Physics Circle home Student Login

Home›Chapters›Nuclear Physics PYQs›Nuclear Fission & Fusion›Question 4

JEE Main 2025 (2 Apr, Shift 2) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from JEE Main 2025 (2 Apr, Shift 2)

Q4JEE Main 2025 · 2 APR · SHIFT 2Easy

Nuclear Fission & Fusion

Energy released when two deuterons (1H2)\left({}_{1}\text{H}^{2}\right) fuse to form a helium nucleus (2He4)\left({}_{2}\text{He}^{4}\right) is: (Given: Binding energy per nucleon of 1H2=1.1 MeV{}_{1}\text{H}^{2} = 1.1\text{ MeV} and binding energy per nucleon of 2He4=7.0 MeV{}_{2}\text{He}^{4} = 7.0\text{ MeV})

Show answer and solution

Answer: (c)

  1. Q=4×7.0−2×(2×1.1)Q = 4\times7.0 - 2\times(2\times1.1).
  2. Q=28.0−4.4=23.6 MeVQ = 28.0 - 4.4 = 23.6\text{ MeV}.
  3. So the answer is (c) 23.6 MeV23.6\text{ MeV}.
Report a mistake

More questions on Nuclear Fission & Fusion