Sunday, January 27, 2019
Working out ''classified'' information.
Depleted-boron the new occupationally-safe to machine, neutron reflector, replacing the legacy Beryllium, of old? That is as you can read below, is a ''classified'' material. Boron-11 in the reliable replacement warhead.
You heard it hear first and as an aside, what's the law on working out ''classified'' info anyhow?
With a mean free path of about 4.4 cm. Being ''edible'' and boron finding uses in ''prosthetics''. It seems to fit the criteria.
My friend Evan is after working out for the other contender. Titanium, With - ''The average distance a neutron...will travel in titanium-48 before interacting with a nucleus is 5.74cm [with that specific interaction likely a reflection, though captures do happen at a fair rate]...this is only using cross section data and mass data for Ti-48, which only accounts for 73% of natural Ti.''
Natural boron, similarly contains two prominent isotopes. One of which is so good of a neutron absorber to render natural boron as absolutely useless as a neutron reflector. You probably know that the isotope responsible in natural boron, is frequently enriched for nuclear reactors that need boron as control rods, boric acid in PWRs and that type of thing.
When this is done, depleted boron is produced as a byproduct. Finding a well known use in space micro-chip applications and that's it...or is it?
-Cursory examination as a neutron reflector -
Boron density doesn't vary all that much at the pressures under implosion.
So with a density of 2.34 g/cm cubed, as the worst case scenario.
11 nucleons per atom, and as 1.67 e-24 is the approx mass of a single nucleon, this gives the mass of full boron nucleus as 1.8 e-23 grams.
The number[N] of boron nuclei per cubic cm is therefore. 2.34 g cc/ 1.8 e-23 => 1.27 e-23 nuclei per cubic cm.
From the chart plotted in the below reference, the cross section/impact area, in 'barns' at an incoming neutron energy of 2 MeV [the avg energy from fission] is approximately 1.8.barns.
One barn = 1 e-24 cm squared. Therefore 1.8 times that is 1.8 e -24.
Mean free path, 1/N x a = [12.7 e-22] x [1.8 e -24] = 0.229
1/0.229 = 4.36 cm
A winner? Very possibly. You can 'eat boron' and ''be healthy'',which is one of the criteria laid out in the disclosure on what the ''classified'' material is.
So, that's my hunch.
Also, titanium would not be entirely ionized at low nuclear yield and thus if used as a reflector, would greatly reduce the emitted light from the fission primary. Much needed light to radiatively implode any potential secondary.
It's actually interestingly, one of the reasons why they switched from using Baratol for slow explosive lenses to the similar slow-explosive Boracitol. With the latter used in all the early thermonuclear designs. Ivy Mike, Castle Bravo etc.
Target, B-11, Reaction n,el
Posted by T.G at 10:13 AM