Question

The mass of photon in motion is (given its frequency = x):
A. $\dfrac{{hx}}{{{c^2}}}$
B. $h{x^3}$
C. $\dfrac{{h{x^3}}}{{{c^2}}}$
D. Zero

Answer 1

Hint: Photons are discrete packets of energy emitted by a source of radiation. Calculate the energy of the photon by using Planck’s quantum theory and comparing it with the energy-mass equation, we can find the relation between mass of photon and its frequency and thus get the correct answer.

Complete step by step answer:
Planck’s quantum theory states that,
Radiation is emitted by different atoms or molecules in discrete quantities only. The smallest amount of energy that can be emitted or absorbed in the form of electromagnetic radiation is known as quantum which is associated with particles called photons.
The energy of the radiation absorbed or emitted is directly proportional to the frequency of radiation
Thus the energy of each photon, $E = h\nu $ , where $E = $ energy of photon; $h = $ Planck’s constant and $\nu = $ frequency of photon.
Now, we know that, $E = m{c^2}$ , where $E = $ energy of photon; $m = $ mass of particle and $c = $ speed of light.
This formula states that when an object is at rest, its energy is equal to the product of its mass with the speed of light squared.
On comparing the above two equations we get,
$
  h\nu = m{c^2} \\
  m = \dfrac{{h\nu }}{{{c^2}}} \\
 $
In this case, we are given that frequency = x
So, $m = \dfrac{{hx}}{{{c^2}}}$

So, the correct answer is “Option A”.

Additional Information:
$
  frequency = \dfrac{{speed}}{{wavelength}} \\
\implies \nu = \dfrac{c}{\lambda } \\
 $
Therefore, energy of photon, $E = h\nu = \dfrac{{hc}}{\lambda }$
As we observed that light exists as discrete packets of energy called photons, this confirms the particle nature of light.

Note:
Photons are electrically neutral so they are not deflected in electric or magnetic field. According to the theory of relativity, the mass m of a particle moving with velocity v comparable with velocity of light c is given by,
$m = \dfrac{{{m_0}}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}$
Thus, the rest mass of a photon is zero.