Alpha emission energies from p + 11B

[chrismb]

I have recently been picked up my MSimon for suggesting there is a ~5.7MeV alpha that comes out of p+11B.

viewtopic.php?t=965 page 4

and he suggested I go look at the literature.

Well, I had previously done that, but not someone's interpretation of data, but the actual atomic masses.

I note that other posts also seem to rely on the Bussard presentation which suggests alphas at 3.76MeV and 2x 2.46MeV, such as;

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viewtopic.php?t=629

viewtopic.php?t=792


This data does not appear to be correct.

According to the wiki page on the isotopes of berillium [they reference Isotope masses from Ame2003 Atomic Mass Evaluation by G. Audi, A.H. Wapstra, C. Thibault, J. Blachot and O. Bersillon in Nuclear Physics A729 (2003).] the following masses are relevant to these numbers:

8Be 8.00530510(4)

11B 11.0093054(4)

4He 4.00260325415(6)

1H 1.00782503207(10)

So if we take the mass loss of the first reaction:

[p+11B] - [4He+8Be] = 9.22207792E-3 u

Taking the value of 1 u = 931.494027(23) MeV/c2, so this first reaction has an energy of 8.590310499MeV.

Newtonian kinetics says (as these are non-relativistic) that the 4He shoots off with 5.72687366 MeV, and the 8Be has 2.863436833 MeV.

Then, the next reaction of [8Be] - [2 x 4He] = 9.85917E-5 u.

So the 4He's run off with a combined 91837.57966eV.

Sharing out the 8Be's 2.86MeV's worth of energy and summing with this new energy, we get 3 alphas of energy;

1 x 5.72687366 MeV

and

2 x 1.477637206 MeV

I believe this calculation is correct, please let me know if otherwise. It does hinge critically on this mass of 8Be so I am, actually, relying on that bit of key data. It is possible that this is inacurate (after all, it does only last 6.7E-17 seconds!)

(As I say, I relied on the base data of atomic masses rather than someone else's 2nd hand information). I trust that provides a suitable bit of archival information.

best regards,

Chris MB.

[Art Carlson]

I did this calculation in another thread. I'm not sure, but I think the Be8 comes off in an excited state, which will affect the numbers. In addition, you are not accounting for the fact that the Be8 is moving and can decay in various directions relative to this motion. This results in a continuous spectrum.

[chrismb]

Thanks for that. If it is observed measured data, then it's measured data, and that's all I am interested in - as a 'fact'!

I would [did] stop to wonder how they could take such an accurate measurement of 8Be in such a short time-span. If it carries off in an excited state in which it remains until fission I am, though, still curious why it doesn't appear more massive.

[I recognise the 8Be will have a different inertial frame as a fusion output product, so would be interested to see the actual emission spectrum. Does it shape up to fit an isotropic fission of the 8Be?]

[chrismb]

Just found this online, from "Observation of neutronless fusion reactions in picosecond laser plasmas V. S. Belyaev, A. P. Matafonov, and V. I. Vinogradov":

The energy distribution of α particles is characterized by
two energy ranges: 3–4 and 6–10 MeV. The first range corresponds
to the production of α12 particles in the reactions

11B + p→α0 + 8Be + 8.586 MeV,
11B + p→α1 + 8Be* + 5.65 MeV,
8Be*→α12 + α12 + 3.028 MeV.

The secondary α12 particles are produced at the decay of
the excited nucleus 8Be*. This main channel has a sharp peak
at the proton energy of 163 keV and a broad maximum at the
proton energy of 660 keV. Kinematics of this reaction results
in generation of α12 particles with the energies of 2–4 MeV.
These α12 particles acquire the additional kinetic energy
from the kinetic energies of protons and boron ions. The
probability for the production of α0 particles and of the
ground-state 8Be is very small α4α. A small number of the
observed 10 MeV α particles can be attributed to this channel.
The increase of α-particle energy also takes place from a
small number of high-energy protons and boron ions in the
tail of their energy spectra.

The second range corresponds, in our opinion, to the α1
particles in the second channel. These α1 particles also acquire
the additional kinetic energy from the kinetic energies
of colliding protons and boron ions.

[MSimon]

Thanks!

The thing to remember about all this is that although the spectrum is continuous what matters from an engineering stand point is the populations in each dE.

Re: ash accumulation. To maintain a given pressure at a given pumping rate there has to be a continuous flow of reactants - most of which don't get reacted. So ash build up as in all such flow situations will reach (asymptotically) some steady state value (reaction rate constant - flow constant - pumping rate constant). The question then is: what kind of flow rate do you need to keep the ash at tolerable levels. If you want continuous operation.

For most turbo-pumps the pumping efficiency is greater as the atomic mass of the "gas" pumped goes up so that will help.

But you also have the problem of B11 "gas" condensing on the blades of the turbo pump. Lots of problems. Interesting problems. I'm hoping problems with solutions. We shall see.

