JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2023
- Easy
- MCQ
\({ }_{92}^{238} A \rightarrow{ }_{90}^{234} B +{ }_2^4 D + Q\) In the given nuclear reaction, the approximate amount of energy released will be \(.....\,MeV\) \(\text { [Given, mass of }{ }_{92}^{238} A =238.05079 \times 931.5\,MeV / c ^2\) \(\text { mass of }{ }_{90}^{234} B =234.04363 \times 931.5\,MeV / c ^2\) \(\text { mass of } \left.{ }_2^4 D =4.00260 \times 931.5\,MeV / c ^2\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(4.25\)
Step-by-step Solution
Detailed explanation
\(=\left(m_A-m_B-m_D\right) \times 931.5\,MeV\)
\(=(238.05079-234.04363-4.00260) \times 931.5\)
\(\Rightarrow 4.25\,Mev\)
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