Explanation

(d)
Ionisation energy of nitrogen =energy of photon
$= Nh\frac{c}{\lambda}$
where, $N = 6.02 \times 10^{23}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c = 3 \times 10^{8}ms^{- 1}$
$\lambda = 854\ \mathring{\mathrm{A}} = 854\ \times 10^{- 10}m$
$= \frac{6.02 \times 10^{23} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{854 \times 10^{- 10}}$
$= 1.4 \times 10^{6}\ J\ mol^{- 1}$
$= 1.4 \times 10^{3}\ kJ\ mol^{- 1}$