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

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/ C2) (B)  (E/C2) (C) (E/ C) (D) (2E/ C)

Tardigrade Admin · Accepted Answer

Energy of radiation, E=h v=(h C/λ) Also, its momentum p=(h/λ)=(E/C)=pi pr=-pi=(E/C) So, momentum transferred to the surface pr=-pi=(E/C)-((E/c))=(2 E/C)

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

$\frac{2 E}{C ^{2}}$

$\frac{E}{C ^{2}}$

$\frac{E}{C}$

$\frac{2 E}{C}$

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

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

C22E​

C2E​

CE​