(d)
Energy required for 1 $Cl_{2}$ molecule $= \frac{242 \times 10^{3}}{N_{A}}J$
$E = \frac{hc}{\lambda}$
$or\ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{- 34} \times 3 \times 10^{8} \times 6.02 \times 10^{23}}{242 \times 10^{3}}$
$= 494 \times 10^{- 9}m = 494\ nm$