How Does Recoil Affect Photon Emission and Absorption Frequencies?

[th5418]

Homework Statement

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.

Homework 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}

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?

 

[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.

 

[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...

 

[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.

 

[th5418]

I am still lost. Sorry ><

 

[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.