Question

The band gap in germanium is $E_g=0.64eV$. Assuming that the number of hole-electron pairs is proportional to $e^{\left(-E_g/2kT\right)}$, find the fractional increase in the number of charge carriers in pure germanium as the temperature is increased from $300K$ to $384K$.
[Take $k=1.4\times 10^{-23},\ln (14.4)=\frac{8}{3}$]

Answer 1

Answer: 13.4

If $n_1$ is the number of charge carriers at temperature $T_1$ and $n_2$ at $T_2$, then,
$n_1=n_0{e}^{\left(-E_g/2k{T}_1\right)}$ and $n_2=n_0{e}^{\left(-E_g/2k{T}_2\right)}$
The fractional increase as the temperature is raised from $T_1$ to $T_2$ is, $\Delta n=\frac{n_2-n_1}{n_1}=\left(\frac{n_2}{n_1}-1\right)$
$\therefore \Delta n=\left[e^{\frac{E_g}{2k}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}-1\right]$ ...(i)
Now,
$\frac{E_g}{2k}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)=\frac{0.64\times 1.6\times 10^{-19}}{2\times 1.4\times 10^{-23}}\left(\frac{1}{300}-\frac{1}{384}\right)$
$=\frac{32\times 8\times 10^2}{7}\left(\frac{84}{384\times 300}\right)$
$=\frac{32\times 12\times 8}{384\times 3}=\frac{8}{3}$
$\therefore e^{\frac{E_g}{2k}}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)=e^{\frac{8}{3}}$
Given that, $\ln (14.4)=\frac{8}{3}$
$\therefore e^{\frac{8}{3}}=14.4$
$\therefore \Delta n=(14.4-1)$
$\therefore \Delta n=13.4$