# Elastic collision between a photon and an electron

1. Mar 31, 2013

### Unicorn.

1. The problem statement, all variables and given/known data

Hello everybody ,
a/There's an elastic collision between a photon of energy E and an electron at rest. After the collision, the energy of the photon is E/2 and propagates in a direction making an angle theta=60° with the initial direction. Find E. What kind of photon is this ?

2. Relevant equations

3. The attempt at a solution

I really have difficulties with this problem, for example in question a/ How can I find the energy if i don't have the velocity of the electron ..?

a/c=1
Conservation of energy gives:
E+mo=E/2+gamma*mo
E(1-1/2)=mo(gamma-1)

If I use conservation of momentum I have :
h/λ=h/λ'cosθ+mv

But then ?

Thank you !

Last edited by a moderator: Mar 31, 2013
2. Mar 31, 2013

### vela

Staff Emeritus
Good start, but you made a slight error. You need to take into account the fact that the electron isn't moving along the x-axis (the axis along which the original photon was moving).

3. Apr 1, 2013

### Unicorn.

I don't understant, after the collision the momentum of the electron isn't mv ?

4. Apr 1, 2013

### vela

Staff Emeritus
Momentum is a vector, right? You can't simply ignore the direction of the electron's momentum.

5. Apr 1, 2013

### Curious3141

A couple of hints:

Let the photon's original path represent the horizontal axis in the positive direction. Resolve the linear momenta along the horizontal and vertical axes. Let the electron fly off at an angle of $\theta$ to the horizontal. You should aim to eliminate $\theta$ from your equations.

Don't bring wavelength into it just yet. For the photon, you'll find it easier to use $E_{\gamma} = p_{\gamma}c$. For the electron, you might find it easier to use the equation $E_e^2 = m_e^2c^4 + p_e^2c^2$ where $m_e$ is the (rest) mass of the electron and $p_e$ is the momentum of the electron and $E_e$ is the TOTAL energy of the electron (includes the rest mass-energy). I found this approach easier than introducing $\gamma$ into the equation.

6. Apr 1, 2013

### vela

Staff Emeritus
It would be better to use $\phi$ for the electron since you've already used $\theta$ to denote the angle of the scattered photon. Curious3141 didn't mean to suggest that the photon and electron scatter with the same angle.

7. Apr 1, 2013

### Curious3141

Yes, of course, thanks. In my working, I was using the actual trig ratios for 60 degrees rather than theta.

8. Apr 1, 2013

### Unicorn.

Along the x axis we have:
Pγ+0=P'γ*cosθ+mv*cosϕ

Along the y axis we have:
P'γ*sinθ=mv*sinϕ

For the energy
E(1-1/2)=(m-mo)

Ee is th energy before and after collision ?

As I took c= 1 I can write Pγ=Q and P'γ=Q/2 right ?

9. Apr 1, 2013

### Curious3141

The momentum conservation along each axis looks fine, except I didn't break up the electron momentum into (relativistic) mass and velocity terms.

The energy before collision is the sum of the initial photon energy plus the rest mass-energy of the electron. The energy after collision is the sum of the final photon energy (0.5E) plus the total energy of the electron (rest mass-energy, which hasn't changed, plus the kinetic energy).

I didn't actually introduce the relativistic mass term ($\gamma m_e$) in the way I did it, which is why I had to consider the rest mass-energy of the electron in the energy conservation. You can do it your way, but the algebra might be trickier.

Yes, but you can go further, since you know E = pc. If c = 1, E = p in your system of units.

I didn't take c = 1, but it might simplify things. You should be careful to remain dimensionally consistent.

10. Apr 1, 2013

### Unicorn.

Along the x axis we have:
E=E/2*cosθ+mv*cosϕ

Along the y axis we have:
E/2*sinθ=mv*sinϕ

For the energy
E(1-1/2)=(m-mo)

I'm not seeing which combinaisons I have to go through to eliminate phi, I tried some but i'm always wrong !

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

### Curious3141

You might find it easier to see if you actually put in the trig ratios for 60 degrees instead of leaving it as $\theta$.

Then remember that $\sin^2 \phi + \cos^2 \phi = 1$. Find a way to manipulate the first two equations so you can take advantage of that identity.

BTW, I would caution against your general approach of letting c = 1, because you need to be able to make sense of your final result in order to calculate the frequency of the photon. Is there any harm in leaving the c there?

12. Apr 1, 2013

### Unicorn.

It's ok, I found that the frequency f=2.5*10^20 1/s. Now for the second part of exercise I don't know exactly what i'm searching for
I just saw that the b/ was removed from my first post

b/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

13. Apr 1, 2013

### vela

Staff Emeritus

14. Apr 1, 2013

### Unicorn.

It's just that the a/ and b/ part are in the same exercise.

15. Apr 1, 2013

### vela

Staff Emeritus
That's okay. Open a new thread and show your attempt at solving it.