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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2019
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  • MCQ

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

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Correct Answer

(D) None of these

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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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