Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Home » Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

The momentum of a photon having energy (hv) is given by the elation:

$p=\frac{h v}{c}=\frac{h}{\lambda}$

$\lambda=\frac{h}{p} \ldots \ldots \ldots$ (i)

Where,

$\lambda=$ wavelength of the electromagnetic radiation

$c=$ speed of light

$\mathbf{h}=$ Planck’s constant

De Broglie wavelength of the photon is given by the relation:

$\lambda=\frac{h}{m v}$

But, $\mathbf{p}=\mathbf{m v}$

Therefore, $\lambda=\frac{h}{p} \ldots \ldots \ldots \ldots$ (ii)

Where, $m=$ mass of the photon

$v=$ velocity of the photon

The wavelength of electromagnetic radiation and the de Broglie wavelength of the photon are same, according to equations (i) and (ii).

i

Sign up now and study all subjects for free