A) \[199.51\text{ }kJmo{{l}^{-1}}\]
B) \[189.51\text{ }kJmo{{l}^{-1}}\]
C) \[198.51\text{ }kJmo{{l}^{-1}}\]
D) \[188.51\text{ }kJmo{{l}^{-1}}\]
Correct Answer: A
Solution :
Energy of 1 photon,\[E=hv\] \[\therefore \] Energy of 1 mole of photon, \[E={{N}_{A}}hv\] \[=6.022\times {{10}^{23}}\times 6.626\times {{10}^{-34}}\times 5\times {{10}^{14}}\] \[=199.508\times {{10}^{3}}J\,mo{{l}^{-1}}\] \[=199.51\text{ }kJmo{{l}^{-1}}\]
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