Problem Set 5
Solutions
Problem A
a.
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When x or y is a, z must be 0, so sin(kxa) = 0 and sin(kya) = 0.
Thus kxa = nx, where nx = 1, 2, 3, …, so kx = nx/a.
Similarly, ky = ny/a, where ny = 1, 2, 3, ….
b.
ky
2/a
/a
/a
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2/a
kx
The dot in the lower left represents k = [(/a)2 + (/a)2]1/2 = 21/2/a, in which case
 = v/ = vk/(2) = 2-1/2v/a.
The dot in the lower left represents k = [(2/a)2 + (2/a)2]1/2 = 21/22/a, in which
case  = 21/2v/a.
The other two dots represents k = [(2/a)2 + (/a)2]1/2 = 51/2/a, in which case  =
51/2v/(2a).
c.
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There is one state in every (/a)2 of area.
A ring of thickness dk has area 2kdk, but allowed values of k lie only the first
quadrant, so we want a quarter of this area: (/2)kdk.
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Thus the density of states is (states per area)(area of quarter-ring) =
1/(/a)2(/2)kdk = a2/(2)kdk = g(k)dk.
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Since  = vk/(2), k = 2/v, and dk = (2/v)d.
Substitute these terms into g(k)dk above to find g()d =
[a2/(2)](2/v)(2/v)d = (2a2/v2)d.
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k = 2/ and dk = -(2/2)d, but we'll drop the negative. When computing
density of states, we're interested in absolute values.
Substitute these terms into g(k)dk to find g()d = [a2/(2)](2/)(2/2)d =
(2a2/3)d.
d.
e.
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3-8.
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We want to determine the smallest, indivisible unit of charge.
Therefore, we need to determine the greatest common factor for all the
measured charges.
There seems to be a small difference in charge between the two lowest charges:
14.29  10-19 C – 12.70  10-19 C = 1.59  10-19 C.
Let's see if all the measured charges are approximately integral multiples of this
amount, 1.59  10-19 C. (Why "approximately"? Experimental error.)
25.41  10-19 C / 1.59  10- 19 C = 15.98
20.64  10-19 C / 1.59  10- 19 C = 12.98
17.47  10-19 C / 1.59  10- 19 C = 10.99
19.06  10-19 C / 1.59  10- 19 C = 11.99
12.70  10-19 C / 1.59  10- 19 C = 7.99
14.29  10-19 C / 1.59  10- 19 C = 8.99

All of the measured charges are well within 1% of an integral multiple of
1.59  10-19 C, so this seems like the implied value for the indivisible unit of
charge.
3-9.
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Let's look at the free-body diagrams for rising and free-falling:
qE
6ηrvdown
vup
mg
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6ηrvup
vdown
mg
The free-body diagram for rising gives qE = mg + 6ηrvup.
The free-body diagram for free-falling gives 6ηrvdown = mg.
We want to determine E so that vdown = vup. In this case, the free-falling equation
becomes 6ηrvup = mg.
Then the first equation becomes qE = 2mg, or E = 2mg/q.
3-10.

