A) \[2.5\times {{10}^{30}}atom/c{{m}^{3}}\]
B) \[2.5\times {{10}^{35}}atom/c{{m}^{3}}\]
C) \[1\times {{10}^{13}}atom/c{{m}^{3}}\]
D) \[1\times {{10}^{15}}atom/c{{m}^{3}}\]
E) none of the above
Correct Answer: D
Solution :
Number density of atoms in silicon specimen \[=5\times {{10}^{28}}atoms/{{m}^{3}}=5\times {{10}^{22}}atoms/c{{m}^{3}}\]. Since, 1 atom of indium is doped in\[5\times {{10}^{7}}\]silicon atoms, so total number of indium atoms doped per\[c{{m}^{3}}\]of silicon will be \[n=\frac{5\times {{10}^{22}}}{5\times {{10}^{7}}}={{10}^{15}}atoms/c{{m}^{3}}\]
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