# Homework Help: Compton collision

1. Nov 20, 2013

### bobie

1. The problem statement, all variables and given/known dataThis is not homework, I am trying to understand Compton scattering considering it a normal elatic collision between particles
2. Relevant equations
The Klein–Nishina formula

Suppose a photon (hf= 0.511 MeV) hits an electron and the latter recoils at an angle $\varphi$ 60°.
- Is there a formula to get the scattering angle θ? is it : 180° - 2 $\varphi$
- Does θ depend on the energy hf?
different energies give same θ

Last edited: Nov 20, 2013
2. Nov 20, 2013

### hjelmgart

Well you can use the laws for conservation of energy. You should at some point encounter the formula:

pelectron' * sin φ = pphoton' * sin θ

Which is also typical for other cases of inelastic collision. p being the momentum, and the prime, ', indicates, that it is after collision.

I think this is what you meant? Of course this relates the energy as well, however, the formula for compton scattering is made so that it is possible to determine the scattering angle depending on the energies, which would typically be how you do it (at least from my experience).

Last edited: Nov 20, 2013
3. Nov 21, 2013

### bobie

I was wandering if there is a general formula, good for all instances.

I tried to work out the case where the photon has 51.1 keV : 1.2356*10^19 hv
the electron gets only 1.6335*10^17 hv
and I find
pelectron' * sin φ
impossible to match to
pphoton' * sin θ = 8.167 *10^8

can someone help me?

4. Nov 21, 2013

### Staff: Mentor

Best thing to do is to start from scratch and use energy and momentum conservation.

1. Draw a diagram and identify the energies, momenta, and angles that you know and don't know.
2. Write down the equation for conservation of energy.
3. Write down the equation for conservation of x-component of momentum.
4. Write down the equation for conservation of y-component of momentum.
5. Use the fact that for each particle, incoming and outgoing, E2 = (pc)2 + (mc2)2.
6. Solve the three equations together to eliminate the unknown quantities that you're not interested in, and solve for the unknown quantity that you're interested in.

5. Nov 21, 2013

### hjelmgart

I am not sure I follow you. How did you calculate the energy of the electron after collision? And did you relate the impulses correctly to the energy of each case.

The numbers you are using aren't proper either. A photon with an energy of 51.1 keV does not exist (at least to my knowledge).

Let's start over:

The electron has an energy at rest of:

E0=m0c2

And it's initial impulse is 0, as it isn't moving.

Then the photon E = hv=hc/λ , p=h/λ

Then they collide and you get new energy hc/λ' for the photon and Ee for the electron. You want to calculate the energy of the electron to find the impule, so that you can relate this to the angle?

The energy is found from conservation of energy so

E + E0 = E' + Ee

6. Nov 21, 2013

### voko

How about the original article by Compton? It is very well written and is available online for free.

7. Nov 24, 2013

### bobie

I do not understand that, it is a γ-ray 1/10 the energy of the rest mass of the electron.

I'll start from the easiest case when we know that the scattering angle is 60° and the photon has energy E equal to rest-mass: 511 keV, 1,2356*1020 hf.
Please correct me where I go wrong.
the wavelength λ'[ increases by (1-cos60° = 0.5) 1+0.5] = 1.5

The energy E' after the impact is E /1.5 = 8.237(3)*1019 hf, and momentum
p = 8.237*109 , and
px = p*cos60° = 4.118(6)*109 and
py = p*sin60° =7.13374*109

Is that right?

The energy of the electron is
Ee = E - E' = 4.118(6)*1019
pey = py
pex = 1.2356*1010 - px = 8.237(3)*109

shoud now by pythagoras:
Ee = pex2+pey 2 ?

Last edited: Nov 24, 2013
8. Nov 24, 2013

### hjelmgart

You are of course correct, I made a brain fart considering gamma rays, haha.

I didn't do any of the calculations though, but it seems good to me.

And you should of course not forget the speed of light in the last equation?

Oh, and I found this document, which might explain it better, and it contains all the steps in deriving many different equations, as well as an illustration.

http://www.phys.utk.edu/labs/modphys/Compton Scattering Experiment.pdf

Last edited: Nov 24, 2013
9. Nov 25, 2013

### bobie

If the equation is right, there must be something wrong, as the values do not match.

Can someone show me how to proceed in order to match the momenta if the scattering angle is 180° :
1-cos180 =2

E' = (1.2356/1.2) 1.03 *1019
p = + 1.2356*109

p'x = E'/ 1010= - 1.03*109
Is this + or -?

