Solution :
Energy (E) of one photon $=E=hv$
Where $h= 6.626 \times 10^{-24}\;Js$
$v= 5 \times 1-^{14}s^{-1}$
$E= (6.626 \times 10^{-14}\;s^{-1})$
$\qquad= 3.313 \times 10^{-12}\;J$
Energy of 2 mole of photon $=(3.313 \times 10^{-19}\;J) \times (2 \times 6.022 \times 10^{23}\;mol^{-1})$
$\qquad= 3990.2 \;kJmol^{-1}$