General Physics II, Section 03
Homework Assignment 10 Solutions (26 points)
Paul Shand
Questions
Q28.1. Reason: As the crystal is compressed, the spacing d between planes of atoms
decreases. The formula giving the Bragg condition is Equation 28.1: 2d cosθm = mλ. As d
decreases, cosθm must increase. But cosθ increases as θ decreases. So θ1 decreases. (2 points)
Q28.5. Reason: (a) A positive anode attracts all of the electrons to the anode. Once all of the
electrons reach the anode, a further increase in ΔV does not cause any further increase in the
current. (1)
(b) If an electron has enough kinetic energy when it leaves the cathode it might just reach the
anode. As ΔV decreases, fewer electrons are able to reach the anode until ΔV = −Vstop when no
electrons reach the anode and the current stops. (1)
(c) Electrons are still emitted, but they are not attracted to the anode and are in fact repelled by it.
The electrons that are emitted do not have sufficient kinetic energy to reach the anode. (1) (3
points)
Q28.9. Reason: (a) If the current is zero when ΔV = 1.0 V that means the frequency of the
light is less than the threshold frequency; f < f0. So no electrons are ejected in the first place.
Doubling the intensity of the light means doubling the number of photons hitting the metal, but
the photons still aren’t energetic enough to eject electrons; that is, the energy of the photons is
less than the work function.
E = hf =
⎞
1 eV
hc (6.63 × 10−34 J ⋅s)(3.00 × 108 m/s) ⎛
=
⎜
⎟ = 4.97 eV < E0
−19
250 nm
λ
J⎠
⎝ 1.60 × 10
So no matter how many photons of this energy one shines on gold there won’t be a
photocurrent. (2)
(b) Since no electrons are emitted in the first place, then increasing ΔV won’t make any
difference either. See characteristic 5 in Section 28.2. (1) (3 points)
Q28.21. Reason: The allowed energies of a particle in a box are given in
Equation 28.14: En = (hn)2 /8mL2 .
(a) From the equation we see that for the energies to be large L, the length of the box should be
very small. (1)
(b) Based on the answer to part (a), the system with the smallest dimension would be likely to
have the largest energies; that is, the proton in the nucleus. (1) (2 points)
[Assess: While it is true that the mass of the proton is larger than the mass of the electron and
m is also in the denominator (which would argue the other way), the size difference is much
greater than the mass difference, and the L is squared.]
1
Q28.25. Reason: (a) It represents a changing speed because the wavelength is changing. The
longer wavelength corresponds to a slower speed. (1)
(b) The shorter wavelength on the left means the particle is moving faster on the left; the longer
wavelength on the right means the particle is moving slower there. (1) (2 points)
[Assess: This question reinforces the idea that faster particles have a shorter wavelength.]
Problems
P28.8. Prepare: Equation 28.6 connects the maximum kinetic energy of an ejected electron
with the frequency of the incident radiation and the work function of the metal. The work
function for copper from Table 28.1 is 4.65 eV. Also note that we must convert eV to J using the
conversion: 1 eV = 1.6 × 10−19 J.
Solve: From Equation 28.6, the maximum kinetic energy is
(6.63 × 10−34 J s)(3.0 × 108 m/s)
1 eV
c
hc
K max = hf − E0 = h − E0 ⇒ λ =
=
×
= 216 nm.
E0 + K max
4.65 eV + 1.10 eV
λ
1.6 × 10−19 J
(2 points)
[Assess: λ = 216 nm is the wavelength of light in the ultraviolet region of the spectrum.]
P28.10. Prepare: A metal can be identified by its work function. From Equation 28.7, the
stopping potential is
hf − E0
Vstop =
⇒ E0 = hf – eVstop
e
Solve:
f=
(a) The frequency and energy of the photons are
c 3.00 × 108 m/s
=
= 1.500 × 1015 Hz
λ 200 × 10−9 m
⎛
⎞
1 eV
−19 ⎟ = 6.21 eV
J⎠
⎝ 1.6 × 10
⇒ hf = [(6.63 × 10−34 J s)(1.500 × 1015 Hz)] × ⎜
If the stopping potential is Vstop = 1.93 V, then eVstop = 1.93 eV. Thus, E0 = hf − eVstop = 6.21 eV − 1.93 eV = 4.28 eV .
which we will report as 4.3 eV.
