bei48482_FM

Fermi Energy

The final step is to calculate the value of F, the Fermi energy. As mentioned in

Sec. 9.4, we can do this by filling up the energy states in the metal sample at T 0

with the Nfree electrons it contains in order of increasing energy starting from 0.

The highest state to be filled will then have the energy Fby definition. The num-

ber of electrons that can have the same energy is equal to the number of states that

have this energy, since each state is limited to one electron. Hence

N 

F

0

g()d 

F

0

 d F^3 ^2

and so

F 

2  3

(9.56)

The quantity NVis the density of free electrons.

Example 9.8

Find the Fermi energy in copper on the assumption that each copper atom contributes one free

electron to the electron gas. (This is a reasonable assumption since, from Table 7.4, a copper

atom has a single 4selectron outside closed inner shells.) The density of copper is 8.94 

103 kg/m^3 and its atomic mass is 63.5 u.

Solution

The electron density NVin copper is equal to the number of copper atoms per unit volume.

Since 1 u 1.66  10 ^27 kg,



8.48 1028 atoms m^3 8.48 1028 electrons m^3

The corresponding Fermi energy is, from (9.56),

F

2  3

1.13 10 ^18 J7.04 eV

At absolute zero, T0 K, there would be electrons with energies of up to 7.04 eV in copper

(corresponding to speeds of up to 1.6  106 m/s!). By contrast, allthe molecules in an ideal

gas at 0 K would have zero energy. The electron gas in a metal is said to be degenerate.

9.10 ELECTRON-ENERGY DISTRIBUTION

Why the electrons in a metal do not contribute to its specific heat except at

very high and very low temperatures

(3)(8.48^1028 electrons/m^3 )

8 

(6.63^10 ^34 Js)^2

(2)(9.11 10 ^31 kg /electron)

8.94 103 kg m^3



(63.5 u)(1.66 10 ^27 kg/u)

mass m^3



mass^ atom

atoms



m^3

N



V

3 N



8 V

h^2



2 m

Fermi energy

16  2 Vm^3 ^2



3 h^3

8  2 Vm^3 ^2



h^3

Statistical Mechanics 325

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