Neutron moderation in a Hydrogen medium and scattering

  • Thread starter dRic2
  • Start date

dRic2

Gold Member
529
93
Hi, I'm reading chapter 6-3 of Lamarsh's book "Introduction to Nuclear Reactor Theory". Here it is discussed the very idealistic case of Hydrogen being used as a Moderator (without adsorption).

The moderator is:
- infinte
- homogeneous
- with uniformly distributed source emitting at constant rate ##S## and constant energy ##E_0##
- (of course) made of Hydrogen
- the nuclei are considered at rest.


In the book it is said that the emitted neutrons hit the Hydrogen atoms and so they lose energy (this I understand perfectly). As a consequence the author assumes that the only possible outcome of a collision is the slowing down of neutrons.

I do not fully agree and I was looking for some help to better understand this.

My thought about what is "missing":

Neutrons emitted from the source have an amount ##E_0## of kinetic energy and they hit the Hydrogen nuclei at rest so that they transfer to them some of their staring kinetic energy thus they (neutrons) slow down (to an energy level ##E##). BUT the Hydrogen atoms have gained some kinetic energy (and this isn't something to be neglected because Hydrogen nuclei and neutron have the same mass!) and so they (Hydrogen nuclei) are now moving with some velocity and kinetic energy. Now that the nucleus in no more at rest if it collides with a neutron it (the nucleus) may "give back" some of the energy he stole in the previous collision and in this way the new neutron could be accelerated instead.

My possible explanation:

The author doesn't seem to care about the kinetic energy of the nuclei (ie the temperature of the moderator) so I guess he assumes that once the collision has occurred the recoiling nucleus is somehow "stopped" (ie the moderator is cooled) otherwise there would be a loss of energy in the system.

What do you think? Can you help me figure it out?

PS: I also like to know why neutron-neutron scattering is totally neglected. Is that so rare ?

(I upload the first page of the chapter for completeness)
 

Attachments

mathman

Science Advisor
7,688
389
The assumption is that the nuclei are at rest, or the moderator temperature is at absolute zero. In reality, the practical problem is slowing down fast neutrons so that they are at the same temperature (thermal) as the moderator. The energy that the protons pick up get moderated. The number of neutrons involved is overwhelmingly smaller than the number of nuclei in the moderator.
 

dRic2

Gold Member
529
93
The energy that the protons pick up get moderated
Ok so basically it is constantly cooled down otherwise this is not possible, right ?
 

anorlunda

Mentor
Insights Author
Gold Member
7,512
4,248
You're overthinking it. Thermalizing a high energy particle involves potentially thousands of scattering collisions. On the average, heat flows from high temperature (high KE) to low temperature. But it does not say that each individual collision moves uniformly in that direction.

The whole idea of thermodynamics, and statistical mechanics is based on averages. If you consider particles one-at-a-time, you can't even define temperature.

In an analogous way, current in a conductor is not the same as instantaneous motion of each electron. See below/

upload_2019-2-12_18-16-34.png
 

Attachments

Astronuc

Staff Emeritus
Science Advisor
18,553
1,682
Neutrons emitted from the source have an amount E0E0E_0 of kinetic energy and they hit the Hydrogen nuclei at rest so that they transfer to them some of their staring kinetic energy thus they (neutrons) slow down (to an energy level EEE). BUT the Hydrogen atoms have gained some kinetic energy (and this isn't something to be neglected because Hydrogen nuclei and neutron have the same mass!) and so they (Hydrogen nuclei) are now moving with some velocity and kinetic energy. Now that the nucleus in no more at rest if it collides with a neutron it (the nucleus) may "give back" some of the energy he stole in the previous collision and in this way the new neutron could be accelerated instead.
A proton has a charge so it would quickly lose any kinetic energy to the other hydrogen atoms. Fast neutrons from fission start with energies in the MeV range. It would be a very low probability of a neutron hitting a proton dead on center, so in most collisions, the neutron does not lose all of it's kinetic energy. To slow from MeV to eV usually requires many collisions.