Using Table 22.1, we can identify the metal as aluminum. (2)
(b) The kinetic energy of the electrons and thus the stopping potential are independent of the light
intensity. A more intense light generates more electrons, but the electrons still have the same
kinetic energy. The stopping potential is Vstop = 1.93 V after the intensity is doubled. (1) (3 points)
P28.27. Prepare: The kinetic energy of a particle of mass m is related to its momentum by
E = p 2 /2m. The momentum of the particle is related to its wavelength through λ = h/p.
Solve: Using λ = h/p, for an electron we have
E=
p2
(6.63 × 10−34 J s)2
h2
=
=
= 2.4 × 10−19 J
2m 2mλ 2 2(9.11× 10−31 kg)(1.0 × 10−9 m)2
2
⎛
⎞
⎝
⎠
1eV
In eV, the energy is (2.41× 10 –19 J) ⎜ 1.6 ×10
–19 J ⎟ = 1.5 eV . (2 points)
Important Note: In this problem you need to use h = 6.63 × 10−34 J ⋅ s (SI unit), not h =
4.14 × 10−15 eV⋅ s . This is because the equation involves the mass m, would need to be converted
to its equivalent in eV-based units. Since we have not worked with those mass units in class, it is
always safest to use SI units for all quantities in your calculations and convert to eV afterwards if
necessary.
P28.39. Prepare: For an emitted photon to have a wavelength of λ the energy must be
E = hc/λ .
We are given λ = 420 nm.
Solve:
E=
⎞
1 eV
hc ⎡ (6.63 × 10−34 J ⋅s)(3.0 × 108 m/s) ⎤ ⎛
=⎢
⎥×⎜
⎟ = 3.0 eV
−9
−19
420 × 10 nm
λ ⎢⎣
J⎠
⎥⎦ ⎝ 1.6 × 10
The n = 4 energy level therefore needs to be 3.0 eV above one of the existing energy levels, but
it must also be above the n = 3 level. It can’t be 3.0 eV above the n = 1 ground state because then
n = 4 would be lower in energy than n = 3.
But n = 4 could be 3.0 eV above the n = 2 state; that is, it could have an energy of
1.5 eV + 3.0 eV = 4.5 eV.
Or it could be 3.0 eV above the n = 3 state; that is, it could have an energy of
4.0 eV + 3.0 eV = 7.0 eV.
In summary, the two answers are 4.5 eV and 7.0 eV. (3 points)
[Assess: Given the fact that the spacing between levels is growing with increasing n in the
diagram, we suspect that the n = 4 level would more likely have an energy of 7.0 eV. ]
P28.73. Prepare: A particle confined in a one-dimensional box has discrete energy levels
given by Equation 28.14.
Solve: From Equation 28.14,
h2 2
h2
En =
En +1 =
(n + 1)2
2 n
8mL
8mL2
Taking the ratio,
En +1 ⎛ n + 1⎞ 2
=⎜
En
⎝ n ⎟⎠
Thus,
n +1
4.5 eV
3
=
= 1.5 = ,
n
2.0 eV
2
Which gives n = 2. Thus, the two levels are n = 2 and n = 3. Now,
E2 = (2.0 eV) × (1.6 × 10−19 J/eV) = 3.2 × 10−19 J.
Using n = 2 and the corresponding energy in the equation for the energy gives
3
So, 3.2 × 10−19 J =
(6.63 × 10−34 J s)2
(2)2 ⇒ L = 8.7 × 10 −10 m = 0.87 nm (4 points)
8(9.11× 10−31 kg)L2
4