I'm almost certain I can make D-T work (with tolerable superconductor lifetimes). D-D should not be too much harder (and would be a preferred fuel since breeding T would be avoided). pB11 is going to be "interesting" to say the least.

Now what do you do with the pumped exhaust in a D-D reactor? Burn it. Collect the D2O for electrolyzation/recycling and compress the He for sale or use in magnet cooling.

[chrismb]

I still don't see what the mechanism is to get rid of the 4He that gets re-ionised and re-enters the reciprocation-about-the-centre process. A population will build up which will never be cleared from the reaction regions, if this process is efficient and holds in ions as it is billed to do.

[alexjrgreen]

I still don't see what the mechanism is to get rid of the 4He that gets re-ionised and re-enters the reciprocation-about-the-centre process. A population will build up which will never be cleared from the reaction regions, if this process is efficient and holds in ions as it is billed to do.

Rick explained how to do this:
viewtopic.php?p=10506&highlight=#10506

[chrismb]

OK. yes, I can see that, a resonant 'pull-out' potential on top of the DC field.

Just one small problem compounding that - went the 4He gets pulled out of the reaction volume, won't it immediately head back down the field into the reaction volume? The magrids do not just generate a field that heads inwards, the fields come 'swooping out' of the back faces then back into the apertures of the magrid. i.e. the 4He will still remain locked into the field between the anode and and negative electron regions, whether or not it gains energy. If it gains energy it will eventually collide with the magrid again, neutralise, ionise, repeat...

So if you pull it out of the reaction volume with this method, fine, but it is still an ion and needs to neutralise to be evacuated out. But once a neutral, it can drift into the reaction volume again.

[93143]

The magnetic fields don't affect the ions much. They aren't strong enough. If an ion gets outside the magrid, it will be repelled by the magrid's positive charge and leave.

[TallDave]

/agree with 93143. The alphas are so massive and energetic they'll barely notice the magnetic field on their way by. In fact, the problem (maybe even a bigger problem than managing brem) is going to be alpha sputtering of the Magrid itself. I think Art did a calculation on that once and it wasn't pretty.

So 93143, with your statement above, I'm wondering if you have an opinion on the allegedly recirculating/oscillating mass of electrons outside the magrid, as to whether it's pretty much all electrons or a quasi-neutral plasma. I'm curious because you seem to have a better grasp of electrostatic principles than most of us here.

[hanelyp]

Any positive ions between the magrid and the negative grid around it won't be sticking around long. They'll be attracted to the negative grid and, probably after sweeping through a few times, hit the grid to be neutralized. (Assuming a direct conversion config.) In a thermal reactor config positive ions would be swept direct to the outer wall. In either case I don't see a neutral plasma being maintained outside the magrid. Maybe a mix of free electrons and diffuse neutral atoms.

[93143]

So 93143, with your statement above, I'm wondering if you have an opinion on the allegedly recirculating/oscillating mass of electrons outside the magrid, as to whether it's pretty much all electrons or a quasi-neutral plasma. I'm curious because you seem to have a better grasp of electrostatic principles than most of us here.

Well, ahm, that's complicated. I assume you mean the cusp flows...

I'm not sure my electrostatics expertise warrants your confidence, but I'll take a crack at it:

The scenario of all-electrons derived from superimposing individual loss paths is not plausible, as Art demonstrated. You don't spontaneously get a cloud of electrons forming at a half gigavolt potential unless there's some applied potential difference at least that large in the system.

So what happens? Well, there are two primary effects. First, the buildup of electrons will repel electrons (cusp plugging). Second, it will attract ions (perforating the potential well). In a LTE system, this would most likely result in what Art describes; a quasi-neutral ion loss path, since the presence of ions in the cusp due to the second effect mostly cancels out the first effect.

In a Polywell, the ions are supposed to be very low-energy at the top of the well, and unless the electron buildup manages to completely level the potential well in the cusp area, the ions are going to have more trouble getting up there than they would in a normal plasma. This should cause a shift away from 'perforated potential well' and towards 'plugged cusps'.

There may or may not be a way to easily calculate how large this effect is. It might not make any practical difference... I have to think about it more. I'd try a back-of-the-envelope simulation, but I'm on Christmas holidays and can't access my main computer...

[TallDave]

Thank you, that illuminates things a bit for me anyway.

[93143]

There's also the fact that the cusps are arranged in a spherically symmetric pattern, which means that the radial effect of a nearby cusp on particles at the top of the wiffleball tends to get cancelled by all the other cusps. So the short-range lateral effect becomes more important. I really need to do some math, but I'm too lazy right now...

Also, if you're talking about the low-density bulk plasma between the magrid and the wall, rather than the cusp flow, I wasn't talking about that and I'd have to think about it more. I mention this because I just noticed the discussion in that other thread...