Q 12-11-027NEETJEE MainEasy
The momentum of a photon of energy $3$ eV is ($1$ eV $= 1.6 \times 10^{-19}$ J)
Answer: (C) $1.6 \times 10^{-27}$ kg m/s
$p = \dfrac{E}{c} = \dfrac{3 \times 1.6 \times 10^{-19}}{3 \times 10^8} = 1.6 \times 10^{-27}$ kg m/s.
Solution by Sreeraj P, M.Sc Physics