JEE Mains · Physics · STD 12 - 13. Nuclei
A nucleus with mass number \(242\) and binding energy per nucleon as \(7.6\,MeV\) breaks into fragment each with mass number \(121\) . If each fragment nucleus has binding energy per nucleon as \(8.1\,MeV\), the total gain in binding energy is \(........MeV\)
- A \(120\)
- B \(121\)
- C \(122\)
- D \(159\)
Answer & Solution
Correct Answer
(B) \(121\)
Step-by-step Solution
Detailed explanation
Initial binding energy \(=242 \times 7.6\,MeV\) Final binding energy \(=121 \times 8.1\,MeV +121 \times 8.1\,MeV\) \(=242 \times 8.1\,MeV\) Total gain in binding energy \(=242(8.1-7.6)=121\,MeV\)
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