bei48482_FM

(Barry) #1
With the help of Eqs. (9.29) and (9.55) we have for the number of electrons in an
electron gas that have energies between and d

n() dg()f()d (9.57)

If we express the numerator of Eq. (9.57) in terms of the Fermi energy Fwe get

n() d (9.58)

This formula is plotted in Fig. 9.11 for T0, 300, and 1200 K.
It is interesting to determine the average electron energy at 0 K. To do this, we first
find the total energy E 0 at 0 K, which is

E 0 


F

0

n()d

Since at T0 K all the electrons have energies less than or equal to the Fermi energy
F, we may let

e(F)kTe
 0

and E 0  F^3 ^2 


F

0

^3 ^2 d NF

The average electron energy  0 is this total energy divided by the number Nof elec-
trons present, which gives

 0  F (9.59)

3

5

Average electron
energy at T 0

3

5

3 N

2

(3N2) F^3 ^2 d

e(F)kT 1

Electron energy
distribution

(8 2 Vm^3 ^2 h^3 )d

e(F)kT 1

326 Chapter Nine


Figure 9.11Distribution of electron energies in a metal at various temperatures.

0 K
300 K
1200 K

eeF

n(

e)

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