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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

`10^19 "per m"^3`

B

`10^11 "per m"^3`

C

`10^5 "per m"^3`

D

`10^21 "per m"^3`

Text Solution

Verified by Experts

The correct Answer is:
B

For intrinsic semiconductor, `n_i=10^16//m^3`and `n_h=n_e` and when phosphorus (pentavalent impurity) is added, it becomes an n-type semiconductor.
`because n_e=10^21//m^3`
`therefore n_h=n_i^2/n_e=(10^16xx10^16)/10^21=10^11//m^3`
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