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) $225$
Concept. Photon $p = \dfrac{E}{c}$; electron (non-relativistic) $p = \sqrt{2mE}$.
Step 1. Energy. $E = 20 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-18}$ J.
Step 2. Electron momentum. $\sqrt{2 \times 9 \times 10^{-31} \times 3.2 \times 10^{-18}} = 2.4 \times 10^{-24}$ kg m s$^{-1}$.
Step 3. Ratio.
$$\frac{p_e}{p_{Ph}} = \frac{2.4 \times 10^{-24} \times 3 \times 10^8}{3.2 \times 10^{-18}} = 225$$
Answer. $225$ (Option C).
Solution by Sreeraj P, M.Sc Physics