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A 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 $p_{ph}$, $\frac{p_e}{p_{ph}}$ is :
(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 $= 9 \times 10^{-31}\text{ kg}$)

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
275

Energy Conversion

First, convert the given energy from electron volts (eV) to Joules (J). The conversion factor is $1 \text{ eV} \approx 1.6 \times 10^{-19} \text{ J}$.

Energy, $E = 20 \text{ eV} = 20 \times (1.6 \times 10^{-19} \text{ J}) = 3.2 \times 10^{-18} \text{ J}$.

Electron Momentum Calculation

The electron has kinetic energy $K_e = 3.2 \times 10^{-18}$ J. Its mass is $m_e = 9 \times 10^{-31}$ kg. Using the non-relativistic formula relating kinetic energy ($K_e$) and momentum ($p_e$): $K_e = \frac{p_e^2}{2m_e}$.

Rearranging to solve for momentum: $p_e = \sqrt{2 m_e K_e}$

Substitute the known values: $p_e = \sqrt{2 \times (9 \times 10^{-31} \text{ kg}) \times (3.2 \times 10^{-18} \text{ J})}$ $p_e = \sqrt{57.6 \times 10^{-49} \text{ kg}^2 \text{m}^2/\text{s}^2}$ $p_e = \sqrt{5.76 \times 10^{-48} \text{ kg}^2 \text{m}^2/\text{s}^2}$ $p_e = 2.4 \times 10^{-24} \text{ kg m/s}$

Photon Momentum Calculation

For a photon, energy ($E_{ph}$) and momentum ($p_{ph}$) are related by $E_{ph} = p_{ph}c$, where $c$ is the speed of light. The photon's energy is $E_{ph} = 3.2 \times 10^{-18}$ J, and the speed of light $c = 3 \times 10^8$ m/s.

Solving for momentum: $p_{ph} = \frac{E_{ph}}{c}$

Substitute the values: $p_{ph} = \frac{3.2 \times 10^{-18} \text{ J}}{3 \times 10^8 \text{ m/s}}$ $p_{ph} = \frac{3.2}{3} \times 10^{-26} \text{ kg m/s}$

Ratio of Momenta

Calculate the required ratio $\frac{p_e}{p_{ph}}$.

$ \frac{p_e}{p_{ph}} = \frac{2.4 \times 10^{-24} \text{ kg m/s}}{\frac{3.2}{3} \times 10^{-26} \text{ kg m/s}} $

Simplify the expression: $ \frac{p_e}{p_{ph}} = \frac{2.4 \times 3}{3.2} \times \frac{10^{-24}}{10^{-26}} $ $ \frac{p_e}{p_{ph}} = \frac{7.2}{3.2} \times 10^2 $ $ \frac{p_e}{p_{ph}} = 2.25 \times 100 $ $ \frac{p_e}{p_{ph}} = 225 $

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Important Questions from Modern Physics

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