Explanation

(d) Energy of photon is given by
$E = \frac{hc}{\lambda}\ \ $…(i)
Where $h$ is the Planck’s constant, c the velocity of light and $\lambda$ its wavelength.
de-Broglie wavelength is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{h}{p}\ \ \ \ \ \ \ \ \ \ \ \ $
Where $p$ is being momentum of photon.
From Eqs. (i) and (ii), we get
$E = \frac{hc}{h/p} = pc$
Or $p = E/c$
Given, E = $1\ MeV = 1 \times 10^{6} \times 1.6 \times 10^{- 19}$ J.
$c = 3 \times 10^{8}\ {ms}^{- 1}$
Hence, after putting numerical values, we obtain
$p = \frac{1 \times 10^{6} \times 1.6 \times 10^{- 19}}{3 \times 10^{8}}$
= $5 \times 10^{- 22\ }kg$-${ms}^{- 1}$