Gamma spectroscopy, Compton edge

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Hello, I have a few uncertainties about this, Forgive the long questions, I appreciate your time.

1) Why is the maximum energy photon loses during Compton scattering about 0.238MeV (if maximum angle 180 achieved aka backscatter) irrespective of the incoming photon total energy? It seems counter-intuitive that the incoming photon irrespective of it's energy can lose no more than this amount which puts the "Compton edge" at a maximum of 0.238 MeV from the Photopeak.

2) Why is the electron in Compton scattering considered "free" yet the ones in the photoelectric effect are bound regular electrons , in both cases the material is the same , for example a HPGe detector material. Especially given the energies for the Photoelectric effect and Compton effect overlay and either one can happen.

3) Looking at the Compton edge spectrum , is it true that most photons tend to be scattered at or near their maximum scattering angle/energy, as it seems that the amplitude rises as one get's further away from the photopeak and hits it's highest value at the Compton edge?


4) Before I ask this I want to acknowledge that I know that a detector counts gamma energies as electron current and electron current arises in the detector when a gamma interacts with one which then interacts with others causing a cascade in the presence of an applied E field.
This being said I cannot fully understand why for example the Photoelectric effect causes a photopeak but the Compton scattered electrons cause their peak/edge at a energy which is close to the photopeak instead of the energy maximum that a single photon can transfer during Compton to a electron.
I would suspect there are a distribution of photons namely , those that scatter and then leave the detector VS those that scatter within the detector and then further interact within the detector.
Now the first ones , those that leave after scatter simply produce a recoil electron with an energy up to 0.238MeV , while those that scatter and further interact in the detector like create a Photoelectron get counted with the energy they had after the Compton scatter event.
So I would suspect that the Compton edge is created by the second kind of photons - those that scattered first off an electron and then further interacted within the detector and the energy registered is corresponding to that which they had after the Compton event?
If this is so then why there isn't also a peak for the photons that Compton scattered but then left the detector? Or is the increased amplitude continuum at the very beginning from 0 eV up to about 240KeV corresponding to these very electron whose photons escaped during scatter?


5) Is the reason why in pair production the "single escape" peak is higher than the value of a 0.511 MeV photon because the detector counts the value of the 511 KeV photon-electron interaction + the energy the original gamma imparted to the produced pair before pair annihilation?
 

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  • #2
gleem
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1) Why is the maximum energy photon loses during Compton scattering about 0.238MeV (if maximum angle 180 achieved aka backscatter) irrespective of the incoming photon total energy? It seems counter-intuitive that the incoming photon irrespective of it's energy can lose no more than this amount which puts the "Compton edge" at a maximum of 0.238 MeV from the Photopeak.

The kinematics of Compton Scattering gives the difference in the scattered wavelength and the incident wavelength Δλ = h/mec(1 - cosθ ) where θ is the scattering angle, me is the rest mass of the electron, h is Planck's constant and c is the velocity of light.

As you can see the loss of energy as determined by the increase in Δλ of the photon only depends on the scattering angle. The maximum change in wavelength is 0.04692x10-8 cm. But a constant difference in wavelength does not mean there is a constant difference in energy or a limit on the maximum energy difference.
 
  • #3
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But a constant difference in wavelength does not mean there is a constant difference in energy or a limit on the maximum energy difference.
Do you mean this in the sense that for two equal scattering angles the photon with a higher frequency would give more energy to the electron?
 
  • #4
gleem
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The higher the photon frequency (energy) the more energy that is transferred to the electron like any collision.

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Thanks @gleem for answering so far , maybe @mfb or anyone else can lend their knowledge with respect to the other questions still left in my post #1?


Meanwhile I want to ask one more thing,
The reason Compton scattered gammas get counted is because they interact with an electron after scattering, so if this interaction is the photoelectric effect for example then the whole leftover energy of the gamma gets registered (whatever that energy is) , but I was wondering whether the change in trajectory/direction after the scatter has any impact on the probability of further interaction for the photon? My reasoning is that for a given size, shape and volume detector a straight line trajectory means less material is traversed by the photon and it spends less time in the detector volume while a scattered trajectory means it spends more time within the volume and has a longer path within it, so in theory the probability of interaction increases?
So it would seem to me that if a photon managed to get scattered then it's probability of interaction increases compared to one that wasn't scattered given both had the same incoming trajectory and energy.

