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A photon of wavelength 6630 $\overset{0}{A}$ is incident on a totally reflecting surface. Momentum delivered by the photon is equal to

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6.63 $\times 10^{- 27} k g m / s$

2 $\times 10^{- 27} k g m / s$

$10^{- 27} k g m / s$

4 $\times 10^{- 27} k g m / s$

answer is B.

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Detailed Solution

momentum of the incident photon is $P = \frac{h}{λ}$ as it is a totally reflecting surface the change in momentum of the photon is 2P

so $∆ P = 2 P = 2 \frac{h}{λ} = 2 \frac{6.63 \times 10^{- 34}}{6630 \times 10^{- 10}} = 2 \times 10^{- 27}$

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