A photon of wavelength 6630 Å is incident on a totally reflecting surface. The momentum delivered by the photon is equal to

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10^–27 kg-m/sec

None of these

6.63 x 10^–27 kg-m/sec

2 x 10^–27 kg-m/sec

answer is B.

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Detailed Solution

The momentum of the incident radiation is given as $p = \frac{h}{\lambda }$ . When the light is totally reflected normal to the surface the direction of the ray is reversed. That means it reverses the direction of it’s momentum without changing it’s magnitude
$\therefore \Delta p = 2p = \frac{{2h}}{\lambda } = \frac{{2 \times 6.6 \times {{10}^{ - 34}}}}{{6630 \times {{10}^{ - 10}}}} = {\text{ }}2 \times {10^{--27}}kg - m/sec$

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