
=
=b
=
=
λx =
kx
nx π
3
ky
ny π
2π
h̄2 k 2
E =
2m
h̄2 2
kx + ky2
=
2m
!
h̄2 π 2 n2x n2y
=
+ 2
2m a2
b
h̄2 π 2
=
2m
9
4
+ 2
2
a
b
λx = 2a/3
λy = b
kx = 3π/a
ky = 2π/b
nx = 3
ny = 2
2 2
π
E = h̄2m
9
+ b42
a2
11
(c) (2 points) Is this state degenerate with any other state? If so, give the quantum
numbers of that other state or states. If not, what can be done to the box,
i.e., the potential, to make the state degenerate?
Answer: The state is not degenerate because a 6= b. In order to have degeneracy, we would have to make a square box, i.e., a = b.
12
U=Uo
L
U=Uo
U=0
9. (10 points) Consider an electron in the presence of the one dimensional potential
shown. The electron has energy E between 0 and Uo : 0 < E < Uo .
(a) (2 points) Can E have any value between 0 and Uo for this potential? Explain.
No. The boundary conditions dictate that the wave function decays exponentially outside the box. These boundary conditions lead to quantization of the
No.
energy.
(b) (2 points) Sketch the wavefunction of the electron for the second excited state
in the potential well, assuming that the second excited state exists in the well.
13
(c) (6 points) If L = 6 nm and Uo = 5 eV, how many bound states can exist
inside the well? Assume that the energies of the finite square well states are
the same as those of the infinite square well with Uo = ∞.
Solution: The energy levels for the infinite square well are given by
2 2
2 π h̄
E=n
2mL2
where n = 1, 2, 3, .... We want E ≤ Uo . So find n such that
π 2 h̄2
= Uo
2mL2
2mc2 L2 Uo
8mc2 L2 Uo
2mL2 Uo c2
=
=
n2 =
π 2 (h/(2π))2 c2
h 2 c2
π 2 h̄2 c2
8(0.511 × 106 eV)(6 nm)2 (5 eV)
=
(1240 eV − nm)2
= 479
√
479 = 21.9
n ≤
nmax = 21
n2
nmax = 21
14
(14)
10. (10 points) Consider an electron approaching the one dimensional potential barrier
(shown below) from the right. The electron has energy E between 0 and Uo :
0 < E < Uo .
L
U=Uo
U=0
U=0
(a) (4 points) Sketch the wavefunction of the electron for all regions of space.
(b) (3 points) Can E have any value between 0 and Uo for this potential? Explain.
Yes. Where U = 0 there are free particle states with exponential decay in beYes.
tween.
15
(c) (3 points) What is the ratio of the wavefunction amplitudes at the left and
right edges of the barrier? In other words, what is |ψR /ψL |? Define your
variables.
Solution:
ψR
∼ e+αL
(15)
ψL
where
α=
q
2m(Uo − E)
h̄
(16)
where m is the mass of the electron, Uo is the barrier height, E is the electron
energy, and h̄ is Planck’s constant divided by 2π.
ψR
ψL
= e+αL
16