91Ó°ÊÓ

The given equation, $\mathrm{E}-{\mathrm{E}}_{\mathrm{F}}=∆\mathrm{E}=1.00\mathrm{eV}$

The probability that an electron will occupy a certain level, according to Fermi-Dirac statistics and Boltzmann's statistics, is known as occupancy probability. Fo the Fermi level the occupancy probability is 0.5.

The occupancy probability due to Fermi-Dirac statistics,

${\mathbf{P}}_{\mathbf{FD}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{∆}\mathbf{E}\mathbf{/}\mathbf{kT}}\mathbf{+}\mathbf{1}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The Boltzmann occupation probability,

$$\begin{gathered} {\mathbf{P}}_{\mathbf{B}}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{∆}\mathbf{E}\mathbf{/}\mathbf{kT}} \\ \mathbf{where}\mathbf{}\mathbf{k}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{38}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{K}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Let the fractional difference between the probability differences be f.

Thus, using equations (1) and (2), the fractional difference can be given as follows:

$$\begin{gathered} f=\frac{P_B-P_{FD}}{P_B} \\ =\frac{1/e^{∆E/kT}-1/\left(e^{∆E/kT}+1\right)}{e^{∆E/kT}} \\ =\frac{\left(e^{∆E/kT}+1\right)-e^{∆E/kT}}{\left(e^{∆E/kT}+1\right)e^{∆E/kT}.e^{∆E/kT}} \\ =\frac{1}{\left(e^{∆E/kT}+1\right)} \end{gathered}$$

The above equation can also be written as-

$e^{∆E/kT}=\frac{1}{f}-1$

For f = 0.01,

$$\begin{gathered} e^{∆E/kT}=\frac{1}{0.01}-1 \\ =99 \end{gathered}$$

Taking logarithm on both sides

$$\begin{gathered} \frac{∆\mathrm{E}}{\mathrm{kT}}=\mathrm{In}\left(99\right) \\ \mathrm{T}=\frac{∆\mathrm{E}}{\mathrm{k}\mathrm{In}\left(99\right)} \\ \mathrm{T}=\frac{1.0\mathrm{eV}×\left(1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)×4.6} \\ \mathrm{T}=2520\mathrm{K} \end{gathered}$$

Hence, the value of the temperature is 2520 K.

If f = 0.1

Then, the temperature is given as-

$$\begin{gathered} \mathrm{T}=\frac{∆\mathrm{E}}{\mathrm{k}\mathrm{In}\left(99\right)} \\ =\frac{\left(1.00\mathrm{eV}\right)\left(1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\mathrm{In}\left(9\right)} \\ =5.28×{10}^3\mathrm{K} \end{gathered}$$

Hence, the value of the temperature is $5.28×{10}^3\mathrm{K}$.