The energy released if hydrogen atoms are combined to form 4 2 He is MeV. (Take binding energies per nucleon of hydrogen as 1.1 MeV and of helium as 7.2 MeV).

Physics · Nuclear Physics

JEE Main 2026 — 6 April, Evening Shift — Question 13

The energy released if hydrogen atoms are combined to form $^4_2\mathrm{He}$ is ______ MeV. (Take binding energies per nucleon of hydrogen as 1.1 MeV and of helium as 7.2 MeV).

Option A:
6.1
Option B:
24.4
Correct
Option C:
26.6
Option D:
5

Answer: B

Step-by-step solution

4 hydrogen atoms combine to form He-4. Total BE of H = 4×1.1 = 4.4 MeV. BE of He = 4×7.2 = 28.8 MeV. Energy released = 28.8 - 4.4 = 24.4 MeV.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction