The binding energy per nucleon for Deuteron and Helium 1H2 and 2He4 are 1.1 MeV and 7 MeV respectively. The energy released when two Deuterons fuse to form a Helium nucleus is
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a
1.1 MeV
b
7 MeV
c
23.6 MeV
d
6 MeV
answer is C.
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Detailed Solution
H 2 1 +H 2 1 →H 4 2 For each H 2 1 →BE1=2(1.1)=2.2 MeV For He 4 2 →BE2=4(7)=28 MeV E=BE2–2BE1 = 28–2(2.2) = 23.6 MeV