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Question: What should be the mass of the photon of sodium if its wavelength is 5894\(\mathop {\rm A}\limits^ \...

What should be the mass of the photon of sodium if its wavelength is 5894A∘\mathop {\rm A}\limits^ \circ ? (The velocity of light is 3×1083 \times {10^8} metre/second and the value of h is 6.6252×10−34kg.m2/s6.6252 \times {10^{ - 34}}kg.{m^2}/s.)
(A) 3.75×10−363.75 \times {10^{ - 36}} g
(B) 3.75×10−363.75 \times {10^{ - 36}} kg
(C) 1.25×10−361.25 \times {10^{ - 36}} kg
(D) 1.25×10−361.25 \times {10^{ - 36}} g

Explanation

Solution

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:
λ=hmc\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:
λ=hmc\lambda = \dfrac{h}{{mc}}-------- (1)
Where, λ = de-Broglie wavelength;
hh = Planck’s constant = 6.6252×10−34kg.m2/s6.6252 \times {10^{ - 34}}kg.{m^2}/s;
cc = velocity of light = 3×1083 \times {10^8} metre/second;
mm = mass of particle.
-The question gives us the value of wavelength is 5894A∘\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):
λλ = 5894A∘\mathop {\rm A}\limits^ \circ = 5894×10−105894 \times {10^{ - 10}} m
hh = 6.6252×10−34kg.m2/s6.6252 \times {10^{ - 34}}kg.{m^2}/s
cc = 3×1083 \times {10^8} metre/second
λ=hmc\lambda = \dfrac{h}{{mc}}
5894×10−10=6.6252×10−34m×3×1085894 \times {10^{ - 10}} = \dfrac{{6.6252 \times {{10}^{ - 34}}}}{{m \times 3 \times {{10}^8}}}
m=6.6252×10−343×108×5894×10−10m = \dfrac{{6.6252 \times {{10}^{ - 34}}}}{{3 \times {{10}^8} \times 5894 \times {{10}^{ - 10}}}}
= 6.6252×10−3417682×10−2\dfrac{{6.6252 \times {{10}^{ - 34}}}}{{17682 \times {{10}^{ - 2}}}}
= 3.746×10−363.746 \times {10^{ - 36}} kg
Hence we can now tell that the mass of the photon of sodium will be 3.746×10−363.746 \times {10^{ - 36}} kg.

So, the correct option will be: 3.746×10−363.746 \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:
λ=hmv\lambda = \dfrac{h}{{mv}}