Question

The momentum of a photon of energy $1MeV$ in $kg-m/s$

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