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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 $p_{Ph}$, $\frac{p_e}{p_{Ph}}$ is: [Take speed of light = $3\times10^8$ ms$^{-1}$, charge of electron = $-1.6\times10^{-19}$ C and mass of electron = $9\times10^{-31}$ kg]
Detailed Solution
With $E_e=E_{Ph}=E$: $\frac{p_e}{p_{Ph}}=\frac{\sqrt{2mE}}{E/c}=\sqrt{\frac{2m}{E}}\times c=\sqrt{\frac{2\times9\times10^{-31}}{20\times1.6\times10^{-19}}}\times3\times10^8=\frac{3}{4}\times10^{-6}\times3\times10^8=225$.