A) 100
B) 1000
C) \[3\times {{10}^{20}}\]
D) \[3\times {{10}^{18}}\]
Correct Answer: C
Solution :
\[p=\frac{nhc}{\lambda t}\Rightarrow 100=\frac{n\times 6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{540\times {{10}^{-9}}\times 1}\]Þ\[n=3\times {{10}^{20}}\]
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