The binding energy per nucleon of deuteron (21H) and helium nucleus (42He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
A
13.9 MeV
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B
26.9 MeV
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C
23.6 MeV
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D
19.2 MeV
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Solution
The correct option is C 23.6 MeV 1H2+1H2→2He4+Q
Total binding energy of helium nucleus = 4×7=28MeV
Total binding energy of each deutron = 2×1.1=2.2MeV
Hence energy released = 28−2×2.2=23.6MeV