Can a single photon get scattered twice in a row? I would think it should be able but the probability of that happening within a small volume is little?
 
  • #6
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1) as discussed the energy loss is unlimited.
2) The photoeffect happens at an energy of a few eV where the binding energy of electrons is relevant. The Compton effect is in the MeV range where the binding energy is negligible (pretty much by definition, otherwise it's not called Compton effect).

I don't understand questions 3 and 4 and I think they are based on your misconception about (1).
 
  • #7
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1) as discussed the energy loss is unlimited.
By which I suppose you mean what has already been said that the higher the incoming gamma energy the higher the KE given to the electron even though the total photon energy loss is always limited to the "Compton wavelength" or 0.238 MeV?
I must say this is rather counter intuitive although makes sense given at what energy those 0.238MeV are lost, if they are lost from 10MeV then that gives more energy than if they are lost from a gamma of 5 MeV.


2) The photoeffect happens at an energy of a few eV where the binding energy of electrons is relevant. The Compton effect is in the MeV range where the binding energy is negligible (pretty much by definition, otherwise it's not called Compton effect).
OK I understand, in both cases electron is ejected and it's binding energy is overcome but if that happens at high gamma energy the binding energy being so negligible is simply ignored.


The 4th question was meant like this, if I can clarify, the gamma comes in the detector, Compton scatters and loses some energy, this lost energy is given to an electron , the gamma then either leaves and is not counted further or interacts further within the detector. So if it interacts further , say by photoeffect, then I assume the Compton + resulting photoeffect gets counted together as if it were Photo effect to begin with and added to the photo peak due to the short time scale in which both of these events happen in series (Compton + photo) ?
But in the case where the photon only Compton scattered and then exits the detector without further interaction , why would those Compton electrons get counted at an energy level close to the photopeak?

I think I have an idea , let me see whether it's correct,
The 0.238MeV aka the Compton wavelength is the maximum energy a photon can lose in a single interaction but it is not the maximum energy the electron gains , the electron gains the total incoming gamma energy minus whatever energy the photon lost in the scatter?
So a 1.5MeV gamma that lost say 0.230 MeV would scatter an electron with an energy of 1.270MeV?
 
  • #8
gleem
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. . . . even though the total photon energy loss is always limited to the "Compton wavelength" or 0.238 MeV
Are you saying that the energy loss of the scattered photon is limited to 0.238 MeV? Also, the Compton wavelength is measured in cm, not MeV. The Compton wavelength is the wavelength of a photon whose energy is the same as the rest mass energy of an electron, 0.511 MeV.
 
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  • #9
Astronuc
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Can a single photon get scattered twice in a row? I would think it should be able but the probability of that happening within a small volume is little?
A scattered photon can scatter again.

As initial photon energy increases, the photon is more likely to be forward scattered.

Consider that for low energy photons, the photoelectron effect is more likely, i.e., a photon is more like to be absorbed, which yields the photopeak.
https://www.nrc.gov/docs/ML1122/ML11229A703.pdf

As photon energy is increased Compton scattering competes with the photoelectron (or photoelectric) effect. Beyond a photon energy of 1.022 MeV, one begins to observe pair production. The energy at which the probability of Compton scattering (any angle) equals that of pair production is dependent on the Z of the nucleus of the atom interacting with the photon.

http://rcwww.kek.jp/research/shield/photon_r.pdf
https://courses.physics.illinois.ed...nteraction of x-ray and gamma ray photons.pdf

If one compares the cross-sections for photon interaction with Fe vs Zr vs W, one can see the shift.

Compton scattering from a free electron is reasonable for a valence/conduction electron (binding energy ~ eV) or a light element, where the ionziation energies are low. For heavier (periods 6 or 7) elements, e.g., error would be introduced if the scattering involves a K- or L- electron depending on the photon energy, and one ignores the binding energy of the electron.