If it is minus, as I suppose, values do not match

Last edited: Nov 25, 2013
10. Nov 25, 2013

### hjelmgart

Alright.. Not sure what you want to find anymore, though I think you might just have got your values wrong.

If you take your example with 180 degrees as scattering angle and a photon energy equal to that of the rest energy of the electron, call both of them E you will get:

1/E' - 1/E = 1/E (1-cos(180)) => E' = E/3

And from conservation of energy

E + E = E' + Ee => Ee = E + E - E/3 = 5/3 E

And finally conservation of momentum: p = -p' + pe and negative p', since it's in the -x direction:

pe = E/c + 1/3 E/c = 4/3 E/c

Last edited: Nov 25, 2013
11. Nov 26, 2013

### bobie

Do you mean that the electron gets more than 2/3 of the energy of the γ-ray?
where does the surplus energy come from?
before collision we have a γ-ray 3/3 E + a rest mass
after collision we have a ray with 1/3 E + a rest mass +2/3 of KE + 3/3 KE ?
How is that possible?

12. Nov 26, 2013

### hjelmgart

Yes for this specific case it gets 5/3 the energy of the incident photon (or the rest energy of the electron). There is no surplus energy, as this is calculated by use from conservation of energy.

I am not sure, I understand your question either. Conservation means, that the total energy before collision, needs to be equal to the total energy after collision.

We found by using the formula from compton scattering, that the energy, E', of the photon after being scattered was 1/3 of its initial energy. So there isn't more energy after collision, it's just divided differently between the photon and the electron, however, it still adds up to the total energy before collision, hence conservation of energy.

If my formula is misunderstood, here goes again:

Before:
Etot = Ephoton + E0

We took the photon to have an energy of 511 keV, which is equal to the rest energy of the electron, E0 so:

Etot = 2 E0

After:
Etot' = Ephoton' + Ee

Now note that the electron is in motion, so it gets an impulse (and the mass becomes relativistic). This gives a new energy, Ee, which we want to find. First we found Ephoton' = E0/3 from the compton equation.

Conservation of energy gives: Etot = Etot'

2 E0 = E0/3 + Ee

Thus:

Ee = 2E0 -E0/3 = E0 5/3

As a last note, if you are interested in this with respect to some classical physics course, well, don't put too much time in it. If you are going to have quantum mechanics later on, you will encounter Compton scattering again, and without doubt, it will be explained properly. Many classical physics courses may lack this ability, and well it makes much more sense, when you get to understand the basics of quantum mechanics, for instance the fact that an electron has an energy, when it isn't moving, is prohibited from a classical point of view.

13. Nov 26, 2013

### bobie

Thanks, hjelmgart,
I did not misunderstand your formula, but we are talking here of high energies, therefore I do not know if what you say applies only to relativistic speeds.
Suppose we have a photon E = 1.2*1015 that hits an electron. The electron gets speed in the order od 105, right so it is not relativistic.
Before collision we have E 1015 + rest mass
after collision we have E' ≈ 1015 + rest mass + KEe 10^10 + 10^20 ?
What is the impulse you mentioned?
and if I may ask, how do you measure the momentum of the photon 55 ?
is it kg*m, gr *cm or what?

14. Nov 26, 2013

### hjelmgart

Well, whether something is relativistic or not, does not depend on the speed of the object alone. The thing is, we cannot classically describe a collision between a massless object and an object with mass. So we need to use relativistic formulas for the energy of a photon and an electron, thus whenver you study a case like this, it is relativistic.

I am getting confused with all the numbers, by the way. Nonetheless, the photon has momentum given by p=h/λ , and momentum always have the same units (kg*m/s), which are embedded in the planck constant.
Also why do you want to add the rest mass after collision?

Now this said, as I noted in the end of my last reply, to describe these things accurately, we need to use quantum mechanics. Here we cannot apply our regular equations for conservation of energy or momentum, because these values are discrete, and applies to both the photon and the electron. Also an electron initially at rest, is quite a bad example. The thing is, we can never know the position of the electron (same goes for the photon).

If we, however, have enough photons, we can assign every photon an average energy, which is satisfactory for our needs. Besides it is very hard to generate a single photon, I even doubt it's possible yet, and thus you always have a smaller wavepack of photons.

15. Nov 27, 2013

### bobie

The scattered photon has p'x = -1/3 E/1010
the original photon has px = E +3/3 1010
the electron should have pe = 4/3 E and not 5/3, am I wrong?

