1
PHYSICS 5520 (SPRING 2011); HOMEWORK #3 SOLUTIONS
Problem 1; conduction band of Ge ( 10 points)
(a) The number of states per unit volume between the bottom of conduction band and energy, E, is
N=
1 X
gj (2m∗j )3/2 (E − Ej )3/2 ,
3π 2 ~3 j
(1)
where gj is the number of valleys j, m∗j = (m∗l m∗t1 m∗t2 )1/3 is the average effective mass, and Ej is the energy at the
bottom of a j-valley. For the conduction band of germanium we have the following parameters:
h111i :
Ej = 0,
gj = 4,
h000i :
h100i :
Ej = 0.15eV,
Ej = 0.18eV,
m∗l = 1.58m,
gj = 1,
gj = 6,
m∗t = 0.082m,
m∗l = m∗t = 0.036m,
m∗l = 0.19m. m∗t = 0.97m,
(2)
To find the position of the Fermi level at zero temperature and for N = 2 × 1018 cm−3 , one should solve Eq. (1) for
EF ,
2 × 1018 =
1 X
gj (2m∗j )3/2 (EF − Ej )3/2 ,
3π 2 ~3 j
with parameters Eq. (2). Converting ergs into eVs (1erg = 6.24 × 1011 eV ) we find that for such a low density only
the lowest four h111i valleys are filled up to EF ≈ 11meV .
Problem 2; absorption in InP (10 points)
The threshold Eo , includes the energy ∆E separating the valleys h000i and h100i, the phonon energy E p involved
in the indirect transition, and the position ξ of the Fermi level above the bottom of the valley h000i:
Eo = ∆E + Ep − ξ.
(3)
The electron density, on the other hand, is related to ξ as
n=
1
(2m∗ ξ)3/2 .
3π 2 ~3
(4)
Thus for the two values of ξ we have:
ξ1 =
(3π 2 n1 )2/3 ~2
,
2m∗
ξ2 =
(3π 2 n2 )2/3 ~2
.
2m∗
By subtracting these relations and using Eq. (3) we get:
Eo,2 − Eo,1 = ξ1 − ξ2 =
(3π 2 )2/3 ~2 2/3
2/3
n
−
n
,
1
2
2m∗
which yields the answer
m∗ =
(3π 2 )2/3 ~2 2/3
2/3
n1 − n 2
≈ 10−28 g.
2 (Eo,2 − Eo,1 )
Problem 3; pressure effects (10 points total)
(a) Absorption coefficient for the two cases reads:
α1 = A(E1 − Eg )1/2 ,
α2 = A(E2 − Eg )1/2 .
(5)
2
From this system of equations we find:
Eg =
E1 − E2(α1 /α2 )2
= 1.44eV.
1 − (α1 /α2 )2
Using this value with any of the relations Eq. (5) the coefficient A is found to be A = 10 4 eV−1/2 cm−1 , so that α is
given by
α = 104 (hν − 1.44)1/2 cm−1 .
(b) Converting the relation for absorption coefficient, α = A(hν − Eg )1/2 , to the relation for Eg yields
Eg = hν −
α2
.
A2
From this relation, variation of Eg due to a small change in α can be expressed as ∆Eg = −2α∆α/A2 = 1.2 × 10−5 eV.
Thus the coefficient for change of energy gap with pressure is
∆Eg
= 1.2 × 10−9 eV − cm2 /Kg.
∆P
(c) For a specimen with a thickness, d, an absorption coefficient, α, and a reflectivity, R, the transmission is given by
T =
(1 − R)2 e−αd
.
1 − R2 e−2αd
In our problem αd 1, so that the above relation simplifies to
T ≈ (1 − R)2 e−αd .
Using this relation one finds:
∆T
= −∆αd = 0.01 = 1%.
T
Problem 4; Sb impurity band in Ge (10 points total)
(a) A Mott transition occurs when
a . 4.5 aH ,
where a is the mean spacing between the donor atoms, and aH is the effective Bohr radius. For an atom immersed in
the high dielectric constant of the Ge crystal one has:
aH =
~2 ≈ 8 × 10−7 cm.
e 2 m∗
Thus the critical donor density Nd0 , at which a transition from Mott insulator to metal occurs is
Nd0 = a−3 = 2.78 × 1019 cm−3 .
(b) From the symmetry of the Ge crystal it is reasonable to suggest that donor lattice is a simple cubic lattice. Then
in tight-binding approximation, the bandwidth, Λ, is given by Λ = 2γ. Expressing the distance via the density,
a = N −1/3 , we get:
Λ = (E2 − Eg )1/2 .
(6)
(c) Impurity band touches the bottom of the conduction band when the impurity bandwidth reaches the ionization
energy, Ed . For Sb donors in Ge, Ed = 9.6meV. According to Eq. (6), Nd1 can be found from the relation
Ed = Λ = f (Nd1 ).
(7)