# Compton Scattering Effect: Max Kinetic Energy

1. Mar 23, 2008

### dakillerfishy

Okay, so I have two questions, both pertaining to the Compton Effect.

1. The problem statement, all variables and given/known data
Evaluate the maximum kinetic energy for a recoiling electron that is struck by a photon with momentum 0.04MeV/c

2. Relevant equations
Maximum Kinetic energy: E$$_{k}$$ = $$\frac{hf}{1 + (mc^{2}/2hf}$$
Compton wavelength of an electron: h/mc = .00243 nm
Compton's equation: $$\lambda$$2 - $$\lambda$$1 = (h/mc)(1-cos$$\theta$$)

3. The attempt at a solution
I tried to rearrange the above maximum kinetic energy and solve for f using the quadratic formula, but then I realized that the mass of the photon was close to zero and got worried because I'd end up dividing by zero.

Also:

1. The problem statement, all variables and given/known data
The maximum kinetic energy given to the electron in a Compton scattering event plays a role in the measurement of gamma-ray spectra using scintillation detectors. The maximum is referred to as the Compton edge. Suppose the Compton edge in a particular experiment is found to be 520 keV. What were the wavelength and energy of the incident gamma rays?

2. Relevant equations
See above question

3. The attempt at a solution
Since we know the maximum kinetic energy, I tried to plug that into the above equation, but ran into the same divide-by-zero problem.

Last edited: Mar 23, 2008
2. Mar 23, 2008

### Shooting Star

For a photon, hf=pc. The p is given, so you can find f. Where are you getting zeros?

3. Mar 23, 2008

### dakillerfishy

For max kinetic energy, the mc^2 term on the bottom looks like it would go to zero because the mass of the photon is negligible.

4. Mar 23, 2008

### Shooting Star

The m is the rest mass of the electron. It seems you haven't got what I said about the momentum of the photon.

The momentum is denoted by p, and p is given in the problem. E of the photon = hf and also E = pc for a particle with a zero rest mass, like the photon. Equate the two to get hf. Use consistent and convenient units, e.g., Mev for energies in this problem.

(Pl note that the denominator does not become zero even if you put m=0, which of course is NOT the case.)

5. Mar 23, 2008

### dakillerfishy

No, I did get what you said about the photon, but perhaps I didn't phrase my question correctly. Why are we using the electron for mc^2?

6. Mar 23, 2008

### Shooting Star

I'm not very sure that I understand what your question is actually...have you done any derivation related to the Compton effect? The mass of the electron comes from that. It seems natural that the mass of the electron should have to be taken into consideration if an electron collides with something else.

You can have a look at these and your book or notes.

http://www.virtualsciencefair.org/2007/pete7o2/Background.html
http://staff.washington.edu/skykilo/compton/compton.html [Broken]

Last edited by a moderator: May 3, 2017
7. Mar 23, 2008

### dakillerfishy

Thanks for the links, shooting star. The derivation wasn't in my book, so I looked at it in the first link and I think I have a better understanding now.

I answered my first question correctly, but I'm still stuck on the second one.

I tried to rearrange the max kinetic energy equation to solve for f and got:
0 = 2h$$^{2}$$f$$^{2}$$ - 2E$$_{k}$$hf - mc$$^{2}$$
then used
$$\lambda$$ = c/f
and ended up with 2.3858 pm, which is incorrect.

8. Mar 23, 2008

### Shooting Star

The Compton edge in the spectrometer occurs when the photon is giving up the maximum amount of energy possible to the electron, and this happens when the photon goes back the way it came.

For that, you should find $$\lambda_{2}$$-$$\lambda_{1}$$, and put $\theta$ = 180 deg, and then convert $$\lambda_{2}$$-$$\lambda_{1}$$ to energy.

There are ready made formulae everywhere, but why don't you do it yourself? You should also deduce (very easy) that $\theta$ = 180 deg for this to happen.

