The average energy released per fission for the nucleus of 92 235 U is 190 MeV . When all the atoms of 47 g pure 92 235 U undergo fission process, the energy released is α × 10 23 MeV . The value of α is \ \ \ \ . (Avogadro Number =6 × 10 23 per mole)

Physics · Nuclear Physics

JEE Main 2026 — 23 January, Evening Shift — Question 42

The average energy released per fission for the nucleus of ${ }_{92}^{235} \mathrm{U}$ is 190 MeV . When all the atoms of 47 g pure ${ }_{92}^{235} \mathrm{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \mathrm{MeV}$ . The value of $\alpha$ is $\_\_\_\_$ . (Avogadro Number $=6 \times 10^{23}$ per mole)

Answer: 228

Numerical answer — enter this value.

Step-by-step solution

Total numbers of U-235 atom is $47 \mathrm{~g}=\frac{47}{235}$ moles $=\frac{1}{5}$ moles ∴ Total energy released $=\frac{1}{5} \times 6 \times 10^{23} \times 190 \mathrm{MeV}$ $=228 \times 10^{23} \mathrm{MeV}$

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Exam: JEE Main 2026
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Nuclear Physics
Topic: Nuclear Fission and Fusion

Which of the following pair of nuclei are isobars of the element?

Answer: 228

Step-by-step solution

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