Question

A photon of wavelength $6600\dot{A}$ 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. $3\times {10}^{-27}kg-m/sec.$

Answer 1

The correct option is B $2\times {10}^{-27}kg-m/sec.$

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
$\therefore \Delta p=2p=\frac{2h}{\lambda }=\frac{2\times 6.6\times {10}^{-34}}{6600\times {10}^{-10}}=2\times {10}^{-27}kg-m/sec.$