Question

The momentum of a photon of energy 1MeV in kg.m/s will be:

• 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 proton is given by,

$E=\frac{hC}{\lambda }$ (1)

debroglie W.L is given by,

$\lambda =\frac{h}{P}$ (2)

From 1 and 2

$E=PC$

$P=\frac{E}{C}$

Given: $E=1MeV=1\times {10}^6\times 1.6\times {10}^{-19}$

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

$\therefore P=\frac{1\times {10}^6\times 1.6\times {10}^{-19}}{3\times {10}^8}Kgm/s$

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