Difference in thermal n cross section for Be-9 and B-10

[Salman2]

I have a question.

From this government reference:

http://www.ncnr.nist.gov/resources/n-lengths/

the thermal neutron cross section for stable isotope Be-9 = 0.0076 barns. This means Be-9 is not expected to absorb a thermal neutron, the probability of this is very, very low. The Be-9 isotope has 5 neutrons and 4 protons.

What I cannot understand is why the next possible stable isotope, B-10, which also has 5 neutrons, has an extremely high neutron cross section = 3,835 barns, compared to Be-9 ? Of course B-10 has 5 neutrons and 5 protons.

Both isotopes would have an unpaired n in the last n energy shell...correct ? The n packing situation for both Be-9 and stable B-10 would be, starting at ground state:

n
nn
nn

So, what aspect of the nuclear shell model explains such a drastic difference in neutron cross section for these two stable light isotopes that both have 5 neutrons ?

Why does B-10 accept the thermal neutron to form completed neutron shells:

nn
nn
nn

but, Be-9 isotope does not accept a neutron ?

I know the shell model is incomplete for heavy isotopes, but it appears that predictive problems begin at Be to B, which are not very heavy.

Thanks for any help with this question.

 

[Salman2]

OK, no comments. Let me offer a shell model hypothesis to try to explain the experiment facts discussed above:

==

Suppose both stable isotopes, Be-9 and B-10, have a stable core with pppp +nnnn.

Be-9 would have one free n in the 'neutron' outer energy shell that would be predicted to interact with an incoming thermal n.

However, B-10 isotope would have both a free p in the 'proton' outer energy shell and a free n in 'neutron' outer energy shell.

So, my hypothesis is that stable B-10 isotope has a higher absorption cross section for an incoming thermal n because there is certain probability that the free p in the outer shell could act as an energy moderator, thus slow the incoming thermal n, and therefore increase the statistical probability the thermal n would enter the non-closed n shell and close it as:

nn
nn
nn

On the other hand, because Be-9 does NOT have a free p available to moderate the incoming thermal n, the probability that the thermal n coming in contact with the free n and thus added to the non-closed n shell is much lower, in fact, at the measured cross section value of 0.0076 barns.

Of course, if this hypothesis holds, it should predict similar such situations in other isotope pairs where they have the same number of neutrons, but differ in number of p by one.

Any comments on this hypothesis I offer ?

==

Ps/ Another hypothesis is that we need to use a type of 'collective model' explanation, in place of the shell model, (as is done for more heavy isotopes where the shell model fails). I have a version of a collective model hypothesis in mind, if it does not violate forum rules to present it.

 

[NLB]

The shell model is not very successful at explaining thermal n cross sections, as you have observed. No one is disagreeing with you about that.

Let me ask you this about your hypothesis: Does your hypothesis make sense for Be7, which has a thermal n cross section of 39,000 barns? Or how about H3 with <10^-6 barns, or He3 with 5330 barns for σp? Or He4 with absolutely zero (well less than H3 anyway) thermal n cross section. Or Na22 with σp at 28,300 barns. (By "σp", I mean it absorbs the thermal n and spits out a p.) Or Cd-113 with σ=20,640 barns. Li6 with σalpha =941 barns. (By "σalpha", I mean it absorbs the thermal n and spits out an alpha.)

If you can make sense of that, with your hypothesis, then it would be worth publishing.

 

[Salman2]

Let me ask you this about your hypothesis: Does your hypothesis make sense for Be7, which has a thermal n cross section of 39,000 barns?

Thanks for reply. I'm looking at the isotopes you suggest, and I have a question about Be-7. This paper puts the n cross section for the Be-7 (n,p)Li-7 reaction as being 18 (+-4) barns. ?

http://link.springer.com/article/10.1007/BF01284487

Do you have another reaction in mind that gives 39,000 ?

 

[NLB]

The reference in which I got the 39,000 number was "Neutron Scattering and Absorption Properties" by Norman Holden, dated 2003.

There is another one, on-line, at:
Here's my reference:

http://ie.lbl.gov/ngdata/sig.htm

This one looks like 1981. It shows 48,000. Either way, it is huge. Yes, it is a sigma-p.

Your other link quoted is at a much higher energy (24.5 keV) than a thermal neutron (25 meV or so). There is a large difference in the absorption of neutrons when they are speeding by, versus when they are drifting by.

