Question

In a fusion reactor the uncharged neutrons produced in some of the reactions will escape the reaction chamber. their energy can be transformed to heat by a surrounding moderator. if the moderator contains ₃Li⁶ . neutron absorption by this isotope will produce an extra source of tritium. Compute the energy available from the reaction per neutron absorbed.

Ans key: 4.76 MeV

Answer 1

Given nuclear reaction

₃Li⁶ + ₀n¹ -----------> ₁H³ + ₂He⁴

Atomic mass of ₃Li⁶ = 6.015122 amu

Atomic mass of ₃n⁰ = 1.008664 amu

Atomic mass of ₁H³ = 3.016049 amu

Atomic mass of ₂He⁴ = 4.002602 amu

Mass defect = Mass of reactants - Mass of products

Mass defect = ( 6.015122 amu + 1.008664 amu )- (3.016049 amu + 4.002602 amu)

Mass defect = 7.023786 amu - 7.018651 = 5.135x 10^-3 amu

1 amu of mass defect released 931 MeV of energy, so 5.135x 10^-3 amu corresponds to

5.135 x 10^-3 amu x 931 MeV/amu = 4.78 MeV