Pure Si at 500 K has equal number of 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)`. The doped semiconductor is of ………

Options

A. n-type semiconductor, electron concentration `n_(e )=5xx10^(22)m^(-3)`
B. p-type semiconductor, electron concentration `n_(e )=2.5xx10^(10)m^(-3)`
C. n-type semiconductor, electron concentration `n_(e )=2.5xx10^(23)m^(-3)`
D. p-type semiconductor, electron concentration `n_(e )=5xx10^(9)m^(-3)`

Answer

D

Explanation

p-type semiconductor, electron concentration `n_(e )=5xx10^(9)m^(-3)`
`n_(i)^(2)=n_(e )n_(h)`
`therefore (1.5xx10^(16))^(2)=n_(e )(4.5xx10^(22))`
`rArr n_(e )=0.5xx10^(10) or n_(e )=5xx10^(9)`
Now `n_(h)=4.5xx10^(22) rArr n_(n) gt gt n_(e )`
`therefore` Semiconductor is p-type and `n_( e)=5xx10^(9)m^(-3)`