Question

The momentum of a photon of energy $1MeV$ in $kg$ $m{s}^{-1}$ , will be

• A. $0.33\times 10^6$
• B. $7\times 10^{-24}$
• C. $10^{-22}$
• D. $5\times 10^{-22}$

Answer 1

Answer: (D)

Energy of photon is given by
$E=\frac{hc}{\lambda }$ ...(i)
Where $h$ is the Planck's constant, c the velocity of light and $\lambda$ its wavelength.
de-Broglie wavelength is given by
$\lambda =\frac{h}{p}$
Where $p$ is being momentum of photon.
From Eqs. (i) and (ii), we get
$p=\frac{E}{c}$
Or $E=p\times c$
Given, E = $1MeV=1\times 10^6\times 1.6\times 10^{-19}$ J.
$c=3\times 10^{8}m{s}^{-1}$
Hence, after putting numerical values, we obtain
$p=\frac{1\times 10^6\times 1.6\times 10^{-19}}{3\times 10^8}$
= $5\times 10^{-22}kg$ $m{s}^{-1}$