I am asking : if the incoming photon has 10^15 hv
Etot = E + E/105
what is the energy of the electron after the collision? do we add the energy of the rest mass E to the energy it gets from the photon?

16. Nov 27, 2013

### hjelmgart

Well almost correct, of course you need to divide by the speed of light as well.
pe = 4/3 E 1/c

Uhm, why do you put it with the 10^15 in front? The energy should just be hv, if your v is the frequency (c/λ), which is pretty high for a gamma ray.

You are probably familiar with the relativistic formula:

E^2 = pe^2 c^2 + m^2 c^4

So for an electron at rest it becomes E = mc^2

Now the electron is in motion, so yes you add the rest mass squared, and by using the momentum, we found:

E = √((4/3 E)^2 + E^2) = √(25/9 E^2) = 5/3 E

17. Nov 27, 2013

### bobie

pe = 4/3 E/c, and Ee = 5/3 E ?

Last edited: Nov 27, 2013
18. Nov 27, 2013

### hjelmgart

Yeah that is it.

19. Nov 27, 2013

### bobie

but pe = 4/3 E/c, and Ee = 5/3 E are related to Em = 511 keV

but I suppose the case of E = 1.23 *1010 is more interesting and can clarify my doubts:
in this case momentum of the electron is 2 (*1.23) and is equal to the energy it gets from the photon 2.47 hv. Is its energy = Em and therefore its speed ≈ c or only ≈ 1 cm/sec ?
Thanks

Shall I start a new thread, or can you tell me what range of energy has been tested and if the recoil angles and speeds of the electron have been accurately measured, too? can you give me a link with the result of the experiments?

20. Nov 27, 2013

### hjelmgart

Well, you can always start over with whatever value. I do not know, for which values the compton scattering is valid, however, I am certain it has been tested for x-rays and gamma rays, but that is the extent of my knowledge, so you could of course try to start a new thread, but I would not do it in the home assignment part of the forum. Probably in the physics it's better.

Also you cannot define a "speed" to an electron. It is wrong. The electron does not "move" it just exists at different positions with respect to time. You shouldn't try to understand it any other way, because it is wrong. So it's better not to, and take some quantum mechanic classes. You can always google "how electrons move", which of course will tell you, that they don't. It's not easy to explain, lol..

21. Nov 28, 2013

### bobie

Thanks , hjelmgart, you've been very helpful!
Do you know if there is a value over which the scattering at 180° is no longer possible, same as in ordinary collision?
And how to proceed if we know only the recoil angle?

22. Nov 28, 2013

### hjelmgart

By value are you referring to energy, and of the incident or the scattered photon?
It would be much easier if you take the Compton formula into a math program and plot it!. You can make a plot of the scattered photon energy versus the scattered angle. And unless you can do a 3d plot, you will probably have to keep the incident photon energy constant.

If we know only the recoil angle, are you then saying we don't know anything about the incident or scattered photon?
There is of course a relationship between the angles, as the momentum needs to be conserved in x-direction. So if you know the recoil angle of the electron, you know how large a fraction, that is scattered in the x-direction. You can use this value to find the scattering angle of the photon.

The momentum also needs to be conserved in the y-direction, so techniqually the electrons y-component must be equal to the negative photon y-component, as the initial momentum in the y-direction is 0

I would, however, postulate, that in most cases, you will know the energy of the incident photon, as this is typically your own decision, when doing experiments.

23. Nov 28, 2013

### bobie

I was referring to the energy of the photon I suppose the above 10^21 the photon cannot rebound, but I was wondering if it can pass through the electron, as it always travels faster

Last edited: Nov 28, 2013
24. Nov 28, 2013

### hjelmgart

Well you can try wolframalpha.com , which is a wonderful math engine, and you can use the free parts of it for a lot of problems. I tried to graph it with the 511eV of the photon and the electron:
http://www.wolframalpha.com/input/?i=f(x)=511-511*(1-cos(x))
I don't know if this link works though, but you can actually draw it in your hand, since it's just a cosine function, although not good for obtaining data though.

Hmm, that is a good question. I am not sure, I can answer it with high precision though.

However, from my knowledge of photons as quantum particles, you usually have certain possibilities for transmission, absorbtion or reflection. If this applies to single photons interaction with single electrons, I don't know though.
Anyway, if it does, it would mean, there is a certain possibility, that it can be transmitted as well.