91Ó°ÊÓ

a) Fermi energy of Germanium,$E_F=0.335\mathrm{eV}$

b) Energy gap of the germanium, E =0.67 eV

c) Temperature, T = 290 K

The probability for an electron to occupy any level in a band is known as the occupancy probability of that level. The occupancy probability of the Fermi level is exactly 0.5.

Formula:

The probability of the condition that a particle will have energy E according to Fermi-Dirac statistics, is-

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

Here, $E_F$is the Fermi energy and T is the absolute temperature.

Using the given data in equation (i), we get-

$$\begin{gathered} P\left(E\right)=\frac{1}{\exp \left[{\displaystyle \frac{\left(0.67-0.335\right)\mathrm{eV}×\left(1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(290\mathrm{K}\right)}}\right]} \\ =1.5×{10}^{-6} \end{gathered}$$

Hence, the value of the probability is $1.5×{10}^{-6}$.

At the top of the valence band $E=0$, the probability that the state is unoccupied is given using equation (i) as follows:

$$\begin{gathered} P\text{'}\left(E\right)=1-\frac{1}{e^{\left(E-E_F\right)/kT}+1} \\ =\frac{1}{e^{\left(E-E_F\right)/kT}+1} \\ =\frac{1}{\exp \left[{\displaystyle \frac{\left(0-0.335\right)\mathrm{eV}×\left(1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(290\mathrm{K}\right)}}\right]} \\ =0.99 \end{gathered}$$

Hence, the value of the probability is 0.99.