JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2023
- Medium
- Numerical
Nucleus a having \(Z=17\) and equal number of protons and neutrons has \(1.2\,MeV\) binding energy per nucleon.Another nucleus \(B\) of \(Z=12\) has total \(26\) nucleons and \(1.8\,MeV\) binding energy per nucleons.The difference of binding energy of \(B\) and \(A\) will be \(...........MeV\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
For \(A\) mass number \(=34\)
Total binding energy \(=1.2 \times 34=40.8\,MeV\)
For \(B\) mass number \(=26\)
total binding energy \(=1.8 \times 26\,MeV\)
\(=46.8\,MeV\)
Difference of \(BE =6\,MeV\)
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