Kinetic energy with compton effect

[StephenD420]

Show that the recoiling electron has kinetic energy:
K = Ephoton^2*(1-cos phi)/ mc^2 + Ephoton(1-cos phi)

Now I have gotten
K = hc/λ - hc/λ' = hc(λ'-λ)/λ'*λ = hc(Δλ)/λ'*λ
and since λ' = λ + Δλ
K = hc *Δλ/(λ(λ+Δλ)) = hf*Δλ/(λ+Δλ) = (Δλ/λ)/(1+Δλ)*hf

and I know that Δλ = h/mc*(1-cos phi)

but how do I go from what I have gotten to K = Ephoton^2*(1-cos phi)/ mc^2 + Ephoton(1-cos phi)?

Thanks.
Stephen

 

[StephenD420]

bump...Please help

 

[vela]

I don't think that expression for the kinetic energy is correct. I can show that
$$K = \frac{E_\gamma^2}{mc^2}(1-\cos\phi) - \left(\frac{K}{mc^2}\right)E_\gamma(1-\cos\phi)$$For the expression you were given to show holds, you'd have to have K/mc² = -1, which can't be true.

 

[StephenD420]

That is the kinetic energy equation given on the homework that my professor gave...

 

[vela]

I'll ask the other helpers to see if they have any insight on this problem. Just to clarify, when you say Ephoton, you're referring to the energy of the incident photon, correct? And phi is the angle of the scattered photon? Are you looking at a specific regime or are you supposed to use some approximation?

 

[StephenD420]

Ephoton is incident photon and we are finding kinetic energy of the recoiling electron... Please help

 

[I like Serena]

Hi StephenD420!

I believe your formula should be (I think you dropped a couple of parentheses):
$$K = {E_{photon}^2 (1-\cos \phi) \over m_e c^2 + E_{photon}(1-\cos \phi)}$$

For the calculation I'd suggest that you use $\lambda = {hc \over E_{photon}}$ and $\lambda' = \lambda + {h \over m_e c} (1 - \cos \phi)$.

 

[StephenD420]

I looked back at the problem and no I did not drop any parentheses, my professor hand rights his homework so any ideas?

 

[I like Serena]

In that case I think your professor dropped the parentheses.

 

[StephenD420]

K = hc *Δλ/(λ(λ+Δλ)) = hf*Δλ/(λ+Δλ) = (Δλ/λ)/(1+Δλ)*hf

so how do I go from above to what you got??

 

[StephenD420]

ok so I have gotten thus far

1/E' = [mc^2 + E(1-cos phi)]/E*mc^2

now how do I go from here to what you got?

 

[I like Serena]

ok so I have gotten thus far

1/E' = [mc^2 + E(1-cos phi)]/E*mc^2

now how do I go from here to what you got?

Calculate E - E', that is, E minus the inverse of what you got there.
From there it is algebraic manipulations to subtract 2 fractions.

 

[StephenD420]

ok now if I wanted to show that for very high energy photons and phi = 180 degrees the kinetic energy approaches K = Ephoton - 255.5KeV??

 

[I like Serena]

ok now if I wanted to show that for very high energy photons and phi = 180 degrees the kinetic energy approaches K = Ephoton - 255.5KeV??

Since K=Ephoton - Ephoton', you will want to show that Ephoton' = 255.5 keV.

The dependencies on Ephoton can be approximated by the fact that 1/Ephoton is relatively close to zero.