Q. A photon of wavelength $6630\ \mathring{\mathrm{A}}$ is incident on a totally reflecting surface. The momentum delivered by the photon is equal to
A) $6.63 \times 10^{- 27}kg$-$m/s$
B) $2 \times 10^{- 27}\ kg$-$m/s$
C) $10^{- 27}kg$-$m/s$
D) None of these
Explanation
(b) 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
$\therefore\mathrm{\Delta}p = 2p = \frac{2h}{\lambda} = \frac{2 \times 6.6 \times 10^{- 34}}{6630 \times 10^{- 10}} = 2 \times 10^{- 27}kg - m/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.