Questions about neutron interactions

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  • #1
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So I am reading a book "Fundamentals of Nuclear reactor physics" by Elmer E. Lewis
On page 37 it talks about neutron interactions. some sentences I feel are left a bit short so let me rephrase here and hopefully get a clear answer.

We have a neutron with a given kinetic energy and bunch of atoms around it. Now one of the following things might happen upon that neutron striking a nucleus
1) Neutron gets scattered, where the scattering can be either
a) Elastic
b) Inelastic
2) Neutron gets absorbed by the nucleus


So my take is that the elastic scattering is when the neutron hits a nucleus but instead of having any interaction with it just changes course and maintains it's kinetic energy , the nucleus is left as is.(this is the impression i got from what was written in the book although it seems weird to me that one side can be left unaffected in a collision)
In inelastic scattering the neutron imparts some of its kinetic energy to the nucleus , the neutron flies off with a lower kinetic energy while the nucleus is now excited and releases energy as gamma rays and gives off a neutron (according ton the book) Is this the case and if it is could you give me an example reaction where this happens and can this happen with any nucleus or are there only certain nucleus that respond in this manner?

And lastly absorption reactions are where the neutron is absorbed by the nucleus creating a different isotope from a previous isotope like U 236 from U 235, which will then later decay to other elements?

Can a stable nucleus with an atomic number lower than lead also absorb a neutron?


thanks
 
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  • #2
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So my take is that the elastic scattering is when the neutron hits a nucleus but instead of having any interaction with it just changes course and maintains it's kinetic energy , the nucleus is left as is.
That would violate conservation of momentum or energy.
The nucleus will get some momentum and therefore energy if the neutron changes its path.
while the nucleus is now excited and releases energy as gamma rays and gives off a neutron (according ton the book)
It will give off something, but it doesn't have to be a photon plus a neutron. It can be just one of them, or a proton, or other combinations. Photons are quite common and neutrons are common at higher energy.
Can a stable nucleus with an atomic number lower than lead also absorb a neutron?
Sure. All the tritium we have comes from that process in nuclear reactors, for example.
 
  • #3
mathman
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Elastic scattering: both momentum and kinetic energy are conserved like billiard balls.
Inelastic scattering: some energy ends up exciting the nucleus - as if the neutron is absorbed and immediately re-emitted.
 
  • #4
artis
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@mfb I did realize after my own post that DT fusion is with tritium and tritium is made from lithium bombardment of high energy neutrons right? Also the scheme for tritium production in tokamak blankets.


@mathman so in elastic scattering the neutron hits the nucleus but is outside the absorption cross sections most likely and so just imparts some of it's energy , so the neutron changes trajectory with less kinetic energy than before and the remaining lost kinetic energy is given to the nucleus that was hit?
If the answer is yes then what happens to the nucleus that now has a higher kinetic energy?
Assuming it did not capture or absorb the neutron there shouldn't be any nuclear processes like gamma emission or transmutation etc so given that a nucleus is not alone but in a material does it simply gain a higher kinetic energy (temp) for a brief moment and then "thermalizes" some time after back down to the average temp/energy of the atoms of the material? Can it be said to be similar to how neutrons themselves "thermalize" in a moderator like water losing their own kinetic energy but raising the energy of the moderator as a consequence?



As for the inelastic scattering I'm not sure i fully got your point, you say "as if" the neutron is absorbed and released, so is it then absorbed and released? From what I read I would gather that it is and with it comes some additional form of either particle or EM radiation?
https://en.wikipedia.org/wiki/Inelastic_scattering

I get that from wiki.
So the nucleus absorbs the neutron briefly then "spits" it out giving it some of the kinetic energy it originally had back? and with it it also emits some other form of radiation,
the final question is, after the nucleus has done all of this it then becomes a different isotope or an isotope so it becomes radioactive and continues to decay much like the split nucleus that underwent fission just in a different decay chain?


thanks
 
  • #5
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@mfb I did realize after my own post that DT fusion is with tritium and tritium is made from lithium bombardment of high energy neutrons right? Also the scheme for tritium production in tokamak blankets.
Yes. There is also the D+n->T reaction in fission reactors that use heavy water.
If the answer is yes then what happens to the nucleus that now has a higher kinetic energy?
It scatters off other atoms/nuclei in the material and is stopped that way. You get a bit of localized radiation damage (in a solid material) and a bit of heat.
 
