Physics - Modern Physics Question with Solution | TestHub

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 =$1 \mathrm{M}\mathrm{e}\mathrm{V}=1\times {10}^6\times 1.6\times {10}^{-19}$J.

$c=3\times {10}^8 {\mathrm{m }\mathrm{s}}^{-1}$

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 }\mathrm{k}\mathrm{g }$${\mathrm{m }\mathrm{s}}^{-1}$