Question
Question: If the binding energy per nucleon in <sub>3</sub>Li<sup>7</sup> and <sub>2</sub>He<sup>4</sup> nucle...
If the binding energy per nucleon in 3Li7 and 2He4 nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction :
p + 3Li7¾® 2 2He4,energy of proton must be:
A
39.2 MeV
B
28.24 MeV
C
17.28 MeV
D
1.46 MeV
Answer
17.28 MeV
Explanation
Solution
DE = BE of RHS – LHS = (2 × 4 × 7.06) – (7 × 5.6)