Page 103 - Electrical Properties of Materials

The density of states and the Fermi–Dirac distribution 85

If we imagine a cube of side L containing the electrons, then we get for the
energy in the same manner

2 2 2 2 h 2 2 2 2
E = k + k + k z = n + n + n z . (6.2) n x , n y , n z are integers.
y
x
y
x
2 m 8 mL 2
6.2 The density of states and the Fermi–Dirac distribution
2
2
The allowed energy, according to eqn (6.2), is an integral multiple of h /8 mL .
3
–6
For a volume of 10 m this unit of energy is
(6.62 × 10 –34 2 –33 –15
)
E unit = =0.6 × 10 J = 3.74 × 10 eV. (6.3)
8 × 9.1 × 10 –31 × 10 –4
This is the energy difference between the first and second levels, but since
the squares of the integers are involved, the difference between neighbouring
energy levels increases at higher energies. Let us anticipate the result obtained
in the next section and take for the maximum energy E = 3 eV, which is a
2
2
2
typical figure. Taking n = n = n , this maximum energy corresponds to a
x y z
~
7
value of n x = 1.64×10 . Now an energy level just below the maximum energy
can be obtained by taking the integers n x –1, n x , n x . We get for the energy
difference
~
–7
E = 1.22 × 10 eV, (6.4)
We can therefore say that, in a macroscopically small energy interval d E, Even at the highest energy, the dif-
there are still many discrete energy levels. So we can introduce the concept of ference between neighbouring en-
–7
density of states, which will simplify our calculations considerably. ergy levels is as small as 10 eV.
The next question we ask is how many states are there between the energy
levels E and E +dE? It is convenient to introduce for this purpose the new
variable n with the relationship
2
2
2
2
n = n + n + n . (6.5)
y
x
z
Thus n represents a vector to a point n x , n y , n z in three-dimensional space. In
this space every point with integer coordinates specifies a state; that is, a unit
cube contains exactly one state. Hence, the number of states in any volume is
just equal to the numerical value of the volume. Thus, in a sphere of radius n,
the number of states is
3
4n π
. (6.6)
3
Since n and E are related, this is equivalent to saying that the number of
states having energies less than E is
3
4n π 4π 3/2 3/2 8 mL 2
= K E with K = . (6.7)
3 3 h 2
Similarly, the number of states having energies less than E +d E is
4π 3/2 3/2
K (E +d E) . (6.8)
3