Question

Show that, at$\mathbf{T}\mathbf{=}\mathbf{0}\mathbf{K}$ , the average energy${\mathbf{E}}_{\mathbf{avg}}$ of the conduction electrons in a metal is equal to$\frac{\mathbf{3}}{\mathbf{5}}{\mathit{E}}_{\mathbf{F}}$ . (Hint:By definition of average${\mathit{E}}_{\mathbf{avg}}\mathbf{=}\left(1/n\right)\mathbf{\int }\mathit{E}{\mathit{N}}_{\mathbf{0}}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$ , where nis the number density of charge carriers.)

Short answer

The average energy $E_{avg}$of the conduction electrons in a metal is equal to $\frac{3}{5}{E}_F$ at 0 K.

Step-by-step solution

The given data

1. By definition of average energy,

$E_{avg}=\left(1/n\right)\int E{N}_0\left(E\right)dE.................................\left(a\right)$

where n is the number of charge carriers in unit volume of the metal.

1. The given temperature, T = 0 K

Conduction electrons

When valence electrons get excited by absorbing energy from the surroundings and then jump from the valence band to the conduction band, they become free to move within the substance. These electrons are known as conduction electrons.

Formula:

The density of occupied states,

$N_0\left(E\right)=N\left(E\right)P\left(E\right)......................................\left(1\right)$

Where, N (E) is the density of the states and P (E) is the probability of state occupancy.

The density of states at a given energy level,

$N\left(E\right)=C{E}^{1/2}.............................................\left(2\right)$

The number of charge carriers per unit volume,

$n={\int }_0^{\infty }N\left(E\right)P\left(E\right)dE...................................\left(3\right)$

Calculation of the average energy

Substituting the value of equation (1) in the given average energy equation, we can get the equation of average energy as follows:

$E_{avg}=\left(1/n\right)\int EN\left(E\right)P\left(E\right)dE$

Now, substituting the value of density of states from equation , we can get the value of average energy as follows:

role="math" localid="1661514340254" $$\begin{gathered} E_{avg}=\left(C/n\right){\int }_0^{E_F}{E}^{3/2}P\left(E\right)dE \\ =\frac{2C}{5n}{E}_F^{5/2}....................................\left(4\right) \end{gathered}$$

(Since, the occupation probability is one for energies below the Fermi energy, we get-.${\int }_0^{E_F}P\left(E\right)dE=1$ )

Now, to get the value of the number of conduction electrons, we substitute the value of the equation (2) in (3) equation as follows:

$$\begin{gathered} n=C{\int }_0^{E_F}{E}^{1/2}P\left(E\right)dE \\ =\frac{2C}{3}{E}_F^{3/2} \end{gathered}$$

Thus, using this above value in equation , we can get the average energy as follows:

$$\begin{gathered} E_{avg}=\frac{2C}{5\left(2C/3\right)E_F^{5/2}} \\ =\frac{3}{5}{E}_F \end{gathered}$$

Hence, the value of average energy is proportional to$\frac{3}{5}{E}_F$ .