Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
The De Broglie wavelength of the photon is given as,
λ = h m υ (1)
The momentum of a photon is given as,
p = m υ (2)
Where, the mass of photon is m , the wavelength of electromagnetic radiation is λ , the velocity of the photon is υ , Planck’s constant is h and the momentum of a photon is p .
By substituting the value from equation (2) to equation (1), we get
λ = h p (3)
The momentum of a photon is given as,
p = h ν c (4)
The frequency of a photon is given as,
ν = c λ (5)
Where, the frequency of a photon is ν , the speed of the light is c .
By substituting the value from equation (5) to equation (4), we get
λ = h p (6)
Thus, it can be observed from equation (6) and (3) that the wavelength of de Broglie wavelength of photon is equal to the wavelength of electromagnetic radiation.
The De Broglie wavelength of the photon is given as,
The momentum of a photon is given as,
Where, the mass of photon is
By substituting the value from equation (2) to equation (1), we get
The momentum of a photon is given as,
The frequency of a photon is given as,
Where, the frequency of a photon is
By substituting the value from equation (5) to equation (4), we get
Thus, it can be observed from equation (6) and (3) that the wavelength of de Broglie wavelength of photon is equal to the wavelength of electromagnetic radiation.
