Photoelectric EffectHard
Question
The binding energy per nucleon of deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to from a helium nucleus the energy released in the fusion is
Options
A.23.6 MeV
B.2.2 MeV
C.28.0 MeV
D.30.2 MeV
Solution
1H2 + 1H2 → 2He4 + ᐃE
The binding energy per nucleon of a deuteron = 1.1 MeV
∴ Total binding energy 2 × 1.1 = 2.2 MeV
The binding energy per nucleon of a helium nuclei = 7 MeV
∴ Total binding energy 4 × 7 = 28 MeV
Hence, energy released
ᐃE = (28 - 2 × 2.2) = 23.6 MeV.
The binding energy per nucleon of a deuteron = 1.1 MeV
∴ Total binding energy 2 × 1.1 = 2.2 MeV
The binding energy per nucleon of a helium nuclei = 7 MeV
∴ Total binding energy 4 × 7 = 28 MeV
Hence, energy released
ᐃE = (28 - 2 × 2.2) = 23.6 MeV.
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