> you would need 100% efficiency to be able to breed your own tritium
How so? As I understand it, the breeding reaction is exothermic (Li-6 + n -> He-4 + T) so all you need are neutrons, for which (as someone pointed out to me two years ago[1]) neutron multipliers would be used.
How so? As I understand it, the breeding reaction is exothermic (Li-6 + n -> He-4 + T) so all you need are neutrons, for which (as someone pointed out to me two years ago[1]) neutron multipliers would be used.
[1] https://news.ycombinator.com/item?id=32224650