Physics - Modern Physics Question with Solution | TestHub

${}_4{}^{ }L{i}^7+{{}_1{}^{ }H}^1\to {}_2{}^{ }\left({}_2{}^{ }H{e}^4\right)$
$BE$ of products$=(\left(5.6 \mathrm{M}\mathrm{e}\mathrm{V}\right)\times 7)+0$
$=39.2 \mathrm{M}\mathrm{e}\mathrm{V}$
$E_i=-39.2 \mathrm{M}\mathrm{e}\mathrm{V}$
$\mathrm{B}\mathrm{E}$ of reactant$=\left(7.06\right)\times 4\times 2$
$=56.48 \mathrm{M}\mathrm{e}\mathrm{V}$
$E_f=-56.48 \mathrm{M}\mathrm{e}\mathrm{V}$

As nuclear energy decreases, some energy will be released.
$Q_{\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{e}}=E_i-E_f=\left(-39.2\right)-\left(-56.48\right)=17.28 \mathrm{M}\mathrm{e}\mathrm{V}$