JEE Mains · Physics · STD 12 - 13. Nuclei
Nucleus A is having mass number \(220\) and its binding energy per nucleon is \(5.6 \,MeV\). It splits in two fragments '\(B\)' and '\(C\)' of mass numbers \(105\) and \(115\) The binding energy of nucleons in '\(B\)' and '\(C\) ' is \(6.4 \,MeV\) per nucleon. The energy \(Q\) released per fission will be............\(MeV\)
- A \(0.8\)
- B \(275\)
- C \(220\)
- D \(176\)
Answer & Solution
Correct Answer
(D) \(176\)
Step-by-step Solution
Detailed explanation
\(Q=(B \cdot E)_{p}-(B . E)_{R}\) \(=(105+115)(6.4)-(220)(5.6)\) \(=176 \,MeV\)
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