Question

The band gap in germanium is $\Delta E=0.68$$\text{eV}$. Assuming that the number of hole-electron pairs is proportional to $e^{\frac{-\Delta E}{2kT}}$, find the percentage increase in the number of charge carries in pure germanium as the temperature is increased from $300\text{K}$ to $320\text{K}$.
$(\text{Take}k=8.62\times {10}^{-5}{\text{eVK}}^{-1},e^{0.82}=2.27)$

• A. $35\%$
• B. $72\%$
• C. $85\%$
• D. $127\%$

Answer 1

The correct option is D $127\%$

If $N_1$ be the number of charge carries at temperature $T_1$ and $N_2$ at $T_2$, we have

$N_1=N_0{e}^{\frac{-\Delta E}{2k{T}_1}}$ and

$N_2=N_0{e}^{\frac{-\Delta E}{2k{T}_2}}$

The percentage increase as the temperature is raised from $T_1$ to $T_2$ is

$f=\frac{N_2-N_1}{N_1}\times 100=(\frac{N_2}{N_1}-1)\times 100$

$\Rightarrow f=100[e^{\frac{\Delta K}{2k}(\frac{1}{T_1}-\frac{1}{T_2})}-1].$

Now,

$\frac{\Delta K}{2k}(\frac{1}{T_1}-\frac{1}{T_2})=\frac{0.68\text{eV}}{2\times 8.62\times {10}^{-5}{\text{eVK}}^{-1}}(\frac{1}{300\text{K}}-\frac{1}{320\text{K}})$

$=0.82$

Thus, $f=100\times [e^{0.82}-1]=100\times [2.27-1]\approx 127$

Thus, the number of charge carries increases by about $127\%$.

Hence, option $\left(D\right)$ is correct.

Why this question? Tip:In an intrinsic semiconductor, the number of conducting electrons is equal to the number of holes. So total charge carriers is twice of number of conduction electrons or holes.