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A radiation of energy $E$ falls on a perfectly reflecting surface. The momentum transferred to the surface is (let $c \equiv$ speed of light)
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The correct answer is:
$\frac{2 E}{c}$
The momentum carried by a photon of energy
$E$ and wavelength $\lambda$ is
$p=\frac{h}{\lambda}=\frac{h c}{\lambda c}=\frac{h c}{\lambda} \times \frac{1}{c}=E \times \frac{1}{c}=\frac{E}{c} \quad\left[\because \frac{h c}{\lambda}=E\right]$
where, $h=$ Planck's constant.
When the photon hits on a perfectly reflecting surface, it reflects back in the opposite direction and goes with the same energy and hence the same momentum.
Hence, the change in momentum of the photon
$\Delta p=\left(\frac{E}{c}\right)-\left(\frac{-E}{c}\right)=\frac{E}{c}+\frac{E}{c}=\frac{2 E}{c}$
This is equal to the momentum transferred by the photon to the surface.
$E$ and wavelength $\lambda$ is
$p=\frac{h}{\lambda}=\frac{h c}{\lambda c}=\frac{h c}{\lambda} \times \frac{1}{c}=E \times \frac{1}{c}=\frac{E}{c} \quad\left[\because \frac{h c}{\lambda}=E\right]$
where, $h=$ Planck's constant.
When the photon hits on a perfectly reflecting surface, it reflects back in the opposite direction and goes with the same energy and hence the same momentum.
Hence, the change in momentum of the photon
$\Delta p=\left(\frac{E}{c}\right)-\left(\frac{-E}{c}\right)=\frac{E}{c}+\frac{E}{c}=\frac{2 E}{c}$
This is equal to the momentum transferred by the photon to the surface.
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