# Special relativity and quantum mech

1. Nov 19, 2013

### bs vasanth

An atom can exist in two states , ground state of M and excited state of mass M+Δ , it goes to excited state by absorption of an photon, what is the photon frequency in the in the lab frame of reference ( where the atom is initially at rest )?

A simple energy equation gives the answer:
E=( M+Δ -M)$c^{2}$=hγ
γ=Δ$c^{2}$/h
But my intuition says that I am overlooking something and can't be that simple..

2. Nov 20, 2013

### PAllen

It is that simple.

3. Nov 20, 2013

### Meir Achuz

You should include the recoil of the excited atom in the final state.
Use Energy^2=(M+K)^2=(M+Delta)^2+K^2, and solve for K (the photon energy). (with c=1)

4. Nov 20, 2013

### PAllen

Yes, I ignored recoil because its contribution is small for common cases. I, of course, agree with the above formula. Note that even for as extreme a case as a 10 Mev delta for a hydrogen atom, including recoil only changes the required photon energy to about 10.05 Mev. For Sodium, to about 10.002 Mev. For delta of 1 Mev and hydrogen, including recoil would require a 1.0005 Mev photon.