Download Chapter 11 The solid State, Home Work Solutions

Chapter 11
The solid State, Home Work Solutions
11.1
Problem 11.7
(a) Calculate the ionic cohesive energy for KCl, which has the same crystal structure as
N aCl. Take r◦ = 0.314 nm and m = 9.
(b) Calculate the atomic cohesive energy of KCl by using the facts that the ionization energy
of potassium is 4.34 eV (that is, K + 4.34 eV → K + + e) and that the electron affinity
of chlorine is 3.61 eV (Cl− + 3.61 eV → Cl + e).
Solution
(a) The ionic cohesive energy U◦ is given by:
e2
U◦ = −αk
r◦
1
1−
m
where α is Madelung constant, k is the Coulomb constant, e is the elementary charge,
r◦ is the equilibrium separation, and m is a constant. Since KCl has the same crystal
structure as N aCl we use the value of α for N aCl.
−19 2
)
1
9 (1.602 × 10
U◦ = −1.7476 × 8.988 × 10
1−
0.314 × 10−9
9
−18
= −1.14 × 10 J
1.14 × 10−18
= −
eV
1.602 × 10−19
= −7.12 eV
2
CHAPTER 11. THE SOLID STATE, HOME WORK SOLUTIONS
(b) Atomic cohesive energy = ionic cohesive energy - energy needed to remove an electron
from Cl− + energy gained by adding the electron to K + so,
Atomic cohesive energy = −7.12 − 3.61 + 4.34 = −6.39 eV
11.2. PROBLEM 11.11
11.2
3
Problem 11.11
The contribution of a single electron or hole to the electric conductivity of a semiconductor
can be expressed by an important property called the mobility. The mobility, µ, is defined
as the particle drift speed per unit electric field or, in terms of a formula, µ = vd /E. Note
that mobility describes the ease with which charge carriers drift in an electric field and that
the mobility of an electron, µn , may be different from the mobility of a hole, µp , in the same
material.
(a) Show that the current density may be written in terms of mobility as
J = neµn E
where µn = eτ /me .
(b) Show that, when both electrons and holes are present, the conductivity may be expressed
in terms of µn and µp as
σ = neµn + peµp
where n is the electron concentration and p is the hole concentration.
(c) If a germanium sample has µn = 3900 cm2 /V · s, calculate the drift speed of an electron
when a field of 100V /cm is applied.
(d) A pure (intrinsic) sample of germanium has µn = 3900 cm2 /V ·s and µp = 1900 cm2 /V ·s.
If the hole concentration is equal to 3.0×1013 cm−3 , calculate the resistivity of the sample.
Solution
(a) The current density J (A/m2 ) is the given by the product of the number of charge
carriers per unit volume n (m−3 ), the charge of the charge carriers e (C), and the drift
speed of the charge carriers vd (m/s), i.e.
J = nevd
From the definition of the mobility we get:
vd = µE
the current density then becomes:
J = neµE
4
CHAPTER 11. THE SOLID STATE, HOME WORK SOLUTIONS
In addition the drift velocity was derived in Chapter 11 (Equation 11.10) as:
vd =
eEτ
me
µ=
eτ
me
we can then conclude that:
(b) The generalized form of Ohm’s law is:
J = σE
Under the effect of an electric field electrons and holes move in opposite directions, so
the currents formed by them will add, i.e.
J =
=
σE =
σ =
Jelectrons + Jholes
neµn E + peµp E
neµn E + peµp E
neµn + peµp
where n and p are the concentration of electrons and holes (number of carriers per unit
volume) respectively. µn and µp are the mobility of electrons and holes respectively.
(c) The electron drift velocity in germanium is given by:
v d = µn E
= 3900 × 100
= 3.9 × 105 cm/s
(d) In an intrinsic semiconductor the concentration of electrons is the same as that of holes,
i.e. n = p, so the conductivity σ and resistivity ρ are:
σ =
=
=
=
=
neµn + peµp
pe(µn + µp )
3.0 × 1013 × 1.60 × 10−19 × (3900 + 1900)
0.028 (Ω · cm)−1
2.8 (Ω · m)−1
1
ρ =
σ
= 0.36 Ω · m
11.3. PROBLEM 11.12
11.3
5
Problem 11.12
Gallium arsenide (GaAs) is a semiconductor material of great interest for its high powerhandling capabilities and fast response time.
(a) Calculate the drift speed of electrons in GaAs for a field of 10 V /cm if the electron
mobility is µn = 8500 cm2 /V · s. (See Problem 11 for the definition of µ)
(b) What percent of the electron’s thermal speed at 300 K is this drift speed?
