In pure germanium ni=1018/m3 crystal at 500K, 1024 atoms phosphorus are added per cubic meter. The hole concentration will be

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In pure germanium $(n_{i} = 10^{18} / m^{3})$ crystal at 500K, $10^{24}$ atoms phosphorus are added per cubic meter. The hole concentration will be

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$10^{12}$

$10^{24}$

$10^{14}$

$10^{36}$

answer is C.

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Detailed Solution

${n_{i}}^{2} = n_{e} n_{h} \Rightarrow n_{h} = \frac{{(10^{18})}^{2}}{10^{24}} = \frac{10^{36}}{10^{24}} = 10^{12} m^{- 3}$

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