posted on Feb, 24 2018 @ 03:08 PM
a reply to: BASSPLYR
Some interesting points raised here.
Shielding wise it can easily take 16 inches or so of lead to shield gamma, but that is to to completely shield it. It doesn’t have to be completely
shielded. Our cyclotron vault for example is 6 foot thick concrete and nothing gets out not even thermal neutrons. We use 70 mm lead for the hot cells
and that takes the dose levels down to pretty much background. But in our case we’re only talking about 511keV gamma from 18F decay - shielding
varies with energy.
If you take into account distance from source as well as shielding you may find you need a lot less than 16 inches of shielding to make the dose rate
to crew negligible.
Lead is the most common shielding material, tungsten is also common and approx 25% thinner. Uranium is thinner still approx 40 % thinner than lead.
Still very heavy though. Graphene is showing some interesting properties wrt shielding so that may be a route.
Positron shielding shouldn’t be too much of a worry as they tend to annihilate.
There is definitely interest in positrons for structural analysis
WRT the ramjet idea, not sure how many positrons you’d need to generate to physically heat something.
If you take F18 generated from a cyclotron our average yield would be 600 GBq. If one bequerel is 1 disintegration per second we’re kicking out 600
Billion 511 keV bursts per second and 1.2 trillion positrons and it’s not even warm (and shielded by 70 mm lead)
Given the mass of a position is approx 1/1800th of an anti-proton it would seem the latter would be more efficient.
But anything that’s going to generate the required amounts of positrons etc is not going to be light. But what do I know I’m just a chemist