I did some calculations using the Hyperphysics calculator:

For Cs-137, with incident energy Ei = 0.662 MeV, produces backscattered photon energy Es, and electron energy as follows:
Es 0.18435 MeV
Ke 0.47765 MeV

Similarly for 1 MeV photon, 10 MeV photon

Ei 1.0 MeV
Es 0.203504 MeV
Ke 0.796495 MeV

Ei 10 MeV
Es 0.249134 MeV
Ke 9.750865 MeV

See discussion on relativistic electrons
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/photel.html#c1
 
  • #10
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Are you saying that the energy loss of the scattered photon is limited to 0.238 MeV? Also, the Compton wavelength is measured in cm, not MeV.
Yes, right wavelengths get measured in length measurement units.

I had made a huge mistake it seems, somehow I listened about the Compton wavelength and got the impression that the maximum amount of energy a photon loses is no more than the energy difference between photopeak and Compton edge and that is why it sounded so weird to me why I cam here and asked.

Now I get it and it sounds fine and reasonable finally and logical that the whole Compton spectrum from 0 up to the Compton edge is made by Compton scattered electrons.
Also the link @Astronuc gave explains this clearly.
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/photel.html#c1

So now I see that a photon in a single Compton scatter event can lose almost all of it's energy to the electron except for a tiny bit (the now infamous 0.238MeV) which is why the Compton spectrum falls off after the Compton edge.
I read that for very high energy gammas the "Compton wavelength" energy difference becomes 0.255MeV or half the electron rest mass, what makes this change happen for increased energy ?
 
  • #11
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By which I suppose you mean what has already been said that the higher the incoming gamma energy the higher the KE given to the electron even though the total photon energy loss is always limited to the "Compton wavelength" or 0.238 MeV?
I must say this is rather counter intuitive although makes sense given at what energy those 0.238MeV are lost, if they are lost from 10MeV then that gives more energy than if they are lost from a gamma of 5 MeV.
That makes no sense and you should read the previous answers, in particular post #2. The energy loss is not limited and neither is the electron energy. The wavelength change is limited but that's a completely different statement.
 
  • #12
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That makes no sense and you should read the previous answers, in particular post #2. The energy loss is not limited and neither is the electron energy. The wavelength change is limited but that's a completely different statement.
Agreed , I already pointed out in my post #10 that I had made a mistake and now it's clear.
 
  • #13
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A scattered photon can scatter again.
This make me ask, can a photon lose almost all it's energy in scattering? I would think it cannot as losing all it's energy would mean an absorption, but how low can a photon go and still exist as the "original photon"?
Like for example a 10MeV photon undergoing a backscatter event losing almost all except, say 0.240 MeV of it's initial energy to a electron (I read the probability for high energy photons to backscatter is low) and then that photon becomes a 0.240MeV photon (much lower frequency, since energy=frequency) and say it again backscatters, how much could it possibly lose the second time from those 0.240MeV?
Because the way I understand it, there can be no such thing as a photon with 0 energy/frequency, but can a gamma scatter down from initial MeV's of energy to eV's?


There is one misunderstanding I'm having at the moment.

here in the video from MIT, at 1:35 the lecturer talks about the wavelength shift of the photon.
The claim is that "wavelength shift itself is independent from photon energy, just a matter of scattering angle" The formula seems to show it too.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compton.html
The text says
The wavelength change in such scattering depends only upon the angle of scattering for a given target particle.
Now, since wavelength=frequency= energy of photon, then if a 3MeV photon backscatters , it loses say 2.5MeV while if a 300KeV photon backscatters it loses say 250KeV , clearly 2.5Mev is higher than 250Kev in energy/frequency , so would I be correct in saying that for a fixed angle (whether 90 or 180 or any other) a photon can lose a fixed percentage of it's total wavelength/energy but the amplitude of this wavelength/energy gets proportionally larger as the initial photon energy increases or gets smaller as the initial photon energy decreases?

