Photon atom collision

1. Apr 1, 2013

Unicorn.

1. The problem statement, all variables and given/known data
There's an elastic collision between a photon of energy E and an atom in an excited state. After the collision, the energy of the photon is still E but its direction changed of angle of 180° and the atom is now going back with velocity Bc. If the atom is in his ground state after collision, what was the excitation energy ? Give the answer in fonction of E, Bc, and mass rest mo

2. Relevant equations

3. The attempt at a solution
Conservation of energy:
E+m'c²=E+γmc²
We are supposed to find β=2E/(4E²+c²²m²) But I really don't see how we can get to that result
When we find β we just have m"excited"=m'-m=γm-m

Thanks

2. Apr 1, 2013

Sunil Simha

Here β is the velocity of the atom after collision right? If so, then it can be found using momentum conservation. If not, don't mind me.

3. Apr 1, 2013

Unicorn.

I can't write the conservation of momentum because I don't see exactly how the photon and atom are moving after collision:
Let's say the photon is moving in the +x direction, after it'll be moving in -x and the atom in the +x or -x ?
Thanks

Solved !

Last edited: Apr 1, 2013
4. Apr 1, 2013

Curious3141

Your question is ambiguous. What, exactly are you required to find? If it's the excitation energy, it's simply equal to the kinetic energy of the atom after collision (this only requires the energy conservation statement, and is a trivial proof). If you're asked to determine $\beta$, you need both the momentum and energy conservation statements.

Anyway, there's an error in your expression for $\beta$. It should be $\displaystyle \beta = \frac{2E}{\sqrt{4E^2 + m_0^2c^4}}$. Correct?

5. Apr 1, 2013

Unicorn.

Yes, correct.
I was asked to find the excitation energy, but to calculate it I need γ so β.
I just used the momentum conservation and I found β, I wasn't understanding in which direction goes the atom after collision. But now it's good I solved the problem.
Thank you !

6. Apr 1, 2013

Curious3141

You're welcome. Your first post said you could leave $\beta$ in the expression. But to eliminate it, you need to consider both conservation laws, as you stated.

Anyway, $\displaystyle E_a = \sqrt{4E^2 + m_0^2c^4} - m_0c^2$