Momentum of photon \(p=E / c\)
\(\therefore p = \frac{E}{c} = \) \(\frac{{1 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}{\text{J}}}}{{3 \times {{10}^8}{\text{m}}{{\text{s}}^{ - 1}}}}\) \( = 0.53 \times {10^{ - 21}}\)
\(=5 \times 10^{-22} \mathrm{kg} \mathrm{m} / \mathrm{s}\)
$\left(\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31}\;\mathrm{kg}\right)$