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Physics - Compton scattering
AnteCantelope
Registered User regular
Registered User regular
Just a quick physics question that I'm having trouble with:
Background: I did an experiment on gamma rays and compton scattering, and measured the scattered gamma rays at different angles. Made a graph out of it, where the y-intercept was 1/incident gamma photon's energy, and the gradient was 1/scattered electron's energy. Now, this is a linear graph, so the scattered electron's energy was 500keV.
OK, now the problem I have is that the scattered photon's energy changes at different angles. If the scattered electron's energy is a constant, where is the extra energy going? I'm guessing it's recoil energy in the nucleus or something, but I'm not sure. I have this big red tick next to the part where I gave the scattered electron's energy as a constant, so I doubt that that's wrong, but... maybe it's meant to change at different angles? I don't know.
Can anyone help with that? Thanks.
Background: I did an experiment on gamma rays and compton scattering, and measured the scattered gamma rays at different angles. Made a graph out of it, where the y-intercept was 1/incident gamma photon's energy, and the gradient was 1/scattered electron's energy. Now, this is a linear graph, so the scattered electron's energy was 500keV.
OK, now the problem I have is that the scattered photon's energy changes at different angles. If the scattered electron's energy is a constant, where is the extra energy going? I'm guessing it's recoil energy in the nucleus or something, but I'm not sure. I have this big red tick next to the part where I gave the scattered electron's energy as a constant, so I doubt that that's wrong, but... maybe it's meant to change at different angles? I don't know.
Can anyone help with that? Thanks.
AnteCantelope on
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Maybe it has to do with something in quantum physics? That the properties (both existential and non existential) of a wave and a particle can change depending on how we are viewing them?
Otherwise, due to conservation of energy.. I wonder if your measurements were correct in your experiment?
Other than that, I got nothin.
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Satans..... hints.....
Well, I got big ticks on everything and full marks for the experiment, so I'm pretty sure that I got everything right. It's just I had to write a report on it, and so I'm looking at it thinking that if the incident photon has a constant energy, and the scattered electron has constant energy, how can the scattered photon have variable energy? Recoil energy in the nucleus is about all that I can think of, and it makes a bit of sense because the recoil energy would be lowest when the photon is scattered in its original direction, and highest when it's scattered in the opposite direction. I just didn't want to write that down if it turned out to be wrong.
Also, Blaket, I have no idea what you're talking about.
From the gang called...
Errr, I mean:
Compton scattering is when you get a photon colliding with an electron in an nucleus (Which you consider to be loosely bound). So you've got an electron at rest and a moving photon. So to conserve momentum you need your electron to be moving slightly afterwards.
The change in energy comes from the energy being shared between the two bodies, which should have a fairly obvious angular relation (if the photon carries straight on then there is no energy transfer and you've got an elastic collision but if your photon reverses direction then you've got the maximum energy transfer).
So, yes the other body is an electron and it is receiving the lost energy. Bear in mind though that the electron is not ejected, that would be the photoelectric effect. Or you could get pair production with higher energies.
i have allways just plotted x vs y and used what the gradient comes out to be to find something.
Edit: mayble let us see your data, because right now i have no idea what you are doing.
Abstract:
There are many naturally occurring radionuclides in the environment, and so it is obvious that an understanding of these, and the gamma radiation that they can emit, is of some importance. This experiment demonstrates the presence of background gamma radiation, and investigates one of the properties of gamma radiation: Compton scattering. By scattering gamma rays from a plastic scintillator, and measuring the energy of the scattered photons, the relation between angle and scattered gamma energy was determined.
Introduction:
While the photoelectric effect is the dominant process of photon absorption at energy levels much lower than 1MeV, Compton scattering becomes the dominant process at energy levels around 1MeV (at several MeV, the dominant process is pair production). Given this range of energies at which Compton scattering predominantly occurs, it is largely associated with gamma radiation. As in the photoelectric effect, Compton scattering occurs when an incident photon of energy Eγ interacts with an electron orbiting an atom in an absorber. The result is an emitted electron of energy Ee` and an emitted photon of energy Eγ`. The resultant gamma photon’s energy varies only slowly with varying incident photon energies; as the energy of the incident photon increases, almost all of the extra energy goes to the emitted electron, and very little is given to the emitted photon.
Method:
Gamma rays are detected with a thallium activated sodium iodide crystal scintillator. Incident photons free electrons from the crystal lattice (by the Compton effect), and these high energy electrons cause ionisation tracks through the crystal lattice. This in turn causes the crystal to emit photons of a number proportional to the incident photon’s energy. This crystal scintillator is coupled to a photomultiplier tube, where the emitted photons eject electrons by the photoelectric effect. These electrons then go through the dynode chain, the effect of which is to amplify the number of electrons to a level that can be recorded as a pulse of current. The strength of this current is proportional to the energy of the original incident gamma photon, and so this setup allows us to measure the energy of incident gamma photons, and the relative intensity of different energies.
