Question

How much energy is released in the following reaction:
⁷Li + p → α + α.
Atomic mass of ⁷Li = 7.0160 u and that of ⁴He = 4.0026 u.

Answer 1

Given:
Mass of ⁷Li = 7.0160 u
Mass of ⁴He = 4.0026 u.
Reaction:
$\mathrm{L}{i}^7+p\mathit{\to }\alpha \mathit{+}\alpha \mathit{+}E\mathit{,}$
Energy release $\left(E\right)$ is given by
$$\begin{gathered} E=\left[m\left({}^7\mathrm{Li}\right)+\left(m_p\right)-2\times m\left({}^4\mathrm{He}\right)\right]c^2 \\ =\left[\left(7.0160\mathrm{u}+1.007276\mathrm{u}\right)-2\left(4.0026\mathrm{u}\right)\right]c^2 \\ =(8.023273\mathrm{u}-8.0052\mathrm{u}){\mathrm{c}}^2 \\ =0.018076\times 931\mathrm{MeV} \\ =16.83\mathrm{MeV} \end{gathered}$$