# Homework Help: Frequency shift of emitted photon due to recoil

1. Aug 7, 2010

### Cruikshank

(From Bransden and Joachain Chapter 1 problem 21):
Show that the fractional change in the frequency of a photon absorbed or emitted by an atom initially at rest is (f-f0)/f = +- hf/(2Mc^2) where M is the mass of the atom and f0 is the frequency of the transition uncorrected for the recoil of the atom.

2. Relevant equations
Momentum: 0 = Mv+hf/c
Energy is not conserved; the internal energy of the atom changes (an electron jumps.)
Doppler effect for light: f = f0((1-v/c)/(1+v/c))^(1/2) =f0(Root)

3. The attempt at a solution
f-f0 = f0[Root - 1]

(f-f0)/f = f0[Root -1]/{(f0)(Root)} = 1- 1/Root = 1 - [(1+v/c)/(1-v/c)]^1/2
(1+v/c)/(1-v/c) = 1 + (2v/c)/(1-v/c)
Using binomial expansion and assuming hf<<Mc^2, the square root becomes roughly
1 + 1/2[(2v/c)/(1-v/c) = 1 + (v/c)/(1-v/c), and since v/c =hf/Mc^2
(f-f0)/f ~1 - [1 + (v/c)/(1-v/c)] = (-v/c)/(1-v/c) = -hf/(Mc^2) / [1 - hf/Mc^2]
= -hf/[Mc^2 - hf] ~ -hf/Mc^2
This is double the correct answer.
My best guess is that somehow an average velocity of the atom during the transition
process is what should be used for the v, giving v/2 assuming constant acceleration.
But I don't really understand the details of the emission process and what should go into
the Doppler formula during an acceleration.

My interpretation of a reply by Rajini in the "Gamma Ray Emission" thread:
If one instead assumes that the original energy E=hf of the uncorrected photon loses a
certain fraction to kinetic recoil (1/2mv^2), and that E(recoil)/E = (f-f0)/f, then one
can use momentum AND energy, giving
1/2Mv^2 = 1/2M(E/Mc)^2 = E^2/(2Mc^2). so 1/2Mv^2/E = E/(2mc^2), the right answer.
That looks quick and painless.

My question is, what is wrong with my solution? Is it that the atom accelerates through a distance d which is 1/2vt, where v is the final recoil speed and t the time of transition?

2. Aug 7, 2010

### kuruman

Incorrect. You need to conserve relativistic energy. Initially, you have an excited atom (more massive) and finally you have an atom in a lower state (less massive) and an emitted photon. You cannot do this problem without conserving energy.
You don't want to go there.

3. Aug 9, 2010

### Cruikshank

I think I get it. E0=hf0 = the excitation energy of the stationary atom. It turns into kinetic energy of recoil, and energy of emitted photon, and so E is conserved that way. I thought there was not enough information to use conservation of energy. Thank you.

On the other hand, my approach works just fine if one includes the average velocity during emission.