The most significant drawback to this setup is that fluctuations occur in many of the processes, which causes gamma photons of one energy to be recorded as a spread of energies. The centre of this spread approaches the correct value as the number of counts increases.
The plastic scintillator, used only in the Compton scattering part of the experiment, is connected to a similar photomultiplier tube. However, in this case the incident gamma photons are freeing electrons, and it is these free electrons that are recorded. This scintillator’s output is connected to a gate, which allows us to discount all measurements at the final detector which are not immediately preceded by a detection at the plastic scintillator. This allows us to make more accurate detections, while greatly increasing the time required for an accurate measurement.
First, the computer program that analysed the detector’s output needed to be calibrated. This was done by placing a weak 137Cs source on the mount in front of the detector (see Fig. 1). This is known to produce gamma rays of energy 662KeV. It also excites electrons in the lead shielding, which then emit 76KeV photons to return to ground state. By analysing the spectrum of 137Cs and matching these expected values to the peaks they produced, the system was calibrated to give answers in KeV, rather than channels. This was then made more accurate by analysing the spectrum of 60Co and recalibrating with the previous two peaks, and the peak expected at 1173KeV.
With the system calibrated, the weak Cs source was removed, and the strong Cs source was unshielded. This source did not emit gamma rays directly to the crystal scintillator connected to the detector; it emitted to the plastic scintillator, which scattered gamma photons through 360 degrees. The crystal scintillator and detector were then pivoted about the plastic scintillator, from 20 degrees to 110 degrees, in 10 degree increments. Measurements could not be made below 20 degrees, as photons that passed through the plastic scintillator would drown out the scattered photons, and measurements could not be made above 110 degrees as lead shielding would interfere with the path of the gamma photons. Due to the gate, recording times needed to be long in order to get useful results, and so each angle was recorded for at least 10 minutes. The system was occasionally recalibrated throughout this procedure.
Results and Discussion:
The energies of the scattered gamma photons Eγ` were found at different angles θ, and these are shown in Table 1. After the modifications shown there, they were plotted in Figure 2. The line of best fit and its errors are shown in Table 2.
We can derive the formula 1/ Eγ` = 1/Ee(1-Cos θ) + 1/Eγ (Eqn 1) from equations for conservation of energy and momentum. This equation corresponds to the trendline found for Fig 2, in the form of y = mx + b. Thus, using Table 2, Ee = 500±20 keV, and Eγ = 650±20 keV.
Eγ is the energy of the incident gamma photon. It is known to be 662keV, and so this shows us that the given errors are appropriate, at least in this case. Ee is the energy of the freed electron, and the value given for this is interesting in that it shows most of the energy goes to the electron. If the energy of the incident photon were greatly increased, the energy of the scattered photon would be barely changed, and so the energy of the scattered electron would be greatly increased. Thus in all cases the electron takes the majority of the energy of the incident photon.
Conclusion:
The energy of a photon scattered by Compton scattering was found to be related to the angle at which it is scattered, by Eqn 1. This relation strongly outweighs the relation to the incident photon’s energy, as the majority of the incident photon’s energy goes to the scattered electron. Through this formula, the values of scattered photons and electrons can be determined at various angles. This served as a useful method to understand the behaviour of gamma photons, and to see how the photoelectric effect ceases to accurately describe the behaviour of electrons and incident photons, when these photons have sufficiently high energy.
Sorry the format is a bit wonkier here than in my word document. If there's any mention in there of natural background radiation, ignore it, it's from the other half of the experiment that I tried to cut out of this because it's irrelevant to my question.
Are you sure about this? It's basically the photoelectric effect, but with a lot more energy, so a photon is produced as well. Pair production is entirely different because the pair comes out of the nucleus, not the electrons orbiting the nucleus.
The photoelectric effect, Compton scattering and pair production are all similar ideas on different energy scales. You're right about pair production involving a nucleus though.
Edit: Look the wikipedia article on Compton Scattering (Which it sounds like you should have a quick read of) even has the graph which explains the energy.
However, I got this equation that gives a constant for the electron energy, and the professor ticked it. Was he just not paying attention, or am I missing something here?
Okay, here it is:
[IMG][/img]
If you plot the inverse of the final energy versus (1-cos(theta)), the y intercept is indeed the inverse of the photon's initial energy, and the slope is 1/mc^2, which is the inverse of the rest mass of the electron, which is a constant.