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