What is the excitation energy of an atom after a photon collision?

[Unicorn.]

Homework Statement

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

Homework Equations

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

 

[Sunil Simha]

We are supposed to find β=2E/(4E²+c²²m²)

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.

 

[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 !

 

[Curious3141]

Homework Statement

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

Homework Equations

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

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?

 

[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 !

 

[Curious3141]

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 !

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