# Change in photon emission

1. Sep 18, 2009

### th5418

1. The problem statement, all variables and given/known data
Show that the fractional change in frequency of a photon absorbed or emitted by an atom initially at rest is
$$\frac{\nu - \nu_o}{\nu} = \pm \frac{h\nu}{2Mc^{2}}$$
where M is the mass of the atom, $$\nu_o$$ is the frequency of the transition uncorrected for the recoil of the atom. In the above equation, the plus sign corresponds to absorption and the minus sign to the emission of a photon.

2. Relevant equations
$$E=h\nu$$
$$p=\frac{h\nu}{c}$$
$$KE = \frac{p^{2}}{2M}$$
$$E = \frac{-m_e}{2h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s^{2}}(\frac{Ze^{2}}{4\pi\epsilon_o}\frac{1}{n^{2}})^{2}$$

3. The attempt at a solution
I'm suppose to do some kind of energy conservation, but I cannot figure it out. What exactly is conserving energy? The atom and the photon released? There is also some kind of momentum conservation?

2. Sep 18, 2009

### Dick

Yes, use that momentum and energy are both conserved. If the momentum of the photon is hv/c then the momentum of the atom is hv/c is the opposite direction. Now the total of the photon energy and the atom kinetic energy is hv_0, the energy of the transition.

3. Sep 18, 2009

### th5418

In the case of $$\nu_o$$ momentum isn't conserved right? My main question is if I need to do like two situations, one where we take into account the recoil, and another we dont. And we do a momentum and energy conservation equation for both situations...

4. Sep 18, 2009

### Dick

If you don't pay any attention to the atom recoil, the calculation is easy. All of the energy goes into the photon (hv_0). But it only conserves momentum if the atom is infinitely massive. There is only one correct calculation.

5. Sep 19, 2009

### th5418

I am still lost. Sorry ><

6. Sep 19, 2009

### Dick

Forget the phrase 'uncorrected for the recoil of the atom' if that's what's confusing you. Just forget you ever saw it. v_0 is just the transition energy - the difference in energy between the two energy levels in the atom. Now use conservation to solve for frequency and show the formula holds.