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If n(c ) and n(h) are electron and hole ...

If `n_(c ) and n_(h)` are electron and hole concentralious in an extrinsic semiconductor and `n_(i)` is electron concentrations in an intrinsic semiconductor, then

A

`((n_(e))/(n_(h)))=n_(i)`

B

`(n_(e)+n_(h))=n_(i)`

C

`(n_(e)-n_(h))s=n_(i)`

D

`n_(e)n_(h)=n_(i)^(2)`

Text Solution

Verified by Experts

`n_(e) and n_(h)` are electron and hole concentrations in an extrinsic semiconductor and `n_(j)` is electron concentration in an intrinsic semiconductor, then
`n_(e)n_(h) = n_(1)^(2)`
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