According to the de Broglie relation, the wavelength $\lambda$ associated with a particle of momentum $p$ is given by $\lambda = h/p$. This relation is satisfied by photons as well.

For a photon, the momentum $p$ is given by $p = hv/c$, where $h$ is Planck's constant, $v$ is the frequency, and $c$ is the speed of light. Substituting this into the de Broglie relation, we get:

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

Since the wavelength of electromagnetic radiation is defined as $c/v$, the de Broglie wavelength of the photon is equal to the wavelength of the electromagnetic radiation.