Nuclear Physics and RadioactivityHard
Question
Per nucleon energy of 3Li7 and 2He4 nucleus is 5.60 meV and 7.06 MeV, than in
3Li7 + 1P1 → 22He4 energy release is : -
3Li7 + 1P1 → 22He4 energy release is : -
Options
A.29.6 MeV
B.2.4 MeV
C.8.4 MeV
D.17.3 MeV
Solution
Q = - 7(5.6) + [4(7.06)]2 ⇒ Q = 17.3 MeV
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