Jika massa 2 1 H = 2,009 sma, 3 1 H = 3,016 sma, 4 2 He =

Jika massa 2 1 H = 2,009 sma, 3 1 H = 3,016 sma, 4 2 He = 4,003 sma, dan 1 0 n = 1,009 sma. Jika 1 sma = 931 MeV, hitung energi yang dibebaskan pada reaksi 2 1 He + 3 1 H -> 4 2 He + 1 0 n .