  • #6
snorkack
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As far as conservation of momentum and energy are concerned, the neutron can elastically scatter off nuclei in such a manner that the momentum is not transmitted to one nucleus alone (which would take up a large recoil energy) but to a whole crystal in which the nuclei are.
 
  • #7
mathman
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By accident I posted this yesterday for a different question.

Simplified description for elastic scattering: In general (proton) being the exception) the nucleus receives little kinetic energy, so that it simply bounces little more. Furthe in case of a solid the energy gets spread out (as described previously). Protn scattering could give the proton half the K.E. of the neutron and move freely taking the molecule (water?) with it or possibly getting loose.

Inelastic scattering is more complicated and the target nucleus will be in an excited state, where what happens depends on the nucleus.
 
  • #8
artis
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ok, here is one more thing with regards to elastic scattering.
to put this short , I read that the fraction of kinetic energy a neutron loses upon elastic scattering in a single event is roughly the same for different neutrons irrespective of their initial KE. This fraction is only dependent on the mass of the nucleus it scatters from.

So I take this as such , as an example if I have a fast neutron having 2MeV and it hits a particular nucleus , like that of deuterium , the book says it needs about 25 such events to decrease it's KE to about 0.025eV.
So if I take the 2MeV to represent 100% and 0.025eV to represent 0% of my neutron energy then in case of deuterium the neutron loses about 4% it's energy each time it elastically scatters from a deuterium nucleus.
This percentage of energy loss per collision seems to get lower as the mass of the nucleus in question gets higher.


My take away from this is that the heavier the nucleus the less energy a neutron can impart on it in an elastic collision and more energy is retained by the neutron itself.
Have I understood everything correctly so far?



ps. this property seems to only work at the quantum level as I cannot think of an analogy in the macroscopic world. Although if I took two rubber balls one small one large and heavy the smaller one even traveling very fast would rebound from the larger one and retain much of it's original KE while the larger one would barely move.
 
  • #9
mathman
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Elastic scattering is best understood as if the collision is macroscopic - moving ball colliding with a stationary ball of a (usually) larger mass. The recoil angle is random. The amount of energy transferred depends on the angle using conservation of momentum and kinetic energy.
 
  • #10
snorkack
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ok, here is one more thing with regards to elastic scattering.
to put this short , I read that the fraction of kinetic energy a neutron loses upon elastic scattering in a single event is roughly the same for different neutrons irrespective of their initial KE. This fraction is only dependent on the mass of the nucleus it scatters from.

So I take this as such , as an example if I have a fast neutron having 2MeV and it hits a particular nucleus , like that of deuterium , the book says it needs about 25 such events to decrease it's KE to about 0.025eV.
So if I take the 2MeV to represent 100% and 0.025eV to represent 0% of my neutron energy then in case of deuterium the neutron loses about 4% it's energy each time it elastically scatters from a deuterium nucleus.
No.
If the neutron were to lose 80 keV at each collision then it would lose 4 % of its energy on first collision (80 keV of 2 MeV) but on the last of 25 collision, it would lose 99,99997 % of its energy (80 keV lost, 25 meV remaining).
4 % and 99,99997 % are not equal fraction.
Instead, the energy must diminish about 80 million times (107,9), so on each of the collisions energy must diminish about 100,316 times. Which means about 2,1 times.
This percentage of energy loss per collision seems to get lower as the mass of the nucleus in question gets higher.


