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. $\frac{2E}{ C^2}$
• B. $ \frac{E}{C^2}$
• C. $\frac{E}{ C}$
• D. $\frac{2E}{ C}$

Answer 1

Answer: (D)

Energy of radiation, $E=h v=\frac{h C}{\lambda}$
Also, its momentum $p=\frac{h}{\lambda}=\frac{E}{C}=p_{i}$
$p_{r}=-p_{i}=\frac{E}{C}$
So, momentum transferred to the surface
$p_{r}=-p_{i}=\frac{E}{C}-\left(\frac{E}{c}\right)=\frac{2 E}{C}$