. In the fusion reaction
1H2 + 1H2→ 2He3 + 0n1 The masses of deutrons, helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1 kg of deuterium undergoes complete fusion. Find the amount of total energy release, 1 amu = 931 MeV/C2
A. 6 × 1013 J
B. 5.6 × 1013 J
C. 9 × 1013 J
D. 0.9 × 1013 J
Best Answer
Ans: C
Sol :1H2 + 1H2→ 2He3 + 0n1
mass defect
Δm = 2 × 2.015 – 1 × 3.017 – 1 × 1.009
Δm = 4.030 – 4.026
Δm = 0.004 amu.
E = Δmc2 = 0.004 × 931 = 3.724 MeV
number of atoms in 1 kg deuterium
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