Question

Assume that the number of hole-electron pairs in an intrinsic semiconductor is proportional to ${{e}^{\dfrac{\Delta E}{2KT}}}$. Here $\Delta E=$energy gap and $k=8.62\times {{10}^{5}}eV{{K}^{-1}}$. The energy gap for silicon is $1.1eV$. The ratio of electron hole pairs at $300K$ and $400K$is,
$\begin{align}
  & A.{{e}^{5.104\times {{10}^{-8}}}} \\
 & B.{{e}^{-5}} \\
 & C.e \\
 & D.{{e}^{2}} \\
\end{align}$

Answer 1

Hint: the number of hole-electron pairs in an intrinsic semiconductor is given as proportional to ${{e}^{\dfrac{\Delta E}{2KT}}}$. Therefore substitute the given temperatures in this equation. So we can get the ratio of these two quantities. This ratio will be equal to the number of electron-hole pairs in the intrinsic semiconductor. Hope this will help you to solve this question.

Complete step by step answer:
it is already mentioned in the question that the energy gap of the silicon semiconductor is,
$\Delta E=1.1eV$
And also it is mentioned that,
$k=8.62\times {{10}^{5}}eV{{K}^{-1}}$
As we know that the number of the electron- hole pair ratio is found to be proportional to ${{e}^{\dfrac{\Delta E}{2KT}}}$.
$\dfrac{{{e}^{\dfrac{\Delta E}{2K{{T}_{1}}}}}}{{{e}^{\dfrac{\Delta E}{2K{{T}_{2}}}}}}=\dfrac{{{N}_{1}}}{{{N}_{2}}}$
Therefore let us calculate this quantity instead of the other for the comparison.
For $300K$, we can find the value of ${{e}^{\dfrac{\Delta E}{2KT}}}$ by substituting the other terms also,
${{e}^{\dfrac{\Delta E}{2K{{T}_{1}}}}}={{e}^{\left( \dfrac{1.1}{2\left( 8.62\times {{10}^{5}} \right)\left( 300 \right)} \right)}}$
For $400K$ we can write that,
${{e}^{\dfrac{\Delta E}{2K{{T}_{2}}}}}={{e}^{\left( \dfrac{1.1}{2\left( 8.62\times {{10}^{5}} \right)\left( 400 \right)} \right)}}$
Let us take the ratio of these quantities. This will be similar to taking the ratio of the number of electron-hole pairs. Therefore the ratio of this quantity can be written as,
$\dfrac{\text{number of electron holepair at 300K}}{\text{number of electron holepair at 400K}}=\dfrac{{{e}^{\left( \dfrac{1.1}{2\left( 8.62\times {{10}^{5}} \right)\left( 300 \right)} \right)}}}{{{e}^{\left( \dfrac{1.1}{2\left( 8.62\times {{10}^{5}} \right)\left( 400 \right)} \right)}}}$
Simplifying the above equation will give,
$\dfrac{{{N}_{1}}}{{{N}_{2}}}={{e}^{\left( \dfrac{1.1}{2\left( 8.62\times {{10}^{5}} \right)} \right)\left( \dfrac{1}{300}-\dfrac{1}{400} \right)}}$
$\dfrac{\text{number of electron holepair at 300K}}{\text{number of electron holepair at 400K}}={{e}^{-5.31 \times 10^{-8}}}$

So, the correct answer is “Option A”.

Note: The electron–hole pair is the basic unit for generation and recombination usually seen in the semiconductors. This corresponds to the transition of the electron between the valence band and the conduction band. Basically the generation of the electron is also a transition from the valence band to the conduction band. By the way, their recombination will lead to a transition in the reverse direction.