Question

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

• A. $6.63\times 10^{-27}kg-m/sec$
• B. $2\times 10^{-27}kg-m/sec$
• C. $10^{-27}kg-m/sec$
• D. None of these

Answer 1

Answer: (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 its momentum without changing its magnitude
$\therefore \Delta p=2p=\frac{2h}{\lambda }$
$=\frac{2\times 6.6\times 10^{-34}}{6630\times 10^{-10}}$
$=2\times 10^{-27}kg-m/\sec$