Like for example 2 photons can lose 50% of their energy at a fixed angle of scatter but 50% of 3 MeV is more than 50% of 300KeV.




And one final somewhat unrelated question , since photon's frequency/energy is not dependent on it;s velocity could it be in theory that, for example, if a gamma ray velocity through a known thickness and density material could be slowed down the probability of it's interaction with said material would increase?
 
  • #14
Astronuc
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Now, since wavelength=frequency= energy of photon, then if a 3MeV photon backscatters , it loses say 2.5MeV while if a 300KeV photon backscatters it loses say 250KeV , clearly 2.5Mev is higher than 250Kev in energy/frequency , so would I be correct in saying that for a fixed angle (whether 90 or 180 or any other) a photon can lose a fixed percentage of it's total wavelength/energy but the amplitude of this wavelength/energy gets proportionally larger as the initial photon energy increases or gets smaller as the initial photon energy decreases?
Bear in mind that the frequency of a photon, ν, is given by c/λ, where c = speed of light (taken as constant) and λ is wavelength, and the photon energy, E, is given by E = hν = hc/λ. As E and ν increase, λ decreases.

Try some examples with the Hyperphysics calculator, or use a spreadsheet or python program and do a sensitivity study, or differentiate the formula for the change in energy (or wavelength) of a photon with respect to the initial photon energy. Also, one can the change-in-wavelength formula and rewrite it in terms of energy of the photon before and after. Actually, the after photon is different than the initial photon (different wavelengths/frequency/energy).

I did 3 examples at 250 keV, 511 keV (~rest mass of e) and 2.5 MeV with a scattering angle 90° and 180°

Code:
Ini. Ph   Scatt. Angle 90°     Scatt. Angle 180
  Ei        Ef, keV  Ke, keV     Ef, keV  Ke, keV     
250 keV    167.87    82.12      126.36   123.64
511 keV    255.5    255.5       170.33   340.67
2.5 MeV    424.28  2075.72      231.81  2268.19

One could also try changing the rest mass of the electron and see what effect it has.
there can be no such thing as a photon with 0 energy/frequency, but can a gamma scatter down from initial MeV's of energy to eV's?
At low energies, the photon is more likely to be completely absorbed by an electron, at a greater probability than being scattered (as in the Compton effect). Also, at low energies, a photon may experience Rayleigh scattering (aka, elastic or coherent scattering).

See - https://www.utoledo.edu/med/depts/radther/pdf/RDII - Chapter 7n handout.pdf
https://en.wikipedia.org/wiki/Elastic_scattering#Optical_elastic_scattering

As for 'free' electrons, I like the statement "The predominant inelastic scattering process around 1 MeV is the well known Compton effect in which the gamma ray is scattered by anatomic electron, where the photon energies (Eo) are so large compared with the electron binding energies that the electrons can be considered as free." In other words, Eo >> Ee,b, the latter being the binding energy of the electron in an atom.
Ref: https://scholarworks.wmich.edu/cgi/viewcontent.cgi?article=4117&context=masters_theses

I'm interested in the influence of gammas on structural materials, and mainly the effect on chemical potential of specific elements in an alloy, as well as the atomic displacement, besides the development of a so-called point defect in a crystal lattice. During the university program, in course on radiation dosimetry and protection, the emphasis was on the attenuation effects, rather than the material effects.
 
  • #15
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Bear in mind that the frequency of a photon, ν, is given by c/λ
Right , so the frequency does change as the speed of the photon deviates from c to below that, like a ray of light gets attenuated while passing through water or glass so it's frequency drops and wavelength increases.
So the sun should seem whiter looking at it directly (if one could) than looking at it through bedroom window.

I did 3 examples at 250 keV, 511 keV (~rest mass of e) and 2.5 MeV with a scattering angle 90° and 180°
Thanks, I can see that I was wrong , even for the same angle different photon energies will result in different percentage of energy divided between the scattered photon and electron.
I'm interested in the influence of gammas on structural materials, and mainly the effect on chemical potential of specific elements in an alloy, as well as the atomic displacement, besides the development of a so-called point defect in a crystal lattice. During the university program, in course on radiation dosimetry and protection, the emphasis was on the attenuation effects, rather than the material effects.
Presumably with regards to nuclear reactors and their vessels containment?
 
