Question

A radiation of energy $'E'$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ( $c=$ velocity of light)
$A.\dfrac{2E}{{{c}^{2}}}$
$B.\dfrac{E}{{{c}^{2}}}$
$C.\dfrac{E}{c}$
$D.\dfrac{2E}{c}$

Answer 1

Concept of energy and momentum carried by photon will be used in this problem to get the required answer. In this case the change in momentum during reflection from the surface will be equal to the momentum transferred to the surface. Photon is defined as the small packets of energy of which the light is made up of.

Formula used:
To solve this problem related to photon we will use the following relation:-
$E=pc$.

Complete step-by-step answer:
We know that the energy carried by the photon is given by the following equation:-
$E=pc$ Which can also be written as follows:-
$p=\dfrac{E}{c}$ ……………….. $(i)$
Where $p$denotes momentum, $E$ is energy and $c$ is speed of light.
One important fact is that the momentum incidence on the surface is equal to the momentum reflected from the surface. But reflected momentum is in the opposite direction.
Therefore, change in momentum, $\vartriangle p={{p}_{i}}-{{p}_{r}}$
Where ${{p}_{i}}$ and ${{p}_{r}}$ are incident and reflected momentum respectively.
${{p}_{i}}=\dfrac{E}{c}$ ……………… $(ii)$
Since, ${{p}_{i}}={{p}_{r}}$
Hence ${{p}_{r}}=\dfrac{-E}{c}$ ………………… $(iii)$ ( ${{p}_{r}}$ is in opposite direction of ${{p}_{i}}$ )
But change in momentum is given as follows:-
$\vartriangle p={{p}_{i}}-{{p}_{r}}$ …………. $(iv)$
Putting value of $(ii)$ and $(iii)$ in $(iv)$, we have
$\Rightarrow \vartriangle p=\dfrac{E}{c}-\dfrac{-E}{c}$
$\Rightarrow \vartriangle p=\dfrac{E}{c}+\dfrac{E}{c}$
$\Rightarrow \vartriangle p=\dfrac{2E}{c}$
Hence, option $(D)$ is correct.

So, the correct answer is “Option D”.

Additional Information: Photon is an elementary particle and like other elementary particles they are explained by quantum mechanics. Photons exhibit dual nature. Photons are massless and have no electric charge. Photons move with the speed of light at vacuum. Photon is a stable particle.

Note: Concept of photon should be clear to us in order to solve this problem correctly. It is also very important to know that despite being a massless particle, a photon has momentum because it transports momentum via waving motion not with mass. We know that a photon moves with the velocity of light that’s why for its velocity, $c$ is used. The energy of a photon depends only on its frequency directly and inversely with its wavelength.