Understanding Compton Scattering: Photon-Electron Collisions Explained

[bobie]

Homework Statement

This is not homework, I am trying to understand Compton scattering considering it a normal elatic collision between particles

Homework 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?
In this applet http://www.ndt-ed.org/EducationResources/CommunityCollege/Radiography/Physics/comptonscattering.htm
different energies give same θ

Thanks for your help

 

[hjelmgart]

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

p_electron' * sin φ = p_photon' * 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).

 

[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
p_electron' * sin φ
impossible to match to
p_photon' * sin θ = 8.167 *10^8

can someone help me?

 

[jtbell]

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.

 

[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 E_e 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' + E_e

 

[voko]

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

 

[bobie]

A photon with an energy of 51.1 keV does not exist (at least to

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
p_x = p*cos60° = 4.118(6)*10⁹ and
p_y = p*sin60° =7.13374*10⁹

Is that right?

The energy of the electron is
E^e = E - E' = 4.118(6)*10¹⁹
p^e_y = p_y
p^e_x = 1.2356*10¹⁰ - p_x = 8.237(3)*10⁹

shoud now by pythagoras:
E^e = p^e_x²+p^e_y ² ?

 

[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

 

[bobie]

... it seems good to me.

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

 

[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' + E_e => E_e = E + E - E/3 = 5/3 E

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

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

 

[bobie]

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

And from conservation of energy

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

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?

 

[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:
E_tot = E_photon + E₀

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

E_tot = 2 E₀

After:
E_tot' = E_photon' + E_e

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

Conservation of energy gives: E_tot = E_tot'

2 E₀ = E₀/3 + E_e

Thus:

E_e = 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.

 

[bobie]

If my formula is misunderstood, here goes again:
.

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 + KE_e 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?

 

[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.

 

[bobie]

E_tot = 2 E₀
After:
E_tot' = E_photon' + E_e
2 E₀ = E₀/3 + E_e
E_e = 2E₀ -E₀/3 = E₀ 5/3

.

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

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?
.

I am asking : if the incoming photon has 10^15 hv
E_tot = 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?

 

[hjelmgart]

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

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

I am asking : if the incoming photon has 10^15 hv
E_tot = 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?

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 = p_e^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

 

[bobie]

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

p_e = 4/3 E/c, and E_e = 5/3 E ?

 

[hjelmgart]

Yeah that is it.

 

[bobie]

but p_e = 4/3 E/c, and E_e = 5/3 E are related to E_m = 511 keV

I was asking about the case E = 5.11 eV,
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 = E_m and therefore its speed ≈ c or only ≈ 1 cm/sec ?
ThanksShall 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?

 

[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..

 

[bobie]

It's not easy to explain, lol..

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?

 

[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.

 

[bobie]

I have no access to such programs.
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

 

[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.