The energy released when (7)/(17.13) kg of ⁷₃Li is converted into ⁴₂He by proton bombardment is α × 10³² eV. The value of α is ______. (Nearest integer) (Mass of ⁷₃Li = 7.0183u, mass of ⁴₂He = 4.004u, mass of proton = 1.008u and 1u = 931 MeV/c² and Avogadro number = 6.0 × 10²³) Numerical answer type. Work it out, then open the answer.

Answer: 6. Δ m = 7.0183 + 1.008 - 2(4.004) = 0.0183 u, so Q = 0.0183 × 931 = 17.04 MeV per reaction. Moles of Li = (7000/17.13 g)/(7 g/mol) = 58.38; nuclei = 58.38 × 6.0 × 10²³ = 3.50 × 10²⁵ E = 3.50 × 10²⁵ × 17.04 × 10⁶ eV = 5.97 × 10³² eV So α ≈ 6.

The energy released when (7)/(17.13) kg of ⁷₃Li is converted into… | JEE Main 2026 (6 Apr, Shift 1)

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Nuclear Reactions & Q-value

The energy released when $\frac{7}{17.13}\text{ kg}$ of ${}^{7}_{3}\text{Li}$ is converted into ${}^{4}_{2}\text{He}$ by proton bombardment is $\alpha \times 10^{32}\text{ eV}$ . The value of $\alpha$ is ______. (Nearest integer) (Mass of ${}^{7}_{3}\text{Li} = 7.0183\text{u}$ , mass of ${}^{4}_{2}\text{He} = 4.004\text{u}$ , mass of proton $= 1.008\text{u}$ and $1\text{u} = 931\text{ MeV/c}^{2}$ and Avogadro number $= 6.0 \times 10^{23}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 6

$\Delta m = 7.0183 + 1.008 - 2(4.004) = 0.0183\ \text{u}$ , so $Q = 0.0183 \times 931 = 17.04\ \text{MeV}$ per reaction.
Moles of Li $= \dfrac{7000/17.13\ \text{g}}{7\ \text{g/mol}} = 58.38$ ; nuclei $= 58.38 \times 6.0 \times 10^{23} = 3.50 \times 10^{25}$
$E = 3.50 \times 10^{25} \times 17.04 \times 10^{6}\ \text{eV} = 5.97 \times 10^{32}\ \text{eV}$
So $\alpha \approx 6$ .

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