Question

Calculate the energy released/absorbed (in MeV) in the nuclear reaction:

\[\ce{_1^1H + _1^3H -> _1^2H + _1^2H}\]

Given: \[\ce{m(_1^1H)}\] = 1.007825 μ

\[\ce{m(_1^2H)}\] = 2.014102 μ

\[\ce{m(_1^3H)}\] = 3.016049 μ

Answer 1

\[\ce{_1^1H + _1^3H -> _1^2H + _1^2H}\]

Mass lost × c² = Energy released

M_LHS = 10.07825 + 3.016049

= 4.023874 μ

M_RHS = 2 × 2.014102

= 4.028204 μ

 M_RHS > M_LHS → Energy absorbed

ΔM = 0.00433 μ

∴ E = ΔMc²

= (0.00433 μ) × (931.5 MeV/c²)

= 4.00545 MeV