Answer
(a) $79.8\,kJ/mol$
(b) $239\,kJ/mol$
(c) $798\,kJ/mol$
Work Step by Step
Recall: Energy of one photon= $\frac{hc}{\lambda}$
Energy of one mole of photons=
$\frac{hc}{\lambda}\times6.022\times10^{23}/mol$
(a) $E=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{1500\times10^{-9}\,m}\times6.022\times10^{23}/mol$
$=79.8\times10^{3}\,J/mol=79.8\,kJ/mol$
(b) $E=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{500\times10^{-9}\,m}\times6.022\times10^{23}/mol$
$=239\times10^{3}\,J/mol=239\,kJ/mol$
(c) $E=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{1500\times10^{-9}\,m}\times6.022\times10^{23}$
$=798\times10^{3}\,J/mol=798\,kJ/mol$
