Question

The rest mass of a photon is
(A) zero
(B) $ 1.6 \times {10^{ - 19}}kg $
(C) $ 3.1 \times {10^{ - 30}}kg $
(D) $ 9.1 \times {10^{ - 31}}kg $

Answer 1

Hint:
According to quantum theory by Einstein, a photon is a type of an elementary particle that exhibits wave-particle duality as they have both the features of waves and particles. They are the quantum form of electromagnetic waves such as light waves and are considered to have a rest mass of zero and move at the speed of light.
Formula used: In this problem, we will be using the following formula,
 $\Rightarrow m = \dfrac{{{m_o}}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }} $
where $ m $ is the mass of the particle at its speed $ v $
 $ {m_o} $ is the rest mass of the particle and $ c $ is the speed of light.

Complete step by step answer:
From Einstein’s quantum theory, it is said that light can have both the wave and the particle nature and hence exhibit wave-particle duality. From the particle nature of light, we consider light to travel in the form of small packets of energy or quanta of energy. These small packets are called photons.
The photons are said to be chargeless and massless particles that travel at the speed of light.
So the rest mass of a photon is taken to be zero. This can be proved from Einstein’s special theory of relativity.
From the mass variation formula we have,
 $\Rightarrow m = \dfrac{{{m_o}}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }} $
from here we can write the equation in terms of the rest mass as,
 $\Rightarrow {m_o} = m\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} $
So now photons are considered to be the particle nature of light. So the speed of photons is equal to the speed of light and is given by $ c $ . Therefore if we put the speed of the electrons as $ v = c $ then we will have,
 $\Rightarrow {m_o} = m\sqrt {1 - \dfrac{{{c^2}}}{{{c^2}}}} $
Cancelling the $ {c^2} $ from the numerator and the denominator we have,
 $\Rightarrow {m_o} = m\sqrt {1 - 1} $
 $\Rightarrow {m_o} = m \times 0 $
From here we get the rest mass of photons as
 $\Rightarrow {m_o} = 0 $
Hence the photons have a rest mass of zero.
So the correct answer will be option (A); zero.

Note:
As the photons obey the laws of quantum mechanics, they have both wave-like and particle-like behaviour. It is registered as a particle when a measuring device is used for it. And while detecting the probability, it is considered as a wave.