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