A) \[{{10}^{21}}\,per\,\,{{m}^{3}}\]
B) \[{{10}^{19}}\,per\,\,{{m}^{3}}\]
C) \[{{10}^{11}}\,per\,\,{{m}^{3}}\]
D) \[{{10}^{5}}\,per\,\,{{m}^{3}}\]
Correct Answer: C
Solution :
By using mass-action law \[n_{1}^{2}={{n}_{e}}{{n}_{h}}\] \[\Rightarrow \] \[{{n}_{h}}=\frac{n_{1}^{2}}{{{n}_{e}}}=\frac{{{\left( {{10}^{16}} \right)}^{2}}}{{{10}^{21}}}={{10}^{11}}per\,\,{{m}^{3}}.\]You need to login to perform this action.
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