The energy released when (7)/(17.13) kg of ^ 7 _ 3 Li is converted...

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: $6$