A) \[1.0\times {{10}^{13}}atoms/c{{m}^{3}}\]
B) \[2.5\times {{10}^{20}}atoms/c{{m}^{3}}\]
C) \[2.5\times 10atoms/c{{m}^{3}}\]
D) \[1.0\times {{10}^{15}}atoms/c{{m}^{3}}\]
Correct Answer: D
Solution :
Given: Density of silicon atoms \[=5\times {{10}^{28}}atoms/{{m}^{3}}\] \[=5\times {{10}^{28}}\times {{10}^{-6}}atoms/c{{m}^{3}}\] \[=5\times {{10}^{22}}atoms/c{{m}^{3}}\] \[\therefore \]Number of accepted atoms \[=\frac{density\text{ }of\text{ }silicon\text{ }atoms}{density\text{ }of\text{ }indium\text{ }atoms}\] \[=\frac{5\times {{10}^{22}}}{5\times {{10}^{7}}}\] \[=1\times {{10}^{15}}atoms/c{{m}^{3}}\]
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