(c) Assuming an effective electron mass equal to the free-electron mass, calculate the average
time between electron collisions.
(d) Calculate the electronic mean free path.
Solution
(a) The drift speed in GaAs is :
vd =
=
=
=
µn E
8500 × 10
85, 000 cm/s
850 m/s
(b) The thermal energy is given by:
1
3
2
mvth
=
kB T
2
2
r
3kB T
vth =
r m
3 × 1.4 × 10−23 × 300
=
9.1 × 10−31
= 1.177 × 105 m/s
vd
850
=
vth
1.177 × 105
= 7.224 × 10−3
= 0.72%
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CHAPTER 11. THE SOLID STATE, HOME WORK SOLUTIONS
(c) The average time between collisions, τ , is given by:
vd =
τ =
=
=
=
eEτ
me
m e vd
eE
9.1 × 10−31 (kg) × 850(m/s)
1.602 × 10−19 (C) × 1000(V /m)
4.83 × 10−12 s
4.83 ps
(d) The electrons’ mean free path, L, is given by:
L =
=
=
=
vth × τ
1.177 × 105 × 4.83 × 10−12
5.69 × 10−7 m
0.569 µm
11.4. PROBLEM 11.16
11.4
7
Problem 11.16
From the optical absorption spectrum of a certain semiconductor, one finds that the longest
wavelength of radiation absorbed is 1.85 µm. Calculate the energy gap for this semiconductor.
Solution
The energy gap Eg of a semiconductor is:
Eg = hfmin
hc
=
λmax
1.240 × 103 (eV · nm)
=
1850(nm)
= 0.670 eV
8
11.5
CHAPTER 11. THE SOLID STATE, HOME WORK SOLUTIONS
Problem 11.19
Under high resolution and at threshold current, the emission spectrum from a GaAs laser is
seen to consist of man sharp lines, as shown in Figure (11.1). Although one line generally
dominates at higher currents, let us consider the origin and spacing of the multiple lines, or
modes, shown in this figure. Recall that standing waves or resonant modes corresponding
to the sharp emission lines are formed when an integral number m of half wavelengths fits
between the cleaved GaAs surfaces. If L is the distance between cleaved faces, n is the index
of refraction of the semiconductor, and λ is the wavelength in air, we have
m=
L
λ/2n
(11.1)
Figure 11.1: A high resolution emission spectrum of the GaAs laser, operated just below the
laser threshold.
(a) Show that the wavelength separation between adjacent modes (the change in wavelength
for the case ∆m = +1) is
λ2
|∆λ| =
2L n − λ dn
dλ
11.5. PROBLEM 11.19
9
(Hint: Since m is large and we want the small change in wavelength that corresponds
to ∆m = 1, we can consider m to ee a continuous function of λ and differentiate m =
2Ln/λ).
(b) Using the following typical values for GaAs, calculate the wavelength separation between
adjacent modes. Compare your results to Figure (11.1).
λ
= 837
2L
= 0.60
n
= 3.58
dn/dλ = 3.8 ×10−4
nm
mm
(nm)−1
at λ = 837 nm
(c) Estimate the mode separation for the He-Ne gas laser using λ = 632.8 nm, 2L = 0.6 m,
n ≈1, and dn/dλ = 0. On the basis of this calculation and the result of part (b), what
is the controlling factor in mode separation in both solids and gases?
Solution
(a) Taking the derivative of Equation(11.1) with respect to λ we get:
dm
d 2Ln
=
dλ
dλ
λ
dn n
2L
=
−
λ
dλ λ
The left hand side of the the last equation is the change in m with respect to the change
in λ, so we can change dm and dλ on the left hand side to ∆m and ∆λ, and taking
∆m = 1 we get:
∆m
2L
dn n
=
−
∆λ
λ
dλ λ
2L
dn
=
λ
−n
λ2
dλ
2L
dn
= − 2 n−λ
λ
dλ
2
∆m λ
∆λ = −
2L n − λ dn
dλ
λ2
|∆λ| =
2L n − λ dn
dλ
10
CHAPTER 11. THE SOLID STATE, HOME WORK SOLUTIONS
(b) The wavelength separation for GaAs laser is:
(837 × 10−9 )2
0.6 × 10−3 (3.58 − 837 × 3.8 × 10−4 )
= 3.6 × 10−10 m
= 0.36 nm
|∆λ =
The wavelength separation in Figure (11.1) is about 0.2 nm which is smaller than the
calculated separation.
(c) The wavelength separation for the He-Ne laser is:
(633 × 10−9 )2
0.6
= 6.68 × 10−13 m
= 6.68 × 10−4 nm
|∆λ| =
The controlling factor here is the cavity length L.