A) \[{{10}^{17}}/{{m}^{3}}\]
B) \[{{10}^{15}}/{{m}^{3}}\]
C) \[{{10}^{4}}/{{m}^{3}}\]
D) \[{{10}^{2}}/{{m}^{3}}\]
Correct Answer: A
Solution :
From law of mass-action \[N_{i}^{2}={{n}_{e}}\times {{n}_{h}}\] where \[{{n}_{i}}\], is concentration of electron-hole pair and \[{{n}_{h}}\] is concentration of acceptor or holes. Given, \[{{n}_{i}}={{10}^{19}}\,per\,{{m}^{3}},\,\,{{n}_{h}}={{10}^{21}}\,per\,{{m}^{3}}\] \[{{({{10}^{19}})}^{2}}={{n}_{e}}\times {{10}^{21}}\] \[\Rightarrow \] \[{{n}_{e}}=\frac{{{10}^{38}}}{{{10}^{21}}}={{10}^{17}}\,per\,{{m}^{3}}\]
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