Question

The fusion of deuterium (one proton + one neutron) with tritium (one proton + two neutrons) to form a Helium nucleus (two protons + two neutrons) and one spare neutron releases about 17 MeV of energy. ( 1 eV = 1.6 x 10-19 Joules). Given that the Sun produces about 380 x 1026 Watts of luminous energy, what is the total mass of deuterium and tritium that must be fusing each second to provide this energy output? If we could master the engineering to harness this sort of fusion reaction, how many grams of material would we need to fuse each second to provide power for the citizens of California at 2kW per person (39 million people), assuming 50% efficiency

Answer 1