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Physics · PYQ NEET UG 2026
हिन्दीA photon and an electron, each of \( 20\text{ eV} \) energy, move in free space. The ratio of linear momentum of electron \( p_e \) to that of photon \( p_{\text{ph}} \) (\( \frac{p_e}{p_{\text{ph}}} \)) is: \n(Take speed of light \( = 3 \times 10^8\text{ ms}^{-1} \), charge of electron \( = -1.6 \times 10^{-19}\text{ C} \) and mass of electron \( = 9 \times 10^{-31}\text{ kg} \))
Correct answer · Option 2
\( \frac{1}{250} \)
Explanation
For photon, \( p_{\text{ph}} = \frac{E}{c} \). For electron, \( p_e = \sqrt{2mE} \). Taking the ratio \( \frac{p_e}{p_{\text{ph}}} = \frac{c \sqrt{2mE}}{E} = c \sqrt{\frac{2m}{E}} \), substituting values gives \( \frac{1}{250} \).
