Q. 183.7( 15 Votes )

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

Answer :

we know the relation between energy and momentum P of a electromagnetic radiation having velocity C and energy E is


P = E/C


and we know energy of electromagnetic radiation is given by relation


E = h饾湀


Where E is the energy of Photon, 饾湀 is the frequency of electromagnetic radiation and h is Planck鈥檚 constant


So putting value of E in equation P = E/c


we get


P = h饾湀/c


We know relation between frequency wavelength and velocity of electromagnetic radiation as


c = 饾湀饾渾


Where 饾湀 is the frequency and 饾渾 is wavelength of electromagnetic radiation having velocity c


So re arranging we get


饾湀 = c/饾渾


Putting the value of 饾湀 in equation P = h饾湀/c


We get


P = hc/位c = h/饾渾


Or we can say the wavelength of electromagnetic radiation is given as


饾渾 = h/P


Where h is Planck鈥檚 constant and P is momentum of electromagnetic radiation


Now we know de Broglie wavelength of quantum (photon) can be given by relation


饾渾 = h/mv


Where 饾渾 is de Broglie wavelength of a photon having mass m and moving with velocity v (here the velocity of photon v = C)


But we know momentum is given by relation


P = mv.


So, substituting we get de Broglie wavelength of quantum (photon) as


饾渾 = h/P


This is same as wavelength of electromagnetic radiation


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