A photon and an electron, each of $20$ eV energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{Ph}$, $\dfrac{p_e}{p_{Ph}}$ is :
(Take speed of light $= 3 \times 10^8\ \text{m s}^{-1}$, charge of electron $= -1.6 \times 10^{-19}$ C and mass of electron $= 9 \times 10^{-31}$ kg)
Answer: (C) $225$
$E = 20\ \text{eV} = 20 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-18}$ J.
Photon: $p_{Ph} = \dfrac{E}{c} = \dfrac{3.2 \times 10^{-18}}{3 \times 10^8} = 1.07 \times 10^{-26}\ \text{kg m s}^{-1}$
Electron (non-relativistic): $p_e = \sqrt{2mE} = \sqrt{2 \times 9 \times 10^{-31} \times 3.2 \times 10^{-18}} = \sqrt{5.76 \times 10^{-48}} = 2.4 \times 10^{-24}\ \text{kg m s}^{-1}$
$$\frac{p_e}{p_{Ph}} = \frac{\sqrt{2mE}}{E/c} = c\sqrt{\frac{2m}{E}} = \frac{2.4 \times 10^{-24} \times 3 \times 10^8}{3.2 \times 10^{-18}} = 225$$
Solution by Sreeraj P, M.Sc Physics