Make whatever reasonable assumptions you like (detector close to the wall) and see what comes up.
Sure. I’m happy to work like that. Throw the numbers in, then see what needs modifying…
Treating the numbers I’ve seen as if they went into a regular fusor, I would say;
1000A means that a total of [1000A/e]*250E-6s = 1.5E18 charged particles have been ‘mobilised’ during the pulse. Now, as both acceleration time and velocity are SQRT(m) dependent, so only one in 60 of those will be deuterons flying around, the rest will be electrons. (That is, at any one time, a given space with one deuteron passing through it will also have one electron in it, but the electron clears that space in 1/60 of the time of the deuteron, so another electron comes to replace it. Total transiting electrons = 60 for each transiting deuteron.)
So we have 2.5E16 deuterons to ‘play’ with.
At @10keV, they’ll be going at 1Mm/s. Cross-section for charge exchange, say = 1E-20m^2 (see Massey, “Atomic Collisions”, if you want anything more accurate.) Take the background pressure as 1E17/m^3. So the rate of charge exchange and fast neutral production will be 1E6[m/s] * 1E-20[m^2] * 1E17[/m^3] = 1000/s per particle.
We have 2.5E16 deuterons, so the fast neutral production rate will be 2.5E19/s.
There will be a chance of re-ionisation for the fast neutral as it makes its way to the wall. The cross-section is about the same as for the charge exchange, at these ~10s of keV energies, but it’s only a chance while it’s on its way – a distance a deuteron covers in a us. So the re-ionisation rate would be a probability of 1000/s * 1E-6 = 0.001. A tenth of one percent re-ionise, which I will therefore ignore.
Take the chamber wall surface area as being 25m^2 (?), so that’d be an irradiation rate of 1E14 fast neutrals/cm^2/s, or 2.5E10/cm^2 for the pulse.
Let’s take the fusion cross-section between a fast neutral and an embedded deuteron to be 1millibarn. The probability of one
fast deuteron hitting a single
embedded deuterium molecule (2x nucleii) in a cm^2 is therefore 0.001E-24cm^2/1cm^2 = 2 chances in 1E-27. We have 2.5E10 hitting that area in one pulse, so the chance of a single fusion with that lone embedded deuterium molecule is 5E-17.
Now let’s consider the neutron rate seen. I’m going to make this assumption – it was bolted on to the chamber wall, so it will receive a half of all the neutrons from that wall areas, which is probably around 100cm^2. So the neutron detector will see 2.5E-15 neutrons per pulse, per embedded deuterium/cm^2.
So I do a quick lookup for interstitial hydrogen in steels, and I’ve picked a figure out of those texts of 1ppm. Let’s say the fast deuteron can reach down into the steel by 500 microns, so the total amount of deuterium per cc (per 500 micron depth) is 1E-6 *[8g/cm^3/20]/3.2E-27kg = 1.25E17.
So the neutron rate from fast-neutrals fusing with interstitial deuterium seen by a neutron detector bolted to the wall of one of these devices is, by this estimate, 2.5E-15neutrons/pulse/embedded-deuterium * 1.25E17embedded-deuterium = 312 per pulse.
I do not know if the neutron detectors were sensitive enough to register 3 clicks in response to 300 neutrons passing through them, but there are various points where my own estimates may be up or down, e.g. the background density, and all is proportionate in this calculation.
Whatever the actual variations of assumption that affect the input numbers I’ve used, I’m satisfied I’m looking at the basic mechanism. Therefore, I see no reason to think the idea of fast neutrals being the source of the detected fusion is inconsistent with the neutron count results. In no way does that say this is
what happens, but only that there isn’t an immediate reason to think it is an unjustified possibility that, therefore, needs to be otherwise controlled for in the WB experiments.