Physics Circle home Student Login

Home›Chapters›Nuclear Physics PYQs›Nuclear Fission & Fusion›Question 2

JEE Main 2026 (23 Jan, Shift 2) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from JEE Main 2026 (23 Jan, Shift 2)

Q2JEE Main 2026 · 23 JAN · SHIFT 2Medium

Nuclear Fission & Fusion

The average energy released per fission for the nucleus of 92235U{}^{235}_{92}\text{U} is 190 MeV. When all the atoms of 47 g pure 92235U{}^{235}_{92}\text{U} undergo fission process, the energy released is α×1023 MeV\alpha \times 10^{23}\text{ MeV}. The value of α\alpha is ______. (Avogadro Number =6×1023= 6 \times 10^{23} per mole)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 228

  1. Moles =47235=0.2= \dfrac{47}{235} = 0.2, so N=0.2×6×1023=1.2×1023N = 0.2\times6\times10^{23} = 1.2\times10^{23} nuclei.
  2. E=1.2×1023×190=228×1023 MeVE = 1.2\times10^{23}\times190 = 228\times10^{23}\text{ MeV}.
  3. So α=228\alpha = 228.
Report a mistake

More questions on Nuclear Fission & Fusion