Question

Show that P(E), the occupancy probability in Eq. 41-6, is symmetrical about the value of the Fermi energy; that is, show that $\mathit{P}\mathbf{(}{\mathit{E}}_{\mathbf{F}}\mathbf{+}\mathit{\Delta }\mathit{E}\mathbf{)}\mathbf{+}\mathit{P}\mathbf{(}{\mathit{E}}_{\mathbf{F}}\mathbf{-}\mathit{\Delta }\mathit{E}\mathbf{)}\mathbf{=}\mathbf{1}$.

Short answer

It is shown that$P(E_F+\Delta E)+P(E_F-\Delta E)=1$ that is the occupancy probability is symmetrical about the value of the Fermi energy.

Step-by-step solution

Understanding the concept of occupancy probability

The probability of a state to be occupied by and electron is referred to as the occupancy probability. At 0 K temperature, the states below the Fermi level have occupancy probability equal to and for the states above the Fermi level, its value is .

Formula:

The occupancy probability of the state with energy Eis-

$P\left(E\right)=\frac{1}{e^{\left(E-E_F\right)/KT}+1}$ ( i )

Here $E_F$is the Fermi energy, $k=8.62\times {10}^{-5}eV/K$ and T is the absolute temperature.

Calculation of the given symmetrical condition of probability

Upon expansion in view of equation (i), the LHS value of the given equation can be solved as follows:

$$\begin{gathered} LHS=P\left(E_F+∆E\right)+P\left(E_F-∆E\right) \\ =\frac{1}{e^{\left(E_F+∆E-E_F\right)/KT}+1}+\frac{1}{e^{\left(E_F-∆E-E_F\right)/KT}+1} \\ =\frac{1}{e^{∆E/KT}+1}+\frac{1}{e^{-∆E//KT}+1} \\ =\frac{e^{∆E/KT}+1+e^{∆E/KT}+1}{\left(e^{∆E/KT}+1\right)\left(e^{-∆E/KT}+1\right)} \end{gathered}$$

On further solving,

$$\begin{gathered} L.H.S=\frac{e^{∆E/KT}+1+e^{∆E/KT}+2}{e^{∆E/KT}+e^{∆E/KT}+2} \\ =1 \\ =R.H.S \end{gathered}$$

Hence, the given condition is proved and this implies the symmetrical condition for occupancy probability.