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 = h𝜈

Where E is the energy of Photon, 𝜈 is the frequency of electromagnetic radiation and h is Planck’s constant

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

we get

P = h𝜈/c

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

c = 𝜈𝜆

Where 𝜈 is the frequency and 𝜆 is wavelength of electromagnetic radiation having velocity c

So re arranging we get

𝜈 = c/𝜆

Putting the value of 𝜈 in equation P = h𝜈/c

We get

P = hc/λc = h/𝜆

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

𝜆 = h/P

Where h is Planck’s constant and P is momentum of electromagnetic radiation

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

𝜆 = h/mv

Where 𝜆 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

𝜆 = h/P

This is same as wavelength of electromagnetic radiation

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Calculate the de-Broglie wavelength of the electron orbitting in the n = 2 state of hydrogen atom.

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The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:

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