Question

The momentum of a photon of energy $1MeV$ in kg-m/s, will be equal to :

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

Answer 1

The correct option is D $5\times {10}^{-22}$

Energy of photon is given by
$E=\frac{hc}{\lambda }............\left(i\right)$

where $h$ is the Planck's constant, $c$ the velocity of light and $\lambda$ its wavelength.

de-Broglie wavelength is given by

$\lambda =\frac{h}{p}............\left(ii\right)$

where $p$ is being momentum of photon.

From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get

$E=\frac{hc}{h/p}=pc$

or $p=E/c$

Given $E=1$ $Mev=1\times {10}^6\times 1.6\times {10}^{-19}J$,

$c=3\times {10}^{8}m/s$

Hence, after putting numerical values we obtain

$p=\frac{1\times {10}^6\times 1.6\times {10}^{-19}}{3\times {10}^8}kg-m/s$

$=5\times {10}^{-22}kg-m/s$