The momentum of a photon associated with a radiation of frequency $2 × 10^{13} Hz$ is

(Given: $h = 6.6 × 10^{-34} Js$ and $c = 3×10^8 m/s$)

$4.4 × 10^{-29}\, kg\, m/s$

The correct answer is Option (4) → $4.4 × 10^{-29}\, kg\, m/s$

Given:

Frequency: $f = 2 \times 10^{13}$ Hz

Planck's constant: $h = 6.6 \times 10^{-34}$ Js

Speed of light: $c = 3 \times 10^8$ m/s

Momentum of a photon: $p = \frac{h f}{c}$

$p = \frac{6.6 \times 10^{-34} \times 2 \times 10^{13}}{3 \times 10^8}$

$p = \frac{13.2 \times 10^{-21}}{3 \times 10^8}$

$p = 4.4 \times 10^{-29}$ kg·m/s

Answer: $p = 4.4 \times 10^{-29}$ kg·m/s