The momentum of a photon of energy \(1 \mathrm{MeV}\) in \(\mathrm{kg}-\mathrm{m} / \mathrm{s}\), will be

Energy of photon is given by
\(
E=\frac{h c}{\lambda}
\)
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
\(
E=\frac{h c}{h / p}=p c
\)
Or \(p=E / c\)
\(
\text {Given, } \begin{aligned}
&=1 \mathrm{MeV}=1 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}, \\
c &=3 \times 10^{8} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Hence, after putting numerical values, we obtain
\(
\begin{aligned}
p &=\frac{1 \times 10^{6} \times 1.6 \times 10^{-19}}{3 \times 10^{8}} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\
&=5 \times 10^{-22} \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
\)