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Magister
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Homework Statement
Consider the electron-proton elastic collision. Prove that the expression for the energy of the outgoing electron is
[tex] E_f= \frac{E}{1+(2E/Mc^2)sin^2(\theta /2)} [/tex]
where [itex]E_f[/itex] is the energy of the outgoing electron, [itex]E[/itex] is the energy of the incident electron, [itex]M[/itex] is the proton mass and [itex]\theta[/itex] is the scattering angle of the electron.
Homework Equations
[tex] E = E_f + E_p [/tex]
[tex] P = P_f cos(\theta) + P_p cos(\theta_p) [/tex]
[tex] 0 = P_f sin(\theta) + P_p sin(\theta_p) [/tex]
[tex] E^2 = P^2 c^2 + M^2 c^4[/tex]
The Attempt at a Solution
Well, I have tried the following
[tex] (P - P_f cos(\theta))^2 = (P_p cos(\theta_p))^2 [/tex]
[tex] (P_f sin(\theta))^2 = (P_p sin(\theta_p))^2 [/tex]
Adding both we get
[tex] P_{p}^2 = (P-P_f cos(\theta))^2 + P_f^2 sin^2(\theta) = P^2 + P_i^2 - 2PP_f cos(\theta) [/tex]
Putting this into the energy equation for the proton we get
[tex] E_p^2 = (P^2 + P_{f}^2 - 2PP_f cos(\theta)) c^2 + M^2 c^4[/tex]
Assuming that [itex]E, E_f>>mc^2[/itex], so that we can ignore the electron mass, we get
[tex] E^2 = P^2 c^2 [/tex]
[tex] E_p^2 = E^2 + E_{f}^2 - 2EE_f cos(\theta)+ M^2 c^4[/tex]
Using the fact that
[tex] E_p^2 = (E - E_f)^2 = E^2 + E_f^2 - 2EE_f [/tex]
we get
[tex] 0 = 2EE_f - 2EE_f cos(\theta) + M^2 c^2 = 2EE_f (1-cos(\theta)) + M^2 c^4 [/tex]
[tex] E_f = \frac{- M^2 c^4 }{2E(1-cos(\theta))} = \frac{- M^2 c^4 }{4E sin^2(\theta/2)} [/tex]
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
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