1
Question 12.1
Explain the energy bands of metals, insulators, and semiconductors in regards
to
(a) Metals are opaque to visible light.
(b) Semiconductors are opaque to visible light, but opaque to infrared light.
(c) Insulator are transparent to visible light.
1.1
1.1.1
Solution
Part (a)
Visible light will be absorbed and scattered by all of the free electrons in a
metal.
1.1.2
Part (b)
Visible light is energetic enough to be absorbed by the valence band electrons
to cause them to jump up to the conduction band. This will prevent the light
from passing through. Infrared radiation will not be energetic enough, so will
pass through without issue.
1.1.3
Part (c)
Visible light is not energetic enough to excite electrons across the massive gap
between the conduction and valence bands of an insulator. That means it will
pass through without getting absorbed.
2
Question 12.3
The resistivity of metals increases when temperature increases. The resistivity
of semiconductors decreases when temperature increases. Why?
2.1
Solution
When you heat up a metal, it increases the collision rate of the electrons. This
results in a lower drift velocity when a field is turned on.
When you heat up an insulator, the gives more electrons the energy to make
the jump into the conduction band. This will result in a lower resistivity.
3
Question 12.4
Discuss the band structure of metals, semiconductors, and insulators. How does
this relate to their electrical properties
1
3.1
Solution
Metals have an unfilled valence band. This unfilled band acts as a conduction
band and there are always electrons that are ready to carry current.
Semiconductors have a small energy gap between their filled valence band
and their nearly empty conduction band. This means that electrons may easily
be excited to the conduction band. When the temperature increases, more
electrons will be in this conduction band, giving higher conductivity. At low
temperatures, however, none will be excited to the conduction band and the
semiconductor will not carry much current.
Insulators have a very large gap between their filled valence and empty
conduction bands. This leads to there being almost no electrons to carry a
charge, regardless of temperature.
4
Question 12.17
Why must a lasing medium have at least three energy levels?
4.1
Solution
There must be a ground state, a meta-stable intermediary state, and an unstable
pump state. Incident photons will kick the atoms up to the pump state. Since
this state is unstable, they will quickly decay to the metastable intermediary
state. Repeating this process will lead to population inversion, when there are
more atoms in the intermediary state than the ground state. Once population
inversion has been achieved, the laser is ready to fire.
5
Problem 12.1
Show that the minimum potential energy of an ion pair in an ionic solid is
1
e2
1−
Umin = −αk
r0
m
5.1
Solution
From the text, we have that
e2
B
+ m
r
r
At the minimum, the derivative of the potential energy will be zero.
U (r) = −αk
0=
d
U (r)
dr
2
0=
d
dr
−αk
e2
+ Br−m
r
e2
− mBr−(m+1)
r2
We know that r =
6 0, otherwise the atoms would be on top of each other. So
we can rearrange
0 = αk
αke2 = mBr−m+1
mB
αke2
1−m
mB
r0 =
αke2
rm−1 =
If we go and plug this into the original potential energy formula, we will get
the minimum energy,
Umin = −αk
e2
= −αk
r0
Umin
Umin
e2
= −αk
r0
e2
B
+ m
r0
r0
1
B
1−
2
(m−1)
αke r
!
0
1
1 mB
1−
m αke2 r(m−1)
!
0
But this will cancel out with r0 giving
Umin
6
e2
= −αk
r0
1
1−
m
Problem 12.2
Prove that the angles between the outer atoms in a tetrahedral molecule are
109.5◦ .
6.1
Solution
We can fit a tetrahedral molecule into a cube, with the middle atom at the
center (0.5, 0.5, 0.5) and the others at the corners (0, 0, 0), (0, 1, 1), (1, 1, 0),
and (1, 0, 1). This is shown in Figure 1.
To find the angle between two of the outer atoms, we must first express their
positions as vectors. This can be seen in Figure 2.
We will see that
3
Figure 1: A tetrahedral molecule inscribed into a sqaure. Picture by Chris
Tralie
v~1 = (0, 1, 1) − (0.5, 0.5, 0.5) = (−0.5, 0.5, 0.5)
and
v~2 = (1, 1, 0) − (0.5, 0.5, 0.5) = (0.5, 0.5, −0.5)
Now we can sue the definition of dot product to find the angle between them
v~1 · v~2 = kv~1 kkv~2 k cos(θ)
Solving for θ gives
θ = cos−1
−1
θ = cos
v~1 · v~2
kv~1 kkv~2 k
(−0.5 · 0.5) + (0.5 · 0.5) + (−0.5 · 0.5))
p
p
3 · (0.5)2 3 · (0.5)2
1
θ = cos−1 −
≈ 109.5
3
4
!
Figure 2: Expressing two of the outer atoms as vectors. Picture by Chris Tralie
7
Problem 12.5
Consider a 1D chain of alternating positively and negatively charged ions. Show
that the potential energy is
e2
r
where α = 2 ln 2 and r is the interatomic spacing.
U (r) = −kα
7.1
Solution
If we focus on one atom, it’s closest neighbors will have an opposite charge, so
will be attractive. Also, there will be two of them. Let’s call this term U1 since
is is only 1 · r away
e2
1·r
The second closest atoms will ave the same charge as our atom, so the
potential will be repulsive. TThey will also be a distance of 2r away
U1 = −2k
5
e2
2r
The next closest atoms will also be attractive and will be a distance of 3r
away.
