First slide

Quantum Theory of Light

Question

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

Moderate

A

$6 .63 \times 10^{- 27} kg - m / \sec$
B

$2 \times 10^{- 27} kg - m / \sec$
C

$10^{- 27} kg - m / \sec$
D

none of these

Solution

The momentum of the incident radiation is given as $p = \frac{h}{λ}$ When the light is totally reflected normal to the surface the direction of the ray is reversed. That means it reverses the direction of its momentum without changing its magnitude. $∴Δp=2p= \frac{2 h}{λ} = \frac{2 \times 6 .6 \times 10^{- 34}}{6630 \times 10^{- 10}} = 2 \times 10^{- 27} kg - m / \sec$ .

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