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)

$\frac{2E}{ C}$

Answer 2

Energy of radiation, $E = h \upsilon = \frac{hC}{\lambda}$
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
$p_r = - p_i = \frac{ E}{ C} - \bigg( \frac{E}{c} \bigg) = \frac{ 2E}{C}$