Question

What should be the mass of the photon of sodium if its wavelength is 5894$\mathop {\rm A}\limits^ \circ $? (The velocity of light is $3 \times {10^8}$ metre/second and the value of h is $6.6252 \times {10^{ - 34}}kg.{m^2}/s$.)
(A) $3.75 \times {10^{ - 36}}$ g
(B) $3.75 \times {10^{ - 36}}$ kg
(C) $1.25 \times {10^{ - 36}}$ kg
(D) $1.25 \times {10^{ - 36}}$ g

Answer 1

Hint: The question gives us the value of wavelength, speed and Planck’s constant. We will calculate the mass of the photon using de Broglie’s equation:
$\lambda = \dfrac{h}{{mc}}$

Complete step by step solution:
-First of all, let us talk about the de-Broglie equation for a photon.
The de-Broglie equation describes the wave nature of an electron. An electromagnetic equation exhibits dual nature: of a particle because it has momentum and wave because it has both wavelength and frequency. The de-Broglie equation exhibits the relationship between the momentum of a particle and its wavelength and so the wavelength is known as de-Broglie wavelength. Mathematically this equation for a photon is:
$\lambda = \dfrac{h}{{mc}}$-------- (1)
Where, λ = de-Broglie wavelength;
$h$ = Planck’s constant = $6.6252 \times {10^{ - 34}}kg.{m^2}/s$;
$c$ = velocity of light = $3 \times {10^8}$ metre/second;
$m$ = mass of particle.
-The question gives us the value of wavelength is 5894$\mathop {\rm A}\limits^ \circ $ and we need to calculate the mass of the photon. We will do this using the de-Broglie equation (1):
$λ$ = 5894$\mathop {\rm A}\limits^ \circ $ = $5894 \times {10^{ - 10}}$ m
$h$ = $6.6252 \times {10^{ - 34}}kg.{m^2}/s$
$c$ = $3 \times {10^8}$ metre/second
$\lambda = \dfrac{h}{{mc}}$
$5894 \times {10^{ - 10}} = \dfrac{{6.6252 \times {{10}^{ - 34}}}}{{m \times 3 \times {{10}^8}}}$
$m = \dfrac{{6.6252 \times {{10}^{ - 34}}}}{{3 \times {{10}^8} \times 5894 \times {{10}^{ - 10}}}}$
= $\dfrac{{6.6252 \times {{10}^{ - 34}}}}{{17682 \times {{10}^{ - 2}}}}$
= $3.746 \times {10^{ - 36}}$ kg
Hence we can now tell that the mass of the photon of sodium will be $3.746 \times {10^{ - 36}}$ kg.

So, the correct option will be: $3.746 \times {10^{ - 36}}$ kg.

Note: The mass ‘m’ we calculate here is the relativistic mass and not the rest mass because the rest mass of a photon is always zero (0).
Also if a particle moves with velocity v, the momentum of the particle will be: p = mv and the de-Broglie wavelength will be:
$\lambda = \dfrac{h}{{mv}}$