A) \[1{{0}^{17}}\,/{{m}^{3}}\]
B) \[1{{0}^{15}}\,/{{m}^{3}}\]
C) \[1{{0}^{4}}\,/{{m}^{3}}\]
D) \[1{{0}^{2}}\,/{{m}^{3}}\]
Correct Answer: A
Solution :
From law of mass-action, \[n_{i}^{2}={{n}_{e}}\times {{n}_{h}}\] where \[{{n}_{i}}\] is the concentration of electron-hole pair and \[{{n}_{h}}\] is the concentration of acceptor or holes. Given, \[{{\operatorname{n}}_{i}}=1{{0}^{19}}per\,{{m}^{3}},\,\,{{n}_{h}}=1{{0}^{21}}per\,{{m}^{3}}\] \[{{({{10}^{19}})}^{2}}=\,\,{{n}_{e}}\,\,\times \,\,{{10}^{21}}\] \[\Rightarrow \,\,\,\,\,\,{{n}_{e}}=\frac{{{10}^{38}}}{{{10}^{21}}}={{10}^{17}}\,per\,\,{{m}^{3}}\]
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