JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2024
- Hard
- Numerical
The disintegration energy \(Q\) for the nuclear fission of \({ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}\) is _______ \(\mathrm{MeV}\). Given atomic masses of \({ }^{235} \mathrm{U}: 235.0439 \mathrm{u},{ }^{140} \mathrm{Ce} ; 139.9054 \mathrm{u},\) \({ }^{94} \mathrm{Zr}: 93.9063 \mathrm{u} ; \mathrm{n}: 1.0086 \mathrm{u},\) Given Value of \(\mathrm{c}^2=931 \mathrm{MeV} / \mathrm{u}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(208\)
Step-by-step Solution
Detailed explanation
\({ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}\)
Disintegration energy
\(\mathrm{Q} =\left(\mathrm{m}_{\mathrm{k}}-\mathrm{m}_{\mathrm{r}}\right) \mathrm{c}^2\)
\(\mathrm{~m}_{\mathrm{n}} =235.0439 \mathrm{u}\)
\(\mathrm{m}_{\mathrm{r}} =139.9054 \mathrm{u}+93.9063 \mathrm{u}+1.0086 \mathrm{u}\)
\(=234.8203 \mathrm{u}\)
\(\mathrm{Q} =(235.0439 \mathrm{u}-234.8203 \mathrm{u}) \mathrm{c}^2\)
\(=0.2236 \mathrm{c}^2\)
\(=0.2236 \times 931\)
\(\mathrm{Q} =208.1716\)
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