Question

What should be the mass of the photon of sodium if its wavelength is 5894$\stackrel{\circ }{\mathrm{A}}$? (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:
$\lambda ={\displaystyle \frac{h}{mc}}$

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:
$\lambda ={\displaystyle \frac{h}{mc}}$-------- (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$\stackrel{\circ }{\mathrm{A}}$ and we need to calculate the mass of the photon. We will do this using the de-Broglie equation (1):
$\lambda$ = 5894$\stackrel{\circ }{\mathrm{A}}$ = $5894\times {10}^{-10}$ m
$h$ = $6.6252\times {10}^{-34}kg.m^2/s$
$c$ = $3\times {10}^8$ metre/second
$\lambda ={\displaystyle \frac{h}{mc}}$
$5894\times {10}^{-10}={\displaystyle \frac{6.6252\times {10}^{-34}}{m\times 3\times {10}^8}}$
$m={\displaystyle \frac{6.6252\times {10}^{-34}}{3\times {10}^8\times 5894\times {10}^{-10}}}$
= ${\displaystyle \frac{6.6252\times {10}^{-34}}{17682\times {10}^{-2}}}$
= $3.746\times {10}^{-36}$ 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:
$\lambda ={\displaystyle \frac{h}{mv}}$