91Ó°ÊÓ

a) The energy of the state above the Fermi level, $E-E_F=0.0620\mathrm{eV}$

b) Temperature for the cases,$T=0K\mathrm{and}T=320\mathrm{K}$

Using the probability function, we can see that the probability value at temperature T = 0 K gives unity values of probability function for energies less than or equal to Fermi energy while the probability value tends to zero for energies at a higher state than that to Fermi level. Similarly, by substituting the given data of energy difference $\left(E_1-E_F\right)$and temperature for the case T = 320 K, we can get the value of the occupancy probability using the formula.

$$\begin{gathered} \mathbf{P}\left(E\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{(}{\mathbf{E}}_{\mathbf{1}}\mathbf{-}{\mathbf{E}}_{\mathbf{F}}\mathbf{)}\mathbf{/}\mathbf{kT}}\mathbf{+}\mathbf{1}} \\ \mathbf{where}\mathbf{}\mathbf{k}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{62}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{eV} \end{gathered}$$

At absolute temperature T = 0 K, the probability is zero considering equation (i) that any state with energy above the Fermi energy is occupied (using the concept);

Hence, the probability of the state is 0 .

Now, for the energy difference, $E-E_F=0.0620\mathrm{eV}$we can have the value of the following term as:

$({\mathrm{E}}_1-{\mathrm{E}}_{\mathrm{F}})/kT=\frac{0.0620\mathrm{eV}}{\left(8.62×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(320\mathrm{K}\right)}$

Thus, the occupancy probability of the given state is-

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{2.248}+1} \\ =0.0955 \end{gathered}$$

Hence, the value of the probability is 0.0955.