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Carrier Concentration

In a pure sil...

Question

In a pure silicon $(n_{i} = 10^{16} / m^{3})$ crystal at $300 K, 10^{21}$ atoms of phosphorous area added per cubic meter. The new whole concentration will be

10^{21} p e r m^{3}

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10^{19} p e r m^{3}

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10^{11} p e r m^{3}

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10^{5} p e r m^{3}

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Solution

The correct option is D $10^{11} p e r m^{3}$
Given:-
Temperature = $300 K$
$n_{i} = 10^{16} / m^{3}$
Atoms of phosphorous = $10^{21} a t o m s$
By using mass action law $n_{i}^{2} = n_{e} n_{h}$ $\Rightarrow n_{h} = \frac{n_{i}^{2}}{n_{e}} = \frac{(10^{16})^{2}}{10^{11}} = 10^{11} p e r m^{3}$

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