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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
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