The energy released when kg of is converted into by proton bombardment is . The value of is (Nearest integer)

Question

The energy released when $\frac{7}{17.13}$ kg of ${}_{3}^{7}{Li}$ is converted into ${}_{2}^{4}{He}$ by proton bombardment is $α \times 10^{32} eV$ . 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 $= 6.0 \times 10^{23}$ ) [2026]

Smart Achievers · Accepted Answer

(6)

${}_{3}^{7}L i + {}_{1}^{1}H ⇌ 2 ({}_{2}^{4}H e) + Q$

$Mass of Reactants:$

$M_{L i} + M_{p} = 7.0183 u + 1.008 = 8.0263 u$

$Mass of Product: 2 \times M_{H e} = 2 \times 4.004 u = 8.008 u$

$Mass defect (Δ m) = 8.0263 u - 8.008 u = 0.0183 u$

$Energy (Q) = Δ m c^{2} = 0.0183 \times 931 MeV \approx 17.0373 MeV$

$Number of Li atoms: N = \frac{Mass}{Molar mass} \times N_{A}$

$N = \frac{\frac{7}{17.13} kg}{7 \times 10^{- 3} kg / mol} \times 6.02 \times 10^{23} = \frac{6}{17.13} \times 10^{26}$

$E_{total} = N \times Q$

$E_{total} = 17.0373 \times \frac{1}{17.13} \times 6 \times 10^{26} \times 10^{6} eV$

$E_{total} \approx 6 \times 10^{32} eV$

$Comparing we get α = 6$

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Solution

Q. The energy released when 717.13kg of Li37 is converted into He24 by proton bombardment is α×1032 eV. The value of α is ______. (Nearest integer) (Mass of Li37=7.0183 u, mass of He24=4.004 u, mass of proton = 1.008 u and 1 u = 931 MeV/c2 and Avogadro number =6.0×1023) [2026]

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