My take away from this is that the heavier the nucleus the less energy a neutron can impart on it in an elastic collision and more energy is retained by the neutron itself.
Have I understood everything correctly so far?



ps. this property seems to only work at the quantum level as I cannot think of an analogy in the macroscopic world. Although if I took two rubber balls one small one large and heavy the smaller one even traveling very fast would rebound from the larger one and retain much of it's original KE while the larger one would barely move.
Yes, easy.
Collide two equal rubber balls. You can actually have the first losing all of its energy and transmitting all of it to the second. Only in case of central collision. In case of glancing collision, less is transmitted. On average, a certain fraction.
Collide a rubber ball with a wall. The wall will not move and the ball will lose no energy.
Collide the ball with a heavy object. The heavier the object, the less it moves and the less energy the ball loses.
 
  • #11
artis
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I guess I messed up my numbers with the percentages. So in theory the neutron loses an equal amount of power after each elastic collision but this equal amount is taken from the energy that it had before the collision. Because this energy changes after each collision the equal amount should be then calculated before each collision, I guess I took the starting value at all times , have I got it right now?
 
  • #12
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It loses an equal fraction. The absolute amount lost per collision decreases by orders of magnitude.
 
  • #13
artis
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But this gradual KE loss at an equal fraction per collision is not a 100% given rule ? I get that there might be collisions in which a neutron has a "head on" collision and loses much more of it's KE , and after such a collision it would not need as many other collisions to get "thermalized"?
Although this is probably a statistically rare event
 
  • #14
mathman
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But this gradual KE loss at an equal fraction per collision is not a 100% given rule ? I get that there might be collisions in which a neutron has a "head on" collision and loses much more of it's KE , and after such a collision it would not need as many other collisions to get "thermalized"?
Although this is probably a statistically rare event
The fraction is an average. The actual amount depends on the recoil direction.
 
  • #15
artis
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Ok so now that I also read about inelastic scattering collisions , I guess in a thermal reactor the main scattering is elastic and there is much less inelastic scattering going on right?
I looked at some graphs and it seems that for a neutron to be scattered inelastically the threshold energies for U238 and other materials are up about the 1MeV range and higher so the majority of neutrons lose their energy through elastic scattering in the moderator and so cannot participate in inelastic scattering anymore,

is there a good graph that would compare the percentages of elastic vs inelastic scattering in common thermal rector or materials in general? I suppose comparing elastic to inelastic in a graph should seem like an X as the elastic cross sections go up as the neutron energy decreases while the inelastic cross sections should go down as the neutron energy decreases?


ps. In general is it true that inelastic scattering threshold energies for neutrons are lower for higher atomic mass nuclei and these energies get higher for nuclei of lower mass/ fewer nucleons per nuclei ?

thank you.
 
  • #16
artis
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I hate to do these bump posts but I guess I can use them once in a while since my last post and the questions in it got overlooked, @mfb @snorkack @mathman
 
  • #17
snorkack
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ps. In general is it true that inelastic scattering threshold energies for neutrons are lower for higher atomic mass nuclei and these energies get higher for nuclei of lower mass/ fewer nucleons per nuclei ?
Likely. Larger nuclei can have more low lying excited states.
 
  • #18
artis
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There is a sentence in the book quote
Fast reactor spectra are concentrated in the keV and MeV range with nearly all of the neutrons absorbed before slowing down to energies less than a keV

I also looked at the graph showing neutron flux vs neutron energies and indeed fast reactor spectrum look like a hill with steep sides at the higher energy range .


Now my question is is the reason why in fast cores there are almost no thermal neutrons because fast reactor cores have high atomic weight materials and so neutrons born with fission energy don't get slowed down as fast as they would from scattering against a low atomic weight material like water. So each fission energy neutron can scatter with U238/235 etc many many times losing relatively small energy each time so by having so many collisions eventually the neutron gets either absorbed , captured or absorbed and causing a fission so no neutrons get left do "stay around" for long enough to have their KE decrease to thermal ranges.

Would this be about right?
 
  • #19
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Right. The chance that a neutron gets absorbed before it gets slow is higher.
 
  • #20
Giant_Cannoli
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Applied separation of charge in certain types of materials including crystals and ceramic molecules might make it possible to deflect or absorb a Neutron using a substantial static electrical charge.
And the mass would indeed be lower than lead.
 
  • #21
Vanadium 50
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What? Neutrons are neutral. How do you deflect them with a static charge?
 

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