We must first construct the free-body diagram:
6ηrvt'
qE
mg 
vt'
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The horizontal component of the air resistance must equal qE, so 6ηrvt'sin =
qE, or sin = qE/6ηrvt'. (The textbook defines b = 6ηr.)
3-42.
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We start with the equation eV0 = hf - .
Since f = c/, eV0 = hc/ - .
We know that V01 = 0.52 V when 1 = 450 nm, and V02 = 1.90 V when 2 = 300
nm. So we get a system of two equations in two unknowns (h and ):
eV01 = hc/1 -  and eV02 = hc/2 - .
If we solve each equation for  and set them equal to each other, we find
hc/1 - eV01 = hc/2 - eV02, which yields h = (e/c)(V01 – V02)/(1/1 – 1/2).
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Plugging in values gives
(1.6  10-19 C/3  108 m/s)(0.52 V – 1.90 V)/(1/450 nm – 1/300 nm)
= 6.62  10-34 J s.
I'll solve for  in the first equation:  = hc/1 – eV01 =
(6.62  10-34 J s)(3  108 m/s)/(450 nm) – (1.6  10-19 C)(0.52 V) = 3.58  10-19 J,
or 2.2 eV.
3-57.
a.
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We must assume that 50 keV is the initial kinetic energy of the electron, since its
rest mass energy is actually 10 times larger than this value.
The electron loses energy hc/1 when the first photon is emitted.
The electron loses energy hc/2 when the second photon is emitted.
Since the electron is brought to rest, all of its kinetic energy must be lost and
hc/1 + hc/2 = 50 keV.
The other equation we can use is 2 = 1 + 0.095 nm.
The first equation can be rearranged as 1/1 + 1/2 = 50 keV/hc, or
1/1 + 1/(1 + 0.095 nm) = 50 keV/hc
(1 + 0.095 nm + 1)/(12 + 1.0.095 nm) = 50 keV/hc
The front of the book gives hc = 1240 eV.nm, so our equation becomes
(1 + 0.095 nm + 1)/(12 + 1.0.095 nm) = 40.32 nm-1
21 + 0.095 nm = 1240.32 nm-1 + 3.831
We can use the quadratic equation, where a = 40.32 nm-1, b = 1.83, and
c = -0.095 nm.
We obtain 1 = 0.031 nm.
So 2 = 0.126 nm.
b.
The first photon reduced the electron energy by hc/1 = 1240 eV.nm/0.031 nm =
40 keV, so the electron had 10 keV left.
4-50.
a.
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The energy levels are given by En = –E0Z2/n2 = -4E0/n2 because Z = 2 for helium.
E0 = 13.6 eV, so En = -(54.4 eV)/n2.
So E1 = -54.4 eV, E2 = -13.6 eV, E3 = -6.04 eV, E4 = -3.4 eV, E5 = -2.12 eV, and
E = 0.
E (eV)
0
-3.4 -2.12
-6.04
-13.6
-54.4
b.
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The ionization energy is the energy required to elevate an electron from
ground state (-54.4 eV) to E = 0.
So the ionization energy is 54.4 eV.
c.
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When an electron drops from some level i to some level f, it releases the energy
difference as a photon of energy hc/.
The electron's initial energy is Ei = –E0Z2/i2.
The electron's final energy is Ef = –E0Z2/f2.
So the photon's energy is hc/ = Ei – Ef, or 1/ = Z2E0(1/f2 - 1/i2)/(hc).
E0/hc = R; for precision, we need to use the "reduced mass correction," which is R
that depends on the mass of the nucleus: R = R/(1 + m/M), where m is the mass
of an electron, M is the mass of the nucleus, and R = 1.0974  107 m-1.
So 1/ = Z2(1/f2 - 1/i2)R/(1 + m/M).
The Lyman series has f = 1, so the first two lines (i = 2 and i = 3) of the hydrogen
Lyman series are
 = [12(1/12 – 1/22)(1.0974  107 m-1)/(1 + 9.11  10-31 kg/1.67  10-27 kg)]-1 =
121.57 nm and
 = [12(1/12 – 1/32)(1.0974  107 m-1)/(1 + 9.11  10-31 kg/1.67  10-27 kg)]-1 =
102.57 nm.
The Balmer series has f = 2, so the first two lines (i = 3 and i = 4) of the He+
Balmer series are
 = [22(1/22 – 1/32)(1.0974  107 m-1)/(1 + 9.11  10-31 kg/6.65  10-27 kg)]-1 =
164.05 nm and

 = [22(1/22 – 1/42)(1.0974  107 m-1)/(1 + 9.11  10-31 kg/6.65  10-27 kg)]-1 =
121.52 nm.
Observe the tremendous similarity between the first Lyman hydrogen line and the
second Balmer helium line. The tiny difference is due to the reduced mass
correction.
d.
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Neglecting the reduced-mass correction, 1/ = Z2(1/f2 - 1/i2).
So 1/H = 1/fH2 - 1/iH2 for hydrogen, and 1/He = 4(1/fHe2 - 1/iHe2) for helium.
We saw in part c that fH = 1 and iH = 2 led to almost exactly the same  as fHe = 2
and iHe = 4.
So maybe if we take double the final and initial energy levels in hydrogen, we get
the helium energy levels that yield the same wavelength.
Let's see if this is generally true: if we set fHe = 2fH and iHe = 2iH and plug this
into 1/He = 4(1/fHe2 - 1/iHe2), we get 1/He = 4(1/(2fH)2 - 1/(2iH)2) = 1/fH2 - 1/iH2.
So any wavelength in the hydrogen spectrum can be found in the helium
spectrum; to find the required energy levels in helium, we simply double the
energy levels in hydrogen.
The reduced-mass correction introduces very small wavelength differences, as we
saw in part c.
Notice that while every hydrogen line can be found in the helium spectrum, the
reverse is not true: many lines in the helium spectrum cannot be found in
hydrogen. (We would need to divide helium's energy levels by two to find the
required levels in hydrogen, but this sometimes gives non-integers, which are not
allowed energy levels.)
Problem B
a.
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The Bragg condition for constructive interference is 2dsin = m.
Here,  = 34, d = 0.347 nm, and m = 1.
So  = 2(0.347 nm)sin34 = 0.388 nm.
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The (110) planes make a 45 angle with the (100) planes.
The spacing of the (110) planes is d' = d/21/2, as illustrated below. The distance
between (100) planes is d.
b.
d
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d'
So for the (110) planes and m = 1, 2d'sin = , or  = sin-1[/(21/2d)]
= sin-1[0.388/(21/2  0.347)] = 52.2.
This is the angle between the incident X-rays and the (110) planes, but we're
asked for the angle between the incident X-rays and the surface of the crystal.


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
We've determined , but we want to know . From the figure we see that  is in
a triangle with . The third angle in this triangle is 45 because there is a 45
angle between (100) and (110) planes.
So  +  + 45 = 180, yielding  = 135 - .
So the desired angle is  = 135 - 52.2 = 82.8.
c.
By symmetry, the scattered X-rays also make an angle of 82.8 with the surface.
d.
These are the (210) planes: