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In a pure silicon \mathrm{\left(n_i=10^{16} / m^3\right)} crystal at \mathrm{300\ K, 10^{21}} atoms of phosphorus are added per cubic meter. The new hole concentration will be

Option: 1

\mathrm{10^{21}\ per\ m^{3}}


Option: 2

\mathrm{10^{19}\ per\ m^{3}}


Option: 3

\mathrm{10^{11}\ per\ m^{3}}


Option: 4

\mathrm{10^{5}\ per\ m^{3}}


Answers (1)

best_answer

By using mass action law \mathrm{n_i^2=n_e n_h}
\mathrm{\Rightarrow n_h=\frac{n_i^2}{n_e}=\frac{\left(10^{16}\right)^2}{10^{21}}=10^{11} \text { per } \mathrm{m}^3}

Posted by

HARSH KANKARIA

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