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Q.

The momentum of photon of energy 1MeV will approximately be.

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a

$10^{- 22} k g - m / s$

b

$5 \times 10^{- 22} k g - m / s$

c

$3 \times 10^{- 6} k g - m / s$

d

0

answer is B.

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Detailed Solution

$p = \frac{E}{C} = \frac{1 \times 10^{6} \times 1.6 \times 10^{- 19}}{3 \times 10^{8}}$

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