JEE Main 2026 Apr 6 Shift 1, Physics Q23: Mass-Energy and Nuclear Binding Energy

JEE Main2026Apr 6, Shift 1Physics

The energy released when $\frac{7}{17.13}$ kg of $_{3}^{7} Li$ is converted into $_{2}^{4} He$ by proton bombardment is $α \times 1 0^{32}$ eV. The value of $α$ is ______. (Nearest integer)

(Mass of $_{3}^{7} Li = 7.0183$ u, mass of $_{2}^{4} He = 4.004$ u, mass of proton $= 1.008$ u and $1$ u $= 931$ MeV/c $^{2}$ and Avogadro number $= 6.0 \times 1 0^{23}$ )

Your answer

Solution

Reaction: $_{3}^{7} Li +_{1}^{1} p \to 2_{2}^{4} He$ .

Mass defect per reaction: $Δ m = (7.0183 + 1.008) - 2 (4.004) = 8.0263 - 8.008 = 0.0183$ u.

Energy per reaction: $E_{1} = 0.0183 \times 931$ MeV $= 17.04$ MeV $= 1.704 \times 1 0^{7}$ eV.

Moles of Li: $n = \frac{7 \times 1 0 ^{3} g}{17.13 \cdot 7 g/mol} = \frac{1 0 ^{3}}{17.13}$ mol.

Number of Li atoms (= number of reactions): $N = n \cdot N_{A} = \frac{1 0 ^{3}}{17.13} \cdot 6 \times 1 0^{23} = \frac{6 \times 1 0 ^{26}}{17.13}$ .

Total energy: $E_{tot} = N E_{1} = \frac{6 \times 1 0 ^{26}}{17.13} \times 0.0183 \times 931 \times 1 0^{6}$ eV.

$= \frac{6}{17.13} \times 0.0183 \times 931 \times 1 0^{32} \approx 5.97 \times 1 0^{32}$ eV $\approx 6 \times 1 0^{32}$ eV.

$∴ α = 6$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q23: Mass-Energy and Nuclear Binding Energy

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