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

(N/A) The momentum $p$ of a photon is given by the relation $p = E / c$,where $E$ is the energy of the photon and $c$ is the speed of light.
Since the energy of a photon is $E = h \nu$,where $h$ is Planck's constant and $\nu$ is the frequency,we have $p = h \nu / c$.
We know that the speed of light is related to frequency and wavelength by $c = \nu \lambda$,which implies $\lambda = c / \nu$.
Substituting $c / \nu = \lambda$ into the momentum equation,we get $p = h / \lambda$.
Rearranging this gives the de Broglie wavelength formula: $\lambda = h / p$.
Thus,the de Broglie wavelength of a photon is exactly equal to the wavelength of the electromagnetic radiation of which the photon is the quantum.