Question

A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ($C$ = velocity of light)

• A. $2E/C^2$
• B. $2E/C^2$
• C. $E/C$
• D. $2E/C$

Answer 1

The correct option is D $2E/C$

Energy of radiation, $E=\frac{hc}{\lambda }$
Momentum of incident radiation$p=\frac{h}{\lambda }=\frac{E}{c}=p_i$
Momentum of reflected radiation$p_r=-p_i=\frac{-E}{c}$
So, momentum transfered to the surface
$p_i-p_r=\frac{E}{c}-\left(-\frac{E}{c}\right)=\frac{2E}{c}$