The occupancy probability at certain energy ${\mathit{E}}_{\mathbf{1}}$in the valence band of a metal is 0.60 when the temperature is 300 K. Is${\mathit{E}}_{\mathbf{1}}$ above or below the Fermi energy?

1. The occupancy probability of the metal at energy,$E_1,P\left(E_1\right)=0.60$
2. The temperature value, T = 300 K

Using the equation of the occupancy probability of a certain material at a given energy, we can get the energy difference value, for the given values of probability and temperature. Now, if the net energy difference is positive that indicates that the energy level is above the Fermi energy, while the negative energy difference indicates that the Fermi level is above the given energy.

Formula:

The occupancy probability for a level in a band is given as,

$P\left(E\right)=\frac{1}{e^{\left(E_1-E_F\right)/kT}+1}$ (i)

Here,$E_1$ is the given energy level,$E_F$ is the Fermi level, T is the absolute temperature, and k is the Boltzmann constant.

Let, the energy difference be given as, $E_1-E_F=∆E$.

Now, from equation the energy difference is given as-

$$\begin{gathered} ∆E=kT\ln \left(\frac{1}{P}-1\right) \\ =\left(1.38\times {10}^{-23}J/K\right)\left(300K\right)\ln \left(\frac{1}{0.6}-1\right) \\ =-1.67\times {10}^{-21}J \\ =-0.0104eV \end{gathered}$$

$\begin{array}{l}{E}_1-E_F<0\\ E_1<E_F\end{array}$

Hence, the energy level is below the Fermi energy level of the metal.