# Homework Help: Kinetic energy of a recoiled electron when backscattered

1. Jan 22, 2012

### StephenD420

Show that for very high energy photons, the kinetic energy of the recoiling electron approaches
KE= Ephoton - 255.5 KeV for back-scattering, phi = 180 degrees.

I know that

λ'/hc = λ/hc + (1-cos phi)/m0c^2
and since E = hc/λ
E' = E + m0c^2/(1-cos phi)
when phi is 180 degrees
E' = E + m0c^2/2
and the resting energy of the electron is m0c^2 = .511 MeV
so
E' = E + 255.5KeV
so E = Ephoton - 255.5Kev

Is this right?? I am not sure about the last step where Ephoton = E'????

Thanks.
Stephen

Last edited: Jan 22, 2012
2. Jan 23, 2012

### vela

Staff Emeritus
No, it's not correct because 1/a = 1/b + 1/c doesn't imply a = b + c, which is what you did when you went from wavelength to energy.

3. Jan 23, 2012

### StephenD420

ok so.....what do I do????

4. Jan 23, 2012

### StephenD420

Ok
1/E'=1/E +(1-cos phi)/.511Mev
now if you inverse both sides what do you get then

5. Jan 23, 2012

### StephenD420

ok so I put this into Mathematica and got
In[2]:= simplify ((1/a)^-1 == (1/b)^-1 + (1/c)^-1)

Out[2]= simplify (a == b + c)

so
E' = E/(1+(E/m0c^2)*(1- cos phi))

Which is my answer when phi = 180 degrees and m0c^2 = .511MeV

so what do you think???

6. Jan 24, 2012

### vela

Staff Emeritus
You seem to think E' is the kinetic energy of the electron. It isn't.

7. Jan 24, 2012

### StephenD420

ok so can you give me a hint then......

8. Jan 24, 2012

### vela

Staff Emeritus
If you have $\frac{1}{a} = \frac{1}{b}+\frac{1}{c}$, it follows that $a = \left[\frac{1}{b}+\frac{1}{c}\right]^{-1}$. You had already repeated the mistake I pointed out when you entered the above into Mathematica.

First thing you want to do is get things straight in your head. What does E represent? What does E' represent? What quantity are you trying to solve for? How can that be expressed in terms of E, E', and other variables in the problem?

9. Jan 24, 2012

### StephenD420

I am just trying to show that with high energy photon the recoiling KE approaches KE = Ephoton - 255.5 KeV

E' is Ephoton
and E is E photon before collision

now since I get the right answer...I do not know how else to get the right answer...any help would be appreciated and (Δλ/hc)^(-1) = hc/Δλ = ΔE = E' - E = (1/m0c^2 *(1-cos phi))^(-1) = m0c^2/(1-cos phi).

and if phi = 180 degrees and m0c^2 = .511 Mev
ΔE = .511MeV/2 = .2555 KeV
so I really do not know what you are getting at.....

10. Jan 24, 2012

### vela

Staff Emeritus
What I'm getting at is that you're not doing the basic algebra correctly. You have
$$\frac{hc}{\Delta \lambda} = \frac{hc}{\lambda' - \lambda}$$ and
$$E'-E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}.$$ You're claiming those two are equal because
$$\frac{hc}{\lambda' - \lambda} = \frac{hc}{\lambda'} - \frac{hc}{\lambda}.$$ Note this is akin to saying $\frac{1}{3-1}$ is equal to $\frac{1}{3}-\frac{1}{1}$. It's obviously not correct.

11. Jan 25, 2012

### StephenD420

and im saying ok since I getting the same numbers the professor wants, then if that is not right please point me in the right direction as I do not see how the professor got the numbers otherwise if not the rest energy of the electron over (1-cos phi) = 2