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Assume that the number of hole-electron ...

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 xx 10^(-5) eV//"kelvin"`
The energy gap for silicon is `1.1 eV`. The ratio of electron hole pairs at`300 K` and `400 K` is :

A

`e^(-5.31)`

B

`e^(-5)`

C

`e`

D

`e^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
` (N_(1))/(N_(2))=(e^(-Delta E//2KT_(1)))/(e^(-Delta E//2KT_(2)))`
`= e^((Delta E)/(2K)((1)/(T_(2))-(1)/(T_(1)))) =(1.1)/(e^(2xx8.62 xx10^(-5))) ((1)/(400)-(1)/(300))`.
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