A) \[2.0\times {{10}^{-22}}J\] peratom
B) \[28.6\times {{10}^{-3}}kcal\text{ }mo{{l}^{-1}}\]photon
C) \[12.0\times {{10}^{-2}}kJ\text{ }mo{{l}^{-1}}\]photon
D) All of the above
Correct Answer: D
Solution :
\[v=\frac{1}{\lambda }\] \[10c{{m}^{-1}}=\frac{1}{\lambda }\]or \[\lambda =\frac{1}{10}cm=0.1\,cm=0.001m\] \[E=\frac{hc}{\lambda }=\frac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{0.001}=19.878\times {{10}^{-23}}\] \[\approx 2.0\times {{10}^{-22}}J/atom=12.04\times 10\text{ }J/mol\] \[=12.04\times {{10}^{-2}}kJ\text{ }mo{{l}^{-1}}=28.6\times {{10}^{-3}}kcal\text{ }mo{{l}^{-1}}\]
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