Chemical Kinetics and Nuclear ChemistryEasy
Question
Energy released in the nuclear fusion reaction is
1H2 + 1H3 → 2He4 + 0n1
Atomic mass of
1H2 = 2.014, 1H3 = 3.016
2He4 = 4.003, 0n1 = 1.009 amu.
1H2 + 1H3 → 2He4 + 0n1
Atomic mass of
1H2 = 2.014, 1H3 = 3.016
2He4 = 4.003, 0n1 = 1.009 amu.
Options
A.8.30eV
B.16.758MeV
C.500 J
D.4 × 106 kcal
Solution
Mass defect = (2.014 + 3.016) - (4.003 +1.009) = 0.018a.m.u.
Now 1a.m.u. ≈ 931 MeV
Energy released = 0.018 0.018 × 931 = 16.758MeV.
Now 1a.m.u. ≈ 931 MeV
Energy released = 0.018 0.018 × 931 = 16.758MeV.
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