[Given : $\mathrm{M}(\mathrm{Li})=6.01690\ \mathrm{amu} . \mathrm{M}\left({ }_1 \mathrm{H}^2\right)=2.01471 \ amu.$ $\mathrm{M}\left({ }_2 \mathrm{He}^4\right)=4.00388\ \mathrm{amu}$,$ and\ $$1 \ \mathrm{amu}=931.5$ $\mathrm{MeV}]$
\( { }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^3 \rightarrow{ }_2 \mathrm{He}^4+{ }_0 \mathrm{n}^1\)
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\({ }_3 \mathrm{Li}^6+{ }_1 \mathrm{H}^2 \rightarrow 2\left({ }_2 \mathrm{He}^4\right)\)
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Energy released in process
\( \mathrm{Q}=\Delta \mathrm{mc}^2 \)
\( \mathrm{Q}=\left[\mathrm{M}(\mathrm{Li})+\mathrm{M}\left({ }_1 \mathrm{H}^2\right)-2 \times \mathrm{M}\left({ }_2 \mathrm{He}^4\right)\right] \times 931.5\ \mathrm{MeV} \)
\( \mathrm{Q}=[6.01690+2.01471-2 \times 4.00388] \times 931.5\ \mathrm{MeV} \)
\( \mathrm{Q}=22.216 \ \mathrm{MeV} \)
\( \mathrm{Q}=22.22\ \mathrm{MeV}\)
જનક ન્યુકિલયસની ન્યુકિલયોનદીઠ બંધનઊર્જા $E_1 $ છે અને જનિત ન્યુકિલયસ માટે $E_2 $ છે, તો પછી .......