What is the correction to the wavelength of an emitted photon due to recoil?

[kent davidge]

Homework Statement

When a photon is emitted by an atom, the atom must recoil to conserve momentum. This means that the photon and the recoiling atom share the transition energy. (a) For an atom with mass m, calculate the correction Δλ due to recoil to the wavelength of an emitted photon. Let λ be the wavelength of the photon if recoil is not taken into consideration. (Hint: The correction is very small, so use this fact to obtain an approximate but very accurate expression for Δλ.)

Homework Equations

The Attempt at a Solution

[/B]
E = energy of system
E' = energy of the photon
K = kinetic energy of the atom
Conservation of energy gives: E_initial = E_final
0 = K + E'
0.5 x mv² = - (hc / λ)

By conservation of momentum:
0 = p_photon + p_atom
mv = - (h / λ)

Solving the above equations, we get:
λ = h / 2mc

The book's answer is h / 2mc. The problem is that the speed found here is twice greater than c.

 

[phyzguy]

This isn't correct. It's true that the momenta of the photon and the atom are equal and opposite. But it isn't true that the energy of the photon is equal to the energy of the atom. In fact the energy of the atom is much less than the energy of the photon. You are supposed to calculate how much the emitted energy of the photon is reduced by the momentum carried by the atom as compared to the energy of the photon if the atom doesn't carry away any energy.

 

[kent davidge]

I noticed that the equation for Δλ in this case is the same as for Compton scattering if we take angles such as the direction of our "new" wavelength is the same as the "older" wavelength. I've tried working this problem several different ways and I don't get the Δλ. It's annoying.

 

[haruspex]

I've tried working this problem several different ways.

Perhaps you could post one.
Did you understand phyzguy's comment? Do you have some revisions as a result?

 

[kent davidge]

phyzguy thank you for remembering me that the change in the wavelength is due to the change in the energy.
haruspex I think what I posted below is finally the correct answer.

I got it

E_{atom before} = E_{atom final} + E_photon
(mc² - p_photonc)² = (mc²)² + (p_atomc)²
...
p_photon / 2mc = -(p_atom² / p_photon2mc) + 1
Using the fact that the momentum of the atom is equal to the negative momentum of the photon and p = h / λ, results:
λ = h / mc
Now, the atom doesn't move at all:
p_photon / 2mc = -0 + 1
and
λ' = h / 2mc

|Δλ| is then λ' - λ = h / 2mc.