JEE Main 2023 · 1 February · morning · Single correct · +4 / −1

The mass of proton, neutron and helium nucleus are respectively 1.0073 u,1.0087 u and 4.0015 u. The binding energy of helium nucleus is :

  1. (a) 28.4 MeV
  2. (b) 56.8 MeV
  3. (c) 7.1 MeV
  4. (d) 14.2 MeV
Answer and solution

Answer (a) 28.4 MeV

Solution

Physics sir · 3 steps

Add up the free nucleons, subtract the nucleus, and turn that missing mass into MeV with 931.

  1. Count the nucleons. Helium-4 has 2 protons and 2 neutrons. Their separate masses add to 2(1.0073) + 2(1.0087) = 2.0146 + 2.0174 = 4.0320 u.
  2. Find the mass defect. The nucleus weighs less than its parts. The difference is Δ m = 4.0320 - 4.0015 = 0.0305 u.

    E_b = Δ m × 931 MeV/u = 0.0305 × 931 ≈ 28.4 MeV

  3. So. The binding energy of the helium nucleus is about 28.4 MeV, option (a).

The idea to keep. Binding energy is the mass of the free protons and neutrons minus the nucleus mass, times 931 MeV per u. Divide by A only if they ask per nucleon.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 110 times in JEE Main since 2020. This question is from 2023.