A) \[2.5\times {{10}^{30}}\,atom/c{{m}^{3}}\]
B) \[2.5\times {{10}^{35}}\,atom/c{{m}^{3}}\]
C) \[1.0\times {{10}^{13}}\,atom/c{{m}^{3}}\]
D) \[1.0\times {{10}^{15}}\,atom/c{{m}^{3}}\]
Correct Answer: D
Solution :
Number density of atoms in silicon specimen \[=5\times {{10}^{28}}\text{ }atom/{{m}^{3}}\] \[=5\times {{10}^{22}}\text{ atom/c}{{\text{m}}^{\text{3}}}\] 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 \[n=\frac{5\times {{10}^{22}}}{5\times {{10}^{7}}}\] \[={{10}^{15}}\,\text{atom/c}{{\text{m}}^{\text{3}}}\]
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