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A pure semiconductor has equal electron and hole concentration of `10^(6)m^(-3)`. Dopping by indium increases `n_(h)` to `4.5xx10^(22)m^(0k3)`. What is `n_(e)` in the dopped semiconductor?

A

`10^(6)m^(-3)`

B

`10^(22)m^(-3)`

C

`(10^(32))/(4.5xx10^(22))m^(-3)`

D

`4.5xx10^(22)m^(-3)`

Text Solution

Verified by Experts

The correct Answer is:
A
`n_(e)h_(e)=n^(2)` or `h_(e)=(n^(2))/(n_(h))=(10^(16)xx10^(16))/(4.5xx10^(22))`
`=(10^(32))/(4.5xx10^(22))m^(-3)`
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A2Z-SEMICONDUCTOR ELECTRONICS-EXERCISE
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