Carbon fusion

First, we determine the nuclear radius $R$ . The nuclei are spheres with a volume $V = \frac{4}{3} \pi R^3$ . The mass of a nucleus $M \approx m_p A$ is porportional to the mass number $A = Z + N$ . (The mass defect of the nucleons is neglected because it accounts for less than 1% of the mass.) Since the nuclei all have the same density, it results $\begin{aligned} & & \rho = \frac{M}{V} &= \frac{m_p A}{\frac{4}{3} \pi R^3 } \\ \Rightarrow & & R &= \left[\frac{3 m_p A}{4 \pi \rho }\right]^{1/3}\\ \Rightarrow & & R_\text{C} &\approx 2.88 \cdot 10^{-15} \,\text{m} \\ & & R_\text{Mg} &\approx 3.63 \cdot 10^{-15} \,\text{m} \end{aligned}$ Only two contributions are important for the determination of the released energy:

• Surface energy: A nucleus with radius $R$ has a surface energy $W_ \text{surface} = \sigma \cdot 4 \pi R ^2$ . Since the surface ( $\propto R ^2$ ) grows more slowly than the volume ( $\propto R ^3$ ), the nucleons can minimize surface energy when two small nuclei combine to form a large nucleus.
• Coulomb energy: A homogeneously charged sphere with radius $R$ and charge $Q = + Z e$ generates an electric field $\vec E = \begin{cases} \dfrac{Q r}{4 \pi \varepsilon_0 R^3} \vec e_r & r < R \\ \dfrac{Q}{4 \pi \varepsilon_0 r^3} \vec e_r & r \geq R \end{cases}$ The electrostatic field energy is given by the following volume integral: $\begin{aligned} W_\text{Coulomb} &= \frac