JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2023
- Medium
- Numerical
A common example of alpha decay is \({ }_{82}^{238} U \longrightarrow{ }_{80}^{234} Th +{ }_2 He ^4+ Q\). ( Given : \({ }_{92}^{238} U =238.05060 u ,\) \({ }_{90}^{234} Th =234.04360 u ,\) \({ }_2^4 He =4.00260 u \text {, and }\) \(1 u =931.5 \frac{ MeV }{ c ^2} )\). The energy released \((Q)\) during the alpha decay of \({ }_{92}^{238} U\) is \(......\,MeV\)
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Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
Energy released \(=(\Delta m )_{ amu } \times 931.5\,MeV\)
\(=\left( m _{ u }- m _{ Th }- m _{ He }\right)_{ amu } \times 931.5\,MeV\)
\(=0.0044 \times 931.5\,MeV =4.0986\,MeV\)
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