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

Explanation

For photon de-Broglie wavelength `lambda=(h)/(p)` ,
Momentum of electromagnetic radiation having v frequency and `lambda` wavelength,
`p=(e )/(c )=(hv)/(c )`
`therefore p=(h)/(c ).(c )/(lambda.)[because v=(c )/(lambda.)]`
`therefore p=(h)/(lambda.)`
`therefore lambda.=(h)/(p)`
but `(h)/(p)=lambda` de-Broglie wavelength of photon
`therefore lambda.=lambda`
Thus `lambda.` is wavelength of electromagnetic radiation which is equl to de-Broglie wavelength of phton with wavelength `lambda`.