 

[Salman2]

The reference in which I got the 39,000 number was "Neutron Scattering and Absorption Properties" by Norman Holden, dated 2003.There is another one, on-line, atere's my reference:http://ie.lbl.gov/ngdata/sig.htmThis one looks like 1981. It shows 48,000. Either way, it is huge. Yes, it is a sigma-p.Your other link quoted is at a much higher energy (24.5 keV) than a thermal neutron (25 meV or so). There is a large difference in the absorption of neutrons when they are speeding by, versus when they are drifting by.

Thanks for link and comments. You are of course correct, my link is not valid for thermal n cross section reaction, I did not pick up on the energy of the incoming n.

The cross section for Be-7 is very high for a light isotope. I don't know how the shell model explains it, but let me give a try with a collective type model, and also discuss some of your other examples. I normally work with stable isotopes, but the collective model I study also is predictive for unstable isotopes such as Be-7, which does stay around for ~53 days 1/2 life, so lots of time to capture a thermal n if present nearby.

Because the model I study is collective, it requires that n and p have ability to form nucleon clusters within isotopes, such as alpha cluster {ppnn}... which is well recognized for many isotopes (i.e., C-12, O-16). Energy shells surely exist in this collective model, but the energy equations must consider cluster configuration possibilities, not only free n and p within shells.

==

Consider that the He-3 cluster {ppn} has a very strong thermal n cross section, at ~5,333 b. The reaction is to form a very stable alpha, thus n + {ppn} -> {ppnn} = He-4. But why ? {ppn} is already stable with binding energy =~7.72 MeV, but, {ppnn} is more stable at ~28.3 MeV.

Next, consider {nnp}=H-3. This has a very low cross section for a thermal n because there are no collective cluster structures that can be formed that are more stable, at best we get {nnp} + n -> {np}+{nn}. So, the collective model explains why {ppn} has much higher thermal n cross section than {nnp}.

You ask about stable Li-6 isotope. One possible collective cluster configuration is Li-6 = {ppn}+{pnn}. Now, {pnn} = H-3 has a very low cross section, so any thermal n in contact with Li-6 must react with the {ppn} cluster to form a {ppnn}, resulting in a transformation of stable Li-6 to stable Li-7. The collective cluster structure of stable Li-7 after the reaction would be {ppnn} + {pnn}. Note that the {pnn} in Li-7 was already present in Li-6, the only nuclear transformation was from {ppn} in Li-6 to {ppnn} in Li-7. Other thermal n reactions possible because Li-6 can have multiple collective cluster configurations. This is the reason the thermal n cross section of Li-6 is only ~940 b, reduced from its 5,333 b potential if only a {pnp} + n reaction was involved...the final cross section becomes a statistical average of all possible collective cluster thermal n reactions, a type of Feynman path integral approach.

You ask about He-4 {ppnn}, why it has near zero thermal n cross section ? It is already a double magic isotope with extreme stability. Using the collective model approach, the following reactions would be possible, {ppnn} + n -> {nn} {ppn} or -> {pnn} {np}, but note that none of these collective configurations are physically possible for the 1s nuclear shell, which only allows four mass units (2 p and 2 p). None of the possible cluster by products are more stable than {ppnn}, which has a binding energy = ~28.3 MeV. Thus there is no allowed energy path available for He-4 as a collective cluster {ppnn} to capture a thermal n and become more stable by fission into any two clusters of 2 or 3 mass units, even if they move into 1p energy shell.

--

OK, now Be-7. This is very interesting. It has a very, very high thermal n cross section, much enhanced than the Li-6 example above, so something other than simple n capture by a {ppn} must be going on. Here is my hypothesis.

Consider Be-7 to be a three group collective cluster configuration:

[{pnp} core + {nn} halo and {pp} halo] = Be-7

The two halo resonances rotating far from the {ppn} core. Now, {nn} and {pp} halos are very well known and documented experimentally to be present in many different isotopes. Note that a {pp} and {nn} halo both rotating far from core have identical mass to the alpha {ppnn}, they just are not bonded as a single cluster within Be-7, although I think it possible the two halos {nn} {pp} could be measured as if they were a single {ppnn} cluster, but such would be incorrect nucleon configuration, if we want to explain very high 39,000 -48,000 thermal neutron cross section.

