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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2022
  • Medium
  • Numerical

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: \({ }_{1}^{2} X +{ }_{1}^{2} X ={ }_{2}^{4} Y\) The binding energies per nucleon \({ }_{1}^{2} X\) and \({ }_{2}^{4} Y\) are \(1.1\,MeV\) and \(7.6\,MeV\) respectively. The energy released in this process is \(MeV\).

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Answer & Solution

Correct Answer

(B) \(26\)

Step-by-step Solution

Detailed explanation

Energy released in the given process \(=\) Binding energy of product \(-\) Binding energy of reactants

\(=7.6 \times 4-(1.1 \times 2) \times 2\)

\(=30.4-4.4\)

\(=26\,MeV\)

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