A) \[3.2\times {{10}^{16}}\]
B) \[1.6\times {{10}^{16}}\]
C) \[4\times {{10}^{16}}\]
D) None of these
Correct Answer: A
Solution :
Energy of each photon \[{{E}_{0}}=\frac{hc}{\lambda }\] \[=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{632.2\times {{10}^{-9}}}\] \[=3.15\times {{10}^{-19}}J\] Total energy \[=E=Pt\] \[\therefore \] \[n{{E}_{0}}=Pt\] \[\Rightarrow \] \[n=\frac{pt}{{{E}_{0}}}\] \[=\frac{5\times {{10}^{-3}}\times 2}{3.15\times {{10}^{-19}}}\] \[=3.2\times {{10}^{16}}\]You need to login to perform this action.
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