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

The nuclear reaction for the proton bombardment of Lithium is:

\(^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He}\)

The mass defect (\(\Delta m\)) for the reaction is:

\(\Delta m = m(^{7}_{3}\text{Li}) + m(^{1}_{1}\text{H}) - 2m(^{4}_{2}\text{He})\)

\(\Delta m = 7.0183 + 1.008 - 2(4.004)\)

\(\Delta m = 8.0263 - 8.008 = 0.0183 \text{ u}\)

Energy released per reaction (\(Q\)) is:

\(Q = \Delta m \times 931 \text{ MeV} = 0.0183 \times 931 \text{ MeV} = 17.0373 \text{ MeV}\)

Number of moles of \(^{7}_{3}\text{Li}\) in \(\dfrac{7}{17.13} \text{ kg}\):

\(n = \dfrac{\dfrac{7}{17.13} \times 10^3 \text{ g}}{7 \text{ g/mol}} = \dfrac{1000}{17.13} \text{ mol}\)

Number of atoms of \(^{7}_{3}\text{Li}\):

\(N = n \times N_A = \dfrac{1000}{17.13} \times 6.0 \times 10^{23} = \dfrac{6}{17.13} \times 10^{26}\)

Total energy released (\(E\)) is:

\(E = N \times Q = \left(\dfrac{6}{17.13} \times 10^{26}\right) \times (17.0373 \times 10^6 \text{ eV})\)

\(E = \dfrac{102.2238}{17.13} \times 10^{32} \text{ eV} \approx 5.967 \times 10^{32} \text{ eV}\)

Rounding to the nearest integer, we get \(\alpha = 6\).

Answer: \(6\)