9. Mar 23, 2008

### dakillerfishy

Aha, that makes much more sense than what I was trying to do. Thanks!

I understand why theta has to be 180 degrees, but how would you go about converting $$\lambda$$$$_{2}$$ - $$\lambda$$$$_{1}$$ into energy? I imagine you would add the difference in $$\lambda$$'s to the original $$\lambda$$, then plug back into the max energy equation, but I don't think that's quite right.

10. Mar 23, 2008

### Shooting Star

Use E = hf = hc/$\lambda$, because they are photons.

11. Mar 23, 2008

### dakillerfishy

So then for wavelength you get $$\lambda$$$$_{2}$$ - $$\lambda$$$$_{1}$$ = 4.86 pm, which is still the difference between lambda's. However, wouldn't $$\lambda$$$$_{2}$$ be the wavelength of the photon after it has recoiled and be $$\lambda$$$$_{2}$$ = hc/Ek ?

Is that right, or am I still not understanding something?

Thanks =)

12. Mar 23, 2008

### Shooting Star

$\lambda$ = hc/E. So put $\lambda_{1}$ = hc/E1 etc. I really think you can solve it now. All the h's will cancel out.

(Refer post #8 again.)

Last edited: Mar 23, 2008
13. Mar 23, 2008

### dakillerfishy

So then

$$\lambda_{2}$$ - $$\lambda_{1}$$ = h/mc(2)
hc/$$E_{k}$$ - $$\lambda_{1}$$ = 2h/mc
$$\lambda_{1}$$ = hc/$$E_{k}$$ - 2h/mc
$$\lambda_{1}$$ = (4.136e-15 * 3e8)/520000 - (2*4.136e-15)/(.511e6*3e8) = 2.386e-12 m

$$\lambda$$ = hc/E
E = hc/$$\lambda$$
E = (4.136e-15 * 3e8)/2.386e-12 = 520,000 eV

So the initial energy is the same as the final energy?

14. Mar 23, 2008

### Shooting Star

The RHS is 2h/(mc).

Last edited: Mar 24, 2008
15. Mar 23, 2008

### dakillerfishy

Sorry, I wrote that wrong... the 2 is actually in the numerator and the results don't change.

16. Mar 24, 2008

### Shooting Star

I find it difficult to follow math steps with lots of numerals plugged in. It becomes so messy. Which problem is this?

Lets go back to the original formula you had given right at the beginning. We'll use only that one for these two problems.

If a photon with a frequency 'f' hits an electron at rest, then the maximum KE that can be imparted to the electron is given by Emax = $$\frac{hf}{1 + mc^{2}/2hf}$$.

Problem 1.

For the photon, hf = pc. Here p is given => you can find hf => you can find Emax.

Problem 2.

Here, Emax is given => you can find hf => you can find pc => you can find p.

If you use the value of the h/mc for the e-, the arithmetic is even simpler.

This is nothing but two minutes algebra. Simplify everything first and then plug in the values.

I wanted you to understand the derivation of the formula, but that can come later. It requires a bit of energy momentum calculation.

17. Mar 24, 2008

### dakillerfishy

I'm sorry this is so frustrating... maybe my algebra just sucks. :grumpy:

I already got #1, so this would be #2

So we have:
E = $$\frac{hf}{1 + mc^{2}/2hf}$$
E + $$\frac{Emc^{2}}{2hf}$$ = hf
(Eh/mc) f + Ec = $$\frac{h^{2}}{mc}$$ $$f^{2}$$
$$\frac{h^{2}}{mc}$$$$f^{2}$$ - (Eh/mc) f - Ec = 0
f = -1.257e20 or -4.115e11

I know that's not right, but I can't see where the error is.

18. Mar 24, 2008

### dakillerfishy

I just thought of this...

Are you saying that I should solve the E max equation in terms of momentum, then use
E - (pc)^2 = (mc^2)^2
to solve for energy, then solve for wavelength?