Answer
(a)
$$Q=13.93\ \mathrm{MeV}$$
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(b)
$$\mathrm{KE}_{\text {nucleus }}=4.71\ \mathrm{MeV}$$
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(c)
$$k T= 5.46 \times 10^{10} \mathrm{K}$$
Work Step by Step
(a)
From Eq. $30–2$, the $Q-value$ is the mass energy of the reactants minus the mass energy of the products.
$$_{6}^{12} \mathrm{C}+_{6}^{12} \mathrm{C} \rightarrow_{12}^{24} \mathrm{Mg}$$
$$Q=2 m_{\mathrm{C}} c^{2}-m_{\mathrm{Mg}} c^{2}=$$
$$[2(12.000000 \mathrm{u})-23.985042 \mathrm{u}] c^{2}\left(931.5 \mathrm{MeV} / c^{2}\right)=13.93\ \mathrm{MeV}$$
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(b)
The distance between the two nuclei will be twice the nuclear radius, from Eq. $30–1$. Each nucleus will have half the total kinetic energy.
$$r=\left(1.2 \times 10^{-15} \mathrm{m}\right)(A)^{1 / 3}=\left(1.2 \times 10^{-15} \mathrm{m}\right)(12)^{1 / 3}\ ;$$
$$ \quad \mathrm{PE}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{\text {nucleus }}^{2}}{2 r}$$
$$\mathrm{KE}_{\text {nucleus }}=\frac{1}{2} \mathrm{PE}=\frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{\text {nucleus }}^{2}}{2 r}$$
$$=\frac{1}{2}\left(8.988 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{(6)^{2}\left(1.60 \times 10^{-19} \mathrm{C}\right)^{2}}{2\left(1.2 \times 10^{-15} \mathrm{m}\right)(12)^{1 / 3}} \times \frac{1 \mathrm{eV}}{1.60 \times 10^{-19} \mathrm{J}}$$
$$=4.71\ \mathrm{MeV}$$
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(c)
Approximate the temperature-kinetic energy relationship by $k T=\mathrm{KE},$ as given in Section $33-7$\
$$k T=\mathrm{KE} \rightarrow T=\frac{\mathrm{KE}}{k}$$
$$=\frac{\left(4.71 \times 10^{6} \mathrm{eV}\right)\left(\frac{1.60 \times 10^{-19} \mathrm{J}}{1 \mathrm{eV}}\right)}{1.38 \times 10^{-23} \mathrm{J} / \mathrm{K}} $$
$$= 5.46 \times 10^{10} \mathrm{K}$$