  • #16
Astronuc
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Right , so the frequency does change as the speed of the photon deviates from c to below that, like a ray of light gets attenuated while passing through water or glass so it's frequency drops and wavelength increases.
The speed of light (in a vacuum) is constant. The change in wavelength or frequency is due to loss of momentum (and energy) of a photon to an electron it encounters.
http://hyperphysics.phy-astr.gsu.edu/hbase/ems1.html

Presumably with regards to nuclear reactors and their vessels containment?
Pressure vessel steels are one concern. I am more interested in materials (alloys and ceramics) operating in the core, e.g., stainless steels providing support for in-core instrumentation, e.g., thermocouples and neutron flux measurement, or control elements.

There is a comment in a manual for a code, PENELOPE, that simulates electron/positron and gamma radiation behavior in materials.
Atoms are primarily ionised by photon interactions and by electron or positron impact. There is a fundamental difference between the ionising effects of photons and of charged particles. A photon is only able to directly ionise a single atom. In the case of photoabsorption, when the photon energy is larger than the K-shell binding energy, about 80% of photoabsorptions occur in the K shell, i.e., the resulting ion with a vacancy in the K shell is highly excited. Incoherent scattering is not as highly preferential, but still the probability that an inner subshell is ionised is nearly proportional to the number of electrons in the subshell. Conversely, fast electrons and positrons (and other charged particles) ionise many atoms along their paths; the ionisations occur preferentially in the less tightly bound atomic subshells, or the conduction band in the case of metals, so that most of the produced ions are only weakly excited.
In a nuclear reactor, each fission produces 2 or 3 neutrons, on average about 2.3 +/-, as well as 7 or 8 spontaneous (prompt) gammas, then at least two gammas from the two fission products. To maintain the reaction, one neutron must survive to be absorbed by another fissile atom (235U or 239Pu), and the remaining neutrons are absorbed without causing fission in a process called 'radiative capture'. Both 235U or 239Pu can absorb a neutron without fissioning through gamma emission becoming 236U or 240Pu, respectively. Basically, other nuclides (elements) resident in the core have a chance to absorb neutrons (e.g., 58Ni + n -> 59Ni, 59Ni + n -> 60Ni, 59Ni + n -> 61Ni, . . . What is interesting about these three reactions is the energy of the gamma emitted, ~9 MeV, 11.4 MeV and 7.8 MeV, respectively. 59Ni is not stable, or naturally occurring, so must be created by the first reaction (or by (n,p) reaction with 59Co, which is usually minimized in stainless steels). Radionuclides of 239Pu and 241Pu produce gamma of ~5.5-5.6 MeV, and certain other nuclides produce high energy gammas, E > 3 MeV. It is the effect (through pair production and Compton scattering) of such gammas that is of most interest.
 
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  • #17
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@Astronuc somewhat related question I had in mind, do all radioactive isotopes and elements in general produce gamma radiation so they can be identified by gamma spectroscopy or are there elements that don't produce gamma at all?
 
  • #18
gleem
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Yes. There are a number of isotope decays that do not produce gammas C14, P33, S35, Ar37, , , , , Hg,194, Pb209 to name a few. However, the resulting atoms will still produce x-rays from disrupted electron orbits resulting from the decay processes which can be used to identify them.
 
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  • #19
Astronuc
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@Astronuc somewhat related question I had in mind, do all radioactive isotopes and elements in general produce gamma radiation so they can be identified by gamma spectroscopy or are there elements that don't produce gamma at all?
Yes. There are a number of isotope decays that do not produce gammas C14, P33, S35, Ar37, , , , , Hg,194, Pb209 to name a few. However, the resulting atoms will still produce x-rays from disrupted electron orbits resulting from the decay processes which can be used to identify them.
My bold for emphasis. These are examples of beta decay with no gamma emission. T decay is another example.

