Question:
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Solution:
The momentum of a photon having energy (hν) is given as:
p=\frac{h v}{c}=\frac{h}{\lambda}
$\lambda=\frac{h}{p}$ ...(i)
Where,
λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:
$\lambda=\frac{h}{m v}$
But $p=m v$
$\therefore \lambda=\frac{h}{p}$ ...(ii)
Where,
m = Mass of the photon
v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.