The OP does not specify the state of the hydrogen: metallic (or solid), liquid or gas. The solid would have a greater macroscopic cross-section, the gas the least.

The OP does not specify a value for E0, so it could anything from eV to MeV, or perhaps even μeV to GeV. For practical fission systems, one would focus on thermal energies from about 0.02 eV to 10 MeV.

The cross-section for neutron-neutron collisions is very low.
https://iopscience.iop.org/article/10.1088/0954-3899/28/10/308/meta
 

dRic2

Gold Member
529
93
Thank you for all the replies. My problem is: "where does the energy go?"

When the Hydrogen medium moderates the fast neutron to thermal energies it must get hotter (the protons have to gain kinetic energy). If it gets hotter the cross section will change and it will moderate worse (ie neutrons won't slow down very easily).

In order to carry on the approach used in the book one should assume that the temperature of the medium doesn't change (ie the protons do not gain kinetic energy) so one must specify that the system lose energy somehow.

A proton has a charge so it would quickly lose any kinetic energy to the other hydrogen atoms.
Are you talking about radiation of a moving charge ? That's fine, but if that energy is captured by other Hydrogen atoms then the energy won't leave the system and as time goes by the medium will get hotter.

You're overthinking it. Thermalizing a high energy particle involves potentially thousands of scattering collisions. On the average, heat flows from high temperature (high KE) to low temperature. But it does not say that each individual collision moves uniformly in that direction.

The whole idea of thermodynamics, and statistical mechanics is based on averages. If you consider particles one-at-a-time, you can't even define temperature.

In an analogous way, current in a conductor is not the same as instantaneous motion of each electron. See below/

View attachment 238644
What you are saying is true, but I do not see the connection here. You have a source of neutrons in this case and if you do not manage to lose some energy the system will get hotter hotter no matter what. Then, as I said, the properties of the medium will change (slightly if it a solid medium -but I do not think solid Hydrogen exists- or much more if the medium is a liquid or gas) and the approach used in the book fails.

Summing up: I think there must be a way for the system to lose energy. What is it? Does the author simply assume that heat is removed from the system? This might look like a stupid question, but I really hate when little details are omitted (even if obvious) because I wasted a whole day thinking about this.
 
Last edited:

Astronuc

Staff Emeritus
Science Advisor
18,553
1,682
Are you talking about radiation of a moving charge ? That's fine, but if that energy is captured by other Hydrogen atoms then the energy won't leave the system and as time goes by the medium will get hotter.
Yes, and yes. The system is an infinite, homogenous mass of hydrogen. If a uniformly distributed source of emits neutrons, then the neutron population would reach some thermal equilibrium with the hydrogen. None of that is physically real, just hypothetical. Note that the title of the chapter is "Neutron Moderation Without Absorption", also not physically real. Free neutrons don't appear spontaneously on their own, and furthermore, free neutrons don't survive very long (half-life ~ 10.23 minutes). In a pure hydrogenous mass, neutron would eventually slow down and be bound by a proton to form a deuteron, the nucleus of deuterium atom.

One could take the neutron source strength and divide it among the number of protons for each neutron and determine an equilibrium energy, which would represent a temperature, as an approximation. For example, a 1 MeV neutron scatters among 1 million protons, thus the average energy would be 1 eV, assuming that the protons are at rest. The average energy is equivalent to 11605 K. If the protons had say an equilibrium thermal energy of 0.025 eV, then the average energy of the system would be 1.025 eV, or 11895 K.
 
Last edited:

dRic2

Gold Member
529
93
Yes, and yes. The system is an infinite, homogenous mass of hydrogen. If a uniformly distributed source of emits neutrons, then the neutron population would reach some thermal equilibrium with the hydrogen. None of that is physically real, just hypothetical. Note that the title of the chapter is "Neutron Moderation Without Absorption", also not physically real. Free neutrons don't appear spontaneously on their own, and furthermore, free neutrons don't survive very long (half-life ~ 10.23 minutes). In a pure hydrogenous mass, neutron would eventually slow down and be bound by a proton to form a deuteron, the nucleus of deuterium atom.
I agree.

The things that is killing me is that the author says:
Neutrons arrive in ##dE## as result of scattering collisions that occurs at higher energies.
Why only higher ? I agree that this is a very special/ideal/non-realistic case, but why not taking into account the possibility that neutrons get speeded up by protons? The protons might be at rest at first (ie at the beginning of the "experiment"), but they are no more while the experiment is going on for a while.

Again we could argue that this is a stupid thing to argue about because this is a very very "un-real" but at the end of the paragraph the author says:
(...)
$$ \phi(E) = \frac S {E \Sigma_s (E)}$$
(...)
While [the above eq.] is strictly correct only for infinite Hydrogen medium, it is often used to give rough estimates of energy-dependent flux in finite hydrogenous system such as water-moderated reactors.
I know "rough estimates" means nothing, but still... How should this method come remotely close to a real situation ?

thus the average energy would be 1 eV, assuming that the protons are at rest. The average energy is equivalent to 11605 K.
How did you came up with this temperature ? Did you use Maxwell-Boltzman distributions ?

PS: I read again my previous posts. I noticed some grammar and other kinds of errors. I'm sorry, I hope I made myself clear enough... I'm still learning english :smile:
 

dRic2

Gold Member
529
93
The things that is killing me is that the author says:
Why only higher ?
I'm not finished reading the whole chapter, but it seems this approach is kept also in other cases. As I said, the author assumes that nuclei are at rest "forever".
 

Astronuc

Staff Emeritus
Science Advisor
18,553
1,682
Why only higher ? I agree that this is a very special/ideal/non-realistic case, but why not taking into account the possibility that neutrons get speeded up by protons? The protons might be at rest at first (ie at the beginning of the "experiment"), but they are no more while the experiment is going on for a while.
The idea is that neutrons start at a much higher energy (e.g., MeV) than protons (at ~eV) in a moderating environment. In light water reactors, the moderator temperature is the coolant temperature, ~270-285°C for BWRs, and 285-330°C for PWRs.

The temperature equivalent of 1 eV is 11604.5 K.
https://physics.nist.gov/cgi-bin/cuu/Convert?exp=6&num=160&From=ev&To=k&Action=Convert

I was simply doing an arithmetic average on the 1 MeV divided into 1x106 protons. A Boltzmann distribution has an average value.

Upscattering of neutrons does occur, but only when the neutron energies achieve energies comparable to the thermal energies of the moderator, protons in this case. For example, a 1 eV proton would have would not increase the energy of a 100 eV neutron, let alone a 1 MeV neutron, but a 1 eV proton would certainly increase the energy of a 0.02 eV neutron. However, in LWRs, the thermal energy of the moderator is on the order of 0.05 eV (~580 K). I'll have to look through my library, but I just happen to read a statement last week that thermal fissions are most likely at 0.1 eV in an LWR. On the other hand, in LWRs, fast fissions account for about 7-10% of fissions in the core depending on the burnup of the fuel.
 

dRic2

Gold Member
529
93
Thank you. I think I get it a little bit now. This so an unreal scenario that you have to "neglect" lots of things... I mean this situation is so far aways from reality that it is easy to find contradictions (in my opinion). @anorlunda is right: I'm probably overthinking it. If I take this examples "less seriously" it makes sense and it is quite easy too.
 

Astronuc

Staff Emeritus
Science Advisor
18,553
1,682
Thank you. I think I get it a little bit now. This so an unreal scenario that you have to "neglect" lots of things... I mean this situation is so far aways from reality that it is easy to find contradictions (in my opinion). @anorlunda is right: I'm probably overthinking it. If I take this examples "less seriously" it makes sense and it is quite easy too.
I found such examples annoying, but they are like a gedanken experiment. In a star, one might find hydrogen at high densities, e.g., 1 kg/m3, but then the equilibrium temperature is much hotter, or the hydrogenous cores of Jupiter or Saturn where the equilibrium temperatures are much colder, but such examples would not be useful.

http://www.astronomy.com/magazine/ask-astro/2018/06/metallic-hydrogen-in-gas-giants

The closest we would have to solid hydrogen in a terrestrial nuclear system would be Zr hydride, ZrH2, which would have it's own issues at operating temperature. Zr hydride has also been considered for neutron shielding.

The idea is to isolate some of the physics, which is this case is in the energy domain, and not spatially dependent. In reality, neutrons in a reactor originate in the fuel from fission, and they must traverse the fuel into the coolant to be moderated. It's possible that a neutron could pass from one fuel rod to the adjacent fuel rod and not hit a proton, or a neutron could traverse the column of fuel and be absorbed in the fuel (and never experience the coolant). In developing neutronics or core simulation codes, we have to breakup the core into volumes of fuel, coolant and cladding (and structural materials), and simulate what happens in each volume. One will eventually learn about lattice codes which feed into core simulations codes. We developed simple methods back in the 50s and 60s because the computing power and memory were severely limited. Now, supercomputers or clusters can be used to solve massive problems in much greater detail (smaller volume elements and many more nuclides).
 
15
6
PS: I also like to know why neutron-neutron scattering is totally neglected. Is that so rare ?
You could consider it from a density point of view. Take a moderator like liquid hydrogen and there is around 10##^{28}## H atoms / m##^3##. Take then an example reactor, which might have 10##^{15}## neutrons / m##^3##. The chance of meeting another neutron is then pretty low compared to meeting another H atom.
 

rpp

66
23
You are correct in your understanding, but Chapter 6 is only discussing the slowing down region above 1 eV.
You might want to go back and read the first 2 paragraphs of the chapter.

Fission neutrons start off about 1 MeV and slow down to thermal energies, where they will become in thermal equilibrium with the moderator.
The moderator energy is almost always less than 1 eV. For example, at room temperature the most common energy is 0.0253 (given by Maxwell-Boltzmann distribution).

From 1 MeV down to 1 eV, the neutrons will have much more energy then the moderator and you can assume the moderator is at rest
and no neutron upscatter will occur.
At thermal energies, it is a different story and you have to account for the thermal motion of the moderator and upscatter will occur.
The thermal spectrum is covered in Chapter 8.
 
1,363
135
You could consider it from a density point of view. Take a moderator like liquid hydrogen and there is around 10##^{28}## H atoms / m##^3##. Take then an example reactor, which might have 10##^{15}## neutrons / m##^3##. The chance of meeting another neutron is then pretty low compared to meeting another H atom.
At low energies, Maxwell-Boltzmann distribution has significant phase space component. At high energies, the mere power of energy should become insignificant compared to its exponent.
Then, if the difference of 13 powers of 10 would mean about 30 powers of e.
If the average energy of protons is about 0,05 eV, the density of protons with energy of 1,5 eV as a matter of high energy tail of Maxwell-Boltzmann distribution should be comparable to that of neutrons.
Of course, there should be non-Maxwell distribution.
 

dRic2

Gold Member
529
93
Hi, I studied chapter 8 where lamarsh takes into account the possibility of up-scattering of neutrons because we are in thermal energy regime.

I then tried to do problem 8-1. The exercise gives you the scattering kernel for neutrons in a gas of free hydrogen and it asks to computer various things. (The scattering kernel is computer for "thermal values" of energy)

At the end I found the probability that a neutron of energy E=kT is up-scattered is about 0.57264. Do you think it is reasonable ? Maybe it should be 0.5 exactly ?
The scattering kernel is a pretty "horrible" function so I guess I could have made some mistakes along the way.

Thanks
Ric
 

Want to reply to this thread?

"Neutron moderation in a Hydrogen medium and scattering" You must log in or register to reply here.

Related Threads for: Neutron moderation in a Hydrogen medium and scattering

Replies
4
Views
5K
Replies
1
Views
3K
  • Posted
Replies
22
Views
8K
  • Posted
Replies
4
Views
1K
  • Posted
Replies
10
Views
15K
  • Posted
Replies
1
Views
2K
Replies
2
Views
5K
Replies
3
Views
3K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top