Question

Assume that the number of hole-electron pair in an intrinsic semiconductor is proportional to $e^{\Delta E/2kT}$. Here $\Delta E=$ energy gap and $k=8.62\times {10}^{5}eV/K$. The energy gap for silicon is $1.1eV$. The ratio of electron hole pairs at $300K$ and $400K$ is

• A. $e^{5.104\times {10}^{-8}}$
• B. $e^{-5}$
• C. $e$
• D. $e^2$

Answer 1

Answer: (A)

$e^{5.104\times {10}^{-8}}$

Given that: $\Delta E=1.1eV$
$k=8.62\times {10}^5\frac{eV}{K}$
Also, the number of hole-electron pair in an intrinsic semiconductor is proportional to $\exp \left(\frac{\Delta E}{2kT}\right)$

Hence, for $300K$:
$\exp \left(\frac{\Delta E}{2kT}\right)=\exp \left(\frac{1.1}{2(8.62\times {10}^5)\left(300\right)}\right)$

Similarly for $400K$,
$\exp \left(\frac{\Delta E}{2kT}\right)=\exp \left(\frac{1.1}{2(8.62\times {10}^5)\left(400\right)}\right)$

Now the ratio of electron hole pairs at $300K$ and $400K$ can be calculated.