The linear momentum of a 3 MeV photon is:

0.01

e V s m^{- 1}

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0.02

e V s m^{- 1}

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0.03

e V s m^{- 1}

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0.04

e V s m^{- 1}

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Solution

The correct option is A 0.01 $e V s m^{- 1}$
Energy of photon, $E = 3 M e V = 3 \times 10^{6} e V$

Linear momentum of the photon, $p = \frac{E}{c}$
where c is the speed of light in vacuum

$p = \frac{3 \times 10^{6} e V}{3 \times 10^{8} m s^{- 1}} = 10^{- 2} e V s m^{- 1} = 0.01 e V s m^{- 1}$

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