# Electron-Proton Elastic Collision

1. Oct 9, 2007

### Magister

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

Consider the electron-proton elastic collision. Prove that the expression for the energy of the outgoing electron is

$$E_f= \frac{E}{1+(2E/Mc^2)sin^2(\theta /2)}$$

where $E_f$ is the energy of the outgoing electron, $E$ is the energy of the incident electron, $M$ is the proton mass and $\theta$ is the scattering angle of the electron.

2. Relevant equations

$$E = E_f + E_p$$
$$P = P_f cos(\theta) + P_p cos(\theta_p)$$
$$0 = P_f sin(\theta) + P_p sin(\theta_p)$$

$$E^2 = P^2 c^2 + M^2 c^4$$

3. The attempt at a solution

Well, I have tried the following

$$(P - P_f cos(\theta))^2 = (P_p cos(\theta_p))^2$$

$$(P_f sin(\theta))^2 = (P_p sin(\theta_p))^2$$

Adding both we get

$$P_{p}^2 = (P-P_f cos(\theta))^2 + P_f^2 sin^2(\theta) = P^2 + P_i^2 - 2PP_f cos(\theta)$$

Putting this into the energy equation for the proton we get

$$E_p^2 = (P^2 + P_{f}^2 - 2PP_f cos(\theta)) c^2 + M^2 c^4$$

Assuming that $E, E_f>>mc^2$, so that we can ignore the electron mass, we get

$$E^2 = P^2 c^2$$

$$E_p^2 = E^2 + E_{f}^2 - 2EE_f cos(\theta)+ M^2 c^4$$

Using the fact that

$$E_p^2 = (E - E_f)^2 = E^2 + E_f^2 - 2EE_f$$

we get
$$0 = 2EE_f - 2EE_f cos(\theta) + M^2 c^2 = 2EE_f (1-cos(\theta)) + M^2 c^4$$

$$E_f = \frac{- M^2 c^4 }{2E(1-cos(\theta))} = \frac{- M^2 c^4 }{4E sin^2(\theta/2)}$$

well, this is obvious wrong… it even gives a negative energy. What am I doing wrong? I have spent a lot of time looking for a mistake.

Thanks a lot for any suggestion

Last edited: Oct 9, 2007
2. Oct 9, 2007

### Meir Achuz

It should be E+M=E_p+E_f

3. Oct 10, 2007

### Magister

That's right. Thanks a lot for the suggestion it was very useful.