U2 = +2k
U3 = −2k
e2
3r
We can see that the nth atoms will give a potential of
Un = (−1)n 2k
e2
nr
This means that the net potential will be
∞
X
U (r) =
(−1)n 2k
n=1
e2
nr
Most of these terms can be pulle dout of the sum, leading to the alternating
harmonic series
U (r) = −
∞
2ke2 X
1
(−1)n+1
r n=1
n
There’s a good reason for pulling out that extra minus sign. It will become
apparent if we consider the series
ln(1 + x) =
∞
X
(−1)n+1
n=1
xn
n
If we let x = 1 then we get
ln(2) =
∞
X
(−1)n+1
n=1
1
n
But this is exactly the series from our potential! Let’s plug it in.
U (r) = −
2ke2
ln(2)
r
If we define α = 2 ln(2) then we get
U (r) = −
which is what we wanted.
6
αke2
r
8
Problem 12.9
Consider the number of electrons that have a collision between t and t + dt
N −t
e τ dt
τ
n(t)dt =
(a) Prove that
∞
Z
N=
n(t)dt
0
(b) Prove that t̄ = τ
(c) Prove that t¯2 = 2τ 2
8.1
8.1.1
Solution
Part(a)
Let’s start out with
Z
∞
n(t)dt
0
Plugging in n(t) gives
Z
∞
0
N −t
e τ dt
τ
Integrating gives
i
t ∞
Nh
−τ e− τ
τ
0
But this just simplifies to N , as needed.
8.1.2
Part (b)
Now we can begin with
1
t̄ =
N
∞
Z
tn(t)dt
0
Plugging in n(t) gives
1
N
Z
∞
N −t
e τ ddt
τ
0
t
The N ’s will cancel out. Let’s also let x =
→ dt = τ dx. Plugging these
τ
in gives
Z ∞
t̄ = τ
xe−x ddx
t̄ =
t
0
7
Solving the integral gives
∞
Z
xe−x ddx
1=
0
Which eads to our answer
t̄ = τ
8.1.3
Part (c)
Now we have
∞
Z
1
t¯2 =
N
t2 n(t)dt
0
Plugging in n(t) gives
1
t¯2 =
N
∞
Z
t2
0
N −t
e τ ddt
τ
t
→ dt = τ dx. Plugging these
The N ’s will cancel out. Let’s also let x =
τ
in gives
Z ∞
2
¯
2
t =τ
x2 e−x ddx
0
This integral is
Z
2=
∞
x2 e−x ddx
0
So we get
t¯2 = 2τ 2
9
Problem 12.10
Silver has a density of 10.5 × 103 kg/m3 and a resistivity of 1.6 × 10−8 Ω·m at
room temperature.
(a) What is the mean time between collisions, τ ?
(b) What is the mean free path? The electrons thermal velocity?
(c) How does the mean free path compare to the lattice spacing?
9.1
Solution
This is one of those dumb problems where you have to do Part (b) before you
can do Part (a), so let’s proceed in that order
8
9.1.1
Part (b)
The definition of resistivity is
(3kT me )1/2
ne2 L
Solving for the mean free path yields
ρ=
L=
(3kT me )1/2
ne2 ρ
kg
We need to convert density, 10.5 3 , into electrons/m3 , also known as n.
m
This is done using
n = 10.5
kg
1000g
1 mol
6.0223 atoms 1 electron
×
×
×
×
3
m
1kg
107.87g
mol
atom
g
The 107.87
comes from silver’s atomic weight, given in a periodic table.
mol
This works out to be
n=
electrons
m3
This makes the mean free path
L=
[3(1.38 × 10−23 )(300)(9.1 × 10−31 )]1/2
≈m
()(1.6 × 10−19 )2 (1.6 × 10−8 )
The thermal velocity can be found from
3
1
kT = me v 2
2
2
or
r
v=
3kT
me
Plugging in the numbers gives
r
3(1.38 × 10−23 )(300)
m
v=
≈
9.1 × 10−31
s
Now we can go back and do part (a)
9
9.1.2
Part (a)
We have
L = vτ
or
τ=
9.1.3
L
≈s
v
Part (c)
A quick Google search reveals the lattice spacing to be about 4.1 angstroms
10
Problem 12.18
(a) What are the binding energies of Si (κ = 12) and Ge (κ = 16)? Compare to
the themal energy at room temperature.
(b) Calculate the adjusted Bohr radius in Si and Ge. How does it relate to
the interatomic spacing (2.34 and 2.43 Angstroms respectively)?
10.1
Solution
10.1.1
Part (a)
To get the new binding energy we replace Z in
ke2 Z 2
= −13.6Z 2 eV
2a0
with Z/κ giving
Ebind =
−13.6Z 2 eV
κ2
These will be
ESi =
−13.6142 eV
≈ eV
122
EGe =
−13.6322 eV
≈ eV
162
and
10
10.1.2
Part (b)
The radius is given by
a0
Z
Again, we replace Z with Z/κ giving
r=
r=
a0 κ
Z
So
rSi =
a0 12
≈
14
rSi =
a0 32
≈
16
and
11