Determine the amount of energy released during the reaction given...

The nuclear reaction can be written using atomic masses as:

^{7}\text{Li} + ^{1}\text{H} \rightarrow 2 ^{4}\text{He}$<br><br>Taking the atomic mass of hydrogen ^{1}\text{H}$ as $1.00783 \text{ u}$, the total mass of the reactants is:

$m_{\text{reactants}} = m(^{7}\text{Li}) + m(^{1}\text{H})$

$m_{\text{reactants}} = 7.0160 \text{ u} + 1.00783 \text{ u} = 8.02383 \text{ u}$

The total mass of the products is:

$m_{\text{products}} = 2 \times m(^{4}\text{He})$

$m_{\text{products}} = 2 \times 4.0026 \text{ u} = 8.0052 \text{ u}$

The mass defect $\Delta m$ is:

$\Delta m = m_{\text{reactants}} - m_{\text{products}}$

$\Delta m = 8.02383 \text{ u} - 8.0052 \text{ u} = 0.01863 \text{ u}$

The energy released $Q$ is calculated by converting the mass defect into energy (using $1 \text{ u} \approx 931 \text{ MeV}/c^2$):

$Q = 0.01863 \times 931 \text{ MeV} \approx 17.34 \text{ MeV}$

Answer: $17.34 \text{ MeV}$