However, when a nuclide absorbs a neutron, it will emit a gamma ray, i.e., prompt gamma with a unique signature. In these cases, a nuclide can be identified by the prompt gammas, which is the basis of prompt-gamma activation analysis.
https://www.nist.gov/laboratories/tools-instruments/prompt-gamma-ray-activation-analysis-pgaa
https://link.springer.com/article/10.1023/A:1006796003933

PGAA requires a neutron beam incident upon a sample, whereby some atoms absorb a neutron. Decay gamma-spectroscopy requires one to wait for the radionulide to decay. In the absence of gamma emission, one would detect the beta particle (and energy) or X-rays from internal conversion. https://www.sciencedirect.com/topics/chemistry/internal-conversion

Some examples of gamma emission.
https://www.nndc.bnl.gov/capgam/byTarget/z015_31P.html (31P + n => 32P)
https://www.nndc.bnl.gov/capgam/byTarget/z016_34S.html (34S + n => 35S)
https://www.nndc.bnl.gov/capgam/byTarget/z082_208Pb.html (208Pb + n => 209Pb)

Unless one separates individual isotopes, one will receive several isotopes from activation analysis, depending on the element.
 
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  • #20
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Additionally if I may go back a step with regards to the scattering, I have a question
In the post #14
I did 3 examples at 250 keV, 511 keV (~rest mass of e) and 2.5 MeV with a scattering angle 90° and 180°

Code:
Ini. Ph Scatt. Angle 90° Scatt. Angle 180
Ei Ef, keV Ke, keV Ef, keV Ke, keV
250 keV 167.87 82.12 126.36 123.64
511 keV 255.5 255.5 170.33 340.67
2.5 MeV 424.28 2075.72 231.81 2268.19
If I look at the 2.5MeV photon scattering off an electron with an angle 180° (backscatter) it seems the photon loses most of it's energy to the electron which now has a KE of 2.268MeV even though the photon energy was more than 4 times larger than the electron rest mass of 511KeV ,on the other hand, in elastic scattering between massive particles, IIRC, a incident particle with higher KE than the target particle rest mass can lose no more than the target particle rest mass in a single collision?

This as far as I'm aware is the reason for water as moderator in thermal fission reactors since hydrogen proton has roughly the same rest mass as neutron so a neutron can lose almost all it's energy during a single collision , but what if the neutron KE was so large that it was twice the proton rest mass (neglecting proton KE due to thermal energy) would it then at most lose no more than roughly half it's KE which would sum to be the proton rest mass?
 
  • #21
Astronuc
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but what if the neutron KE was so large that it was twice the proton rest mass (neglecting proton KE due to thermal energy) would it then at most lose no more than roughly half it's KE which would sum to be the proton rest mass?
In a fission reactor, 'prompt' fission neutrons have a spectrum of energy, with a most probably energy somewhere ~0.8 MeV, and an average energy round 1.9 MeV, as compared to a rest energy of a neutron 939.566 MeV, or proton, 938.272 MeV, so a 1.877 GeV neutron would be extreme. Such a high energy neutron would have to come from a highly energetic nuclear spallation reaction or a particle (e.g., p+p) collision, not fission, and not even (d+t) fusion, En ~14.1 MeV.

Even so, the neutron could lose most of its energy to a proton, but would more likely scatter, and perhaps cause some interaction. I am not familiar with the results of such collisions, so I'd have to read the literature on 1.9 GeV neutrons colliding on protons at rest.
 
  • #22
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@Astronuc Yes I agree not a simple question but I was wondering whether macroscopic analogies work well for elastic scattering, they seem to work well for low energies whereby an object colliding with another object of similar mass can transfer as much as all it's KE to the other object , namely , like a billiard ball.
Now as you confirmed fission neutrons have negligible KE compared to the rest mass of either proton or neutron and given both particle rest masses are similar one can transfer most or all it's extra KE (extra meaning above that of it's rest mass) in a single collision so then I was wiling to go a step further and contemplate whether a "superfast" neutron could also give away most or all it's energy in a single collision like with low energies or whether there the rules change?
 

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