2.7. MOBILE CARRIERS 55
EXAMPLE 2.1A particular metal has 1022 electrons per cubic centimeter. Calculate the Fermi energy
and the Fermi velocity (at 0 K).
The Fermi energy is the highest occupied energy state at 0 K and is given by (measured from the con-
duction bandedge)
EF = 2
2 m 0(
3 π^2 n) 2 / 3= (1.^05 ×^10− (^34) ) (^2) [3π (^2) (10 (^28) )] 2 / 3
2(0. 91 × 10 −^30 )
=2. 75 × 10 −^19 J
=1.72 eV
The Fermi velocity is
vF =
m 0
(
3 π^2 n
) 1 / 3
= (1.^05 ×^10
− (^34) J.s)(3π (^2) × 1028 m− (^3) ) 1 / 3
0. 91 × 10 −^30 kg =7.^52 ×^10
(^5) m/s
=7. 52 × 107 cm/s
Thus, the highest energy electron has a large energy and is moving with a very large speed.
2.7.2 Electrons and holes in semiconductors
In pure semiconductors there are no mobile carriers at zero temperature. As temperature is
raised, electrons from the valence band are thermally excited into the conduction band, and in
equilibrium there is an electron densitynand an equal hole densityp,as shown in figure 2.17a
Note that the density of allowed states has the form
N(E)=
√
2(m∗dos)^3 /^2 (E−Ec)^1 /^2
π^2 ^3(2.7.6)
wherem∗dosis the density of states mass andEcis the conduction bandedge. A similar expression
exists for the valence band except the energy term is replaced by(Ev−E)^1 /^2 and the density of
states exist below the valence bandedgeEv. Figure 2.17 shows a schematic view of the density
of states.
It is important to note that the density of states mass has a special term in indirect gap mate-
rials. In direct gap semiconductorsm∗dosis just the effective mass for the conduction band. In
indirect gap materials it is given by (see Appendix C)
m∗dos=(m∗ 1 m∗ 2 m∗ 3 )^1 /^3wherem∗ 1 m∗ 2 m∗ 3 are the effective masses along the three principle axes. For Si counting the six
degenerateX-valleys we have
m∗dos=6^2 /^3(
mm^2 t