Question

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

Answer 1

If $n _{1}$ is the number of charge carriers at temperature $T_{1}$ and $n_{2}$ at $T_{2}$, then,
$n _{1}= n _{0} e ^{\left(- E _{ g } / 2 kT _{1}\right)}$ and $n _{2}= n _{0} e ^{\left(- E _{ g } / 2 kT _{2}\right)}$
The fractional increase as the temperature is raised from $T _{1}$ to $T _{2}$ is, $\Delta n =\frac{ n _{2}- n _{1}}{ n _{1}}=\left(\frac{ n _{2}}{ n _{1}}-1\right)$
$\therefore \Delta n=\left[e^{\frac{E_{g}}{2 k}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)}-1\right]$ ...(i)
Now,
$\frac{E_{ g }}{2 k }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)=\frac{0.64 \times 1.6 \times 10^{-19}}{2 \times 1.4 \times 10^{-23}}\left(\frac{1}{300}-\frac{1}{384}\right)$
$=\frac{32 \times 8 \times 10^{2}}{7}\left(\frac{84}{384 \times 300}\right)$
$=\frac{32 \times 12 \times 8}{384 \times 3}=\frac{8}{3}$
$\therefore e^{\frac{E_{g}}{2 k}}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)=e^{\frac{8}{3}}$
Given that, $\ln (14.4)=\frac{8}{3}$
$\therefore e^{\frac{8}{3}}=14.4$
$\therefore \Delta n=(14.4-1)$
$\therefore \Delta n=13.4$