11.Dual Nature of Radiation and matter
easy

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

A

$\frac{E}{{C\;}}$

B

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

C

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

D

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

(AIPMT-2015) (AIEEE-2004)

Solution

Energy of radiation, $E=m v=\frac{h C}{\lambda}$

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

$=p_{i}-p_{r}=\frac{E}{C}-\left(-\frac{E}{C}\right)=\frac{2 E}{C}$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.