The energy released if hydrogen atoms are combined to form ⁴₂He is ______ MeV. (Take binding energies per nucleon of ²₁H and ⁴₂He as 1.1 MeV and 7.2 MeV, respectively) (a) 6.1 (b) 24.4 (c) 26.6 (d) 5

Answer: (b). The reaction is 2 ²₁H → ⁴₂He; Q = BE(product) - BE(reactants). BE of ⁴He = 4×7.2 = 28.8 MeV; BE of two ²H = 2×2×1.1 = 4.4 MeV. Q = 28.8 - 4.4 = 24.4 MeV. So the answer is (b) 24.4.

The energy released if hydrogen atoms are combined to form ⁴₂He is… | JEE Main 2026 (6 Apr, Shift 2)

Q1JEE Main 2026 · 6 APR · SHIFT 2Easy

Nuclear Fission & Fusion

The energy released if hydrogen atoms are combined to form ${}^{4}_{2}\text{He}$ is ______ MeV. (Take binding energies per nucleon of ${}^{2}_{1}\text{H}$ and ${}^{4}_{2}\text{He}$ as 1.1 MeV and 7.2 MeV, respectively)

Show answer and solution

Answer: (b)

The reaction is $2\,{}^2_1\text{H} \to {}^4_2\text{He}$ ; $Q = \text{BE(product)} - \text{BE(reactants)}$ .
BE of ${}^4\text{He} = 4\times7.2 = 28.8\text{ MeV}$ ; BE of two ${}^2\text{H} = 2\times2\times1.1 = 4.4\text{ MeV}$ .
$Q = 28.8 - 4.4 = 24.4\text{ MeV}$ .
So the answer is (b) $24.4$ .

Report a mistake

Nuclear Fission & Fusion question from JEE Main 2026 (6 Apr, Shift 2)

More questions on Nuclear Fission & Fusion