Ok, my hypothesis is that the thermal n reaction with Be-7 would be in two steps, using this predicted collective structure:

[{pnp} core + {nn} halo and {pp} halo] = Be-7

step 1. Because the {pp} halo is far from {ppn} core it would act as a valance entity for a thermal n, the reaction would be: {pp} + n -> {np} + p by-product. So, in step #1 we see the p released as by-product from the thermal n reaction AND a new collective cluster formed within Be-7, {np}. The cross section for this reaction must be very high, and from experimental data, it is predicted by the collective model to be in the ~33,667 to 42,667 b range based on your two reference links.

step 2. Because a {ppn} cluster is present within the ground state 1s shell of Be-7, there is a empty energy hole for another n to fill the 1s. We predict a {nn} halo is present in the higher 1p energy shell, thus, after the reaction in step 1, one of the n from the 1p shell is predicted to move to the lower 1s shell to fill and complete the ground energy level. This is predicted because the Be-7 core shell is not completely filled, having a {ppn} core. The above energy transition of one n from 1p shell to 1s allows for the {ppn} already present in the core to capture this n and form a stable {ppnn}. The cross section for this reaction {ppn} + n is well known to be 5,333 b.

In summary, the sum of the reaction cross section from step 1 of ~42,667 b and the reaction cross section from step 2 of 5,333 b yields the experimental thermal n cross section for Be-7 of 48,000 b (or 33,667 + 5,333 = 39,000 b). These are the thermal n cross section given in your two links.

===

I would greatly appreciate your review and comments on the above hypothesis. I would predict that similar halo type reactions would explain the very high enhanced cross sections you mention for Na-22 and Cd-113, but I will wait until I get feedback before I tackle these.

---

ps/ I do see one way to falsify my hypothesis for Be-7. A study showing that the thermal n cross section for a {pp} halo resonance, found in Be isotopes, cannot be ~42,667 b, or whatever number needed to add to 5,333 b to yield the experimental Be-7 cross section recorded.

==

EDIT COMMENT: More support for the Ho I present above.

It is known that Be-7 is unstable and has a decay mode of 100% via 'electron capture' (EC), to yield stable Li-7 isotope + neutrino$\upsilon$. Symbolically, this EC mode would be p + e- (captured) -> n + $\upsilon$.

Note in the Ho above I suggest that thermal n capture by a (pp) halo is a similar process, however, the reaction in the outer halo shell is {pp} + thermal n -> (np) + p by-product. But, the end result is exactly the same between EC and thermal n capture, e.g., formation of a stable Li-7 isotope with a very stable alpha collective cluster core {ppnn} in ground state 1s shell.

 

[NLB]

I think you will find that the alpha particle hypothesis is much more likely to product the results you want.

 

[NLB]

May I get in touch with you? If I leave me email address here for you, would you please email me? Would that be OK?

 

[Salman2]

May I get in touch with you? If I leave me email address here for you, would you please email me? Would that be OK?

Yes, I have sent you an off forum message with email contact information. I realize any further discussion most likely would violate forum rules, so it may be better I discuss my thoughts with anyone that has an interest in the topic via private email.

Perhaps further discussion on this thread should end ? I'll let forum Admin. make that call, if yes, please close the thread.

 

[Salman2]

I think you will find that the alpha particle hypothesis is much more likely to product the results you want.

But, this requires a {ppnn} alpha cluster in ground state 1s shell, with pp/n_ (_=unfilled energy hole) in 2p. So, using a collective model approach, the 2p shell takes the form of cluster {ppn}, and this as He-3 has a cross section of 5,333 b...and this is not close to the experimental range 39,000 to 48,000. So, the alpha particle hypothesis cannot be correct, there must be some significant enhancement in cross section the alpha particle hypothesis cannot account for. The two step explanation above using {pp} halo as the initial target for the thermal n can explain this enhancement, e.g., there is a great affinity of the thermal n to interact with an unstable {pp} halo, causing fusion of the n with one p to form {np}, then the second p ejected as a by-product, resulting in stable Li-7. Why is this not a valid hypothesis to consider as a quantum possibility for unstable Be-7 to explain how it can absorb a thermal n and transform into stable Li-7 with release of an energetic p by-product ?

So my question...how does the shell model of M. Mayer which requires that p and n are independent, not forming collective states, explain the high thermal n cross section of Be-7 (in range of 39,000 to 48,000 barns) ?