The energy released when 7/17.13 kg of 7 Li is converted into 4 He by proton bombardment is α × 10 32 eV . The value of α is (Nearest integer). (Mass of 7 Li =7.0183 u , mass of 4 He =4.004 u , mass of proton=1.008u, 1 u =931 MeV/c 2 , Avogadro number= 6.0times10 23 )

Physics · Nuclear Physics

JEE Main 2026 — 6 April, Morning Shift — Question 18

The energy released when $\frac{7}{17.13}\mathrm{kg}$ of $^7\mathrm{Li}$ is converted into $^4\mathrm{He}$ by proton bombardment is $\alpha \times 10^{32}\mathrm{eV}$ . The value of $\alpha$ is (Nearest integer). (Mass of $^7\mathrm{Li}=7.0183\mathrm{u}$ , mass of $^4\mathrm{He}=4.004\mathrm{u}$ , mass of proton=1.008u, $1\mathrm{u}=931\mathrm{MeV/c}^2$ , Avogadro number= $6.0\times10^{23}$ )

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

Reaction: $^7\mathrm{Li} + p \rightarrow 2^4\mathrm{He}$ . Mass defect = (7.0183+1.008) - 2×4.004 = 8.0263 - 8.008 = 0.0183 u. Energy per reaction = 0.0183×931 = 17.0373 MeV. Number of Li atoms = (mass Li / molar mass) × NA = ( (7/17.13) × 1000? Wait given mass = 7/17.13 kg = 7000/17.13 g? Actually 7/17.13 kg = 7000/17.13 g ≈ 408.6 g. Molar mass of Li-7 = 7 g/mol. So moles = 408.6/7 ≈ 58.37, atoms = 58.37×6×10^23 = 3.502×10^25. Total energy = 3.502e25 × 17.0373 MeV = 5.966×10^26 MeV = 5.966×10^32 eV. So α ≈ 6.

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