Question

The momentum of a photon of energy $1 \, MeV$ 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{h c}{\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 = $1 \, MeV=1\times 10^{6}\times 1.6\times 10^{- 19}$ 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 \, $ $m \, s^{- 1}$