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For pure ''Ge'' semiconductor quantity o...

For pure ''Ge'' semiconductor quantity of ''e'' and hole is `10^(19) e//m^(3)` if we doped donor impurity in it with density `10^(23) e//m^(3)` then quantity of hole `(e//m^(3))` in semiconductor is :-

A

`10^(15)`

B

`10^(19)`

C

`10^(23)`

D

`10^(27)`

Text Solution

Verified by Experts

The correct Answer is:
A
`n_(h)=(n_(i)^(2))/(n_(e))=((10^(19))^(2))/(10^(23))=10^(15)`
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ALLEN-SEMICONDUCTORS-Part-3(Exercise-1)
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