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In a pure silicon (n(i)=10^(16)//m^(3)) ...

In a pure silicon `(n_(i)=10^(16)//m^(3))` crystal at `300K, 10^(21)` atoms of phosphorus are added per cubic meter. The new hole concentration will be

A

`10^(21)" per "m^(3)`

B

`10^(19)" per "m^(3)`

C

`10^(11)" per "m^(3)`

D

`10^(5)" per "m ^(3)`

Text Solution

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The correct Answer is:
C
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