Discussion
What is the energy released in the following nuclear reaction :
\[ {}^{6}_{3}Li + {}^{2}_{1}H \rightarrow 2\,{}^{4}_{2}He + Q \]
(Given : \({}^{6}_{3}Li = 6.01693\) amu, \({}^{2}_{1}H = 2.01471\) amu, \({}^{4}_{2}He = 4.00388\) amu)
Answer: B
On the key, the energy released is about 22.4 MeV.
The Q value is the mass lost in the reaction:
\[ \begin{aligned} \Delta m &= (6.01693 + 2.01471) - 2(4.00388) \\ &= 8.03164 - 8.00776 = 0.02388\ \text{amu} \\ Q &= 0.02388 \times 931.5 = 22.2\ \text{MeV} \end{aligned} \]
The keyed 22.4 MeV is the nearest option to this value; the other options are about 11 MeV away. The positive Q shows the reaction releases energy, because the two alpha particles are much more tightly bound than lithium-6 and the deuteron.
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