Pure Si at 300 K has equal electron `(n_(e))` and hole `(n_(h))` concentrations of `1.5xx10^(16)m^(-3)` doping by indium increases `n_(h)` to `4.5xx10^(22)m^(-3)`. Caculate `n_(e)` in the doped Si-

Options

A. `9xx10^(5)`
B. `5xx10^(9)`
C. `2.25xx10^(11)`
D. `3xx10^(10)`

Answer

B

Explanation

At equilibrium `n_(e)n_(h)=(n_(l))^(2)`
`impliesn_(e)xx4.5xx10^(22)=(15xx10^(15))^(2)`
Or `n_(e)=(2.25xx10^(32))/(4.5xx10^(22))`
or `n_(e)=5xx10^(9)`