JEE Mains · Physics · STD 12 - 13. Nuclei
Consider the nuclear fission \(Ne^{20} \to 2He^4 + C^{12}\) .Given that the binding energy / nucleon of \(Ne^{20}, He^4\) and \(C^{12}\) are, respectively, \(8.03\, MeV\), \(7.07\, MeV\), and \(7.86\, MeV\), identify the correct statement
- A energy of \(12.4\, MeV\) will be supplied
- B \(8.3\, MeV\) energy will be released
- C energy of \(3.6\, MeV\) will be released
- D None of these
Answer & Solution
Correct Answer
(D) None of these
Step-by-step Solution
Detailed explanation
\({\text{Q}} = {({\text{B}}.{\text{E}})_{\text{R}}} - {({\text{B}} \cdot {\text{E}} \cdot )_P}\) \( = 20 \times 8.03 - (8 \times 7.07 + 12 \times 7.86)\) \( = 160.6 - (56.56 + 94.32)\) \(\therefore {\text{Q}} = + 9.72\,{\text{meV}}\) \(9.72\,MeV\) released.
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