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The momentum of a photon of energy 1 MeV in kg m/s, will be -
A
$0.33 \times 10^{6}$
B
$7 \times 10^{-24}$
C
$10^{-22}$
D
$5 \times 10^{-22}$
Detailed Solution
For a photon, $E = pc$
$\Rightarrow p = \dfrac{E}{c}$
$E = 1$ MeV $= 10^{6} \times 1.6 \times 10^{-19}$ J $= 1.6 \times 10^{-13}$ J
$p = \dfrac{10^{6} \times 1.6 \times 10^{-19}}{3 \times 10^{8}}$
$p = 5.33 \times 10^{-22}$ kg m s$^{-1}$
$p \approx 5 \times 10^{-22}$ kg m/s
$\Rightarrow p = \dfrac{E}{c}$
$E = 1$ MeV $= 10^{6} \times 1.6 \times 10^{-19}$ J $= 1.6 \times 10^{-13}$ J
$p = \dfrac{10^{6} \times 1.6 \times 10^{-19}}{3 \times 10^{8}}$
$p = 5.33 \times 10^{-22}$ kg m s$^{-1}$
$p \approx 5 \times 10^{-22}$ kg m/s