In a pure Silicon (ni = (10)16 m- 3) crystal at 300 K, 1021 atoms of phosphorus are added per cubic meter. The new hole concentration will be

Question

In a pure Silicon (ni = (10)16 m- 3) crystal at 300 K, 1021 atoms of phosphorus are added per cubic meter. The new hole concentration will be (A) 1021 textperm3 (B) 1019 textperm3 (C) 1011 textperm3 (D) 105 textperm3

Tardigrade Admin · Accepted Answer

By using mass action law ni2=nenh ⇒ nh=(ni2/ne)=(((10)16)2/(10)21)=(10)11 per m3

Q. In a pure Silicon $(n_{i} = (10)^{16} m^{- 3})$ crystal at $300 K, 1 0^{21}$ atoms of phosphorus are added per cubic meter. The new hole concentration will be

$1 0^{21} per m^{3}$

$1 0^{19} per m^{3}$

$1 0^{11} per m^{3}$

$1 0^{5} per m^{3}$

By using mass action law $n_{i}^{2} = n_{e} n_{h}$
$\Rightarrow n_{h} = \frac{n _{i}^{2}}{n _{e}} = \frac{( ( 10 ) ^{16} ) ^{2}}{( 10 ) ^{21}} = (10)^{11} p er m^{3}$

Q. In a pure Silicon (ni​=(10)16m−3) crystal at 300K,1021 atoms of phosphorus are added per cubic meter. The new hole concentration will be

1021perm3

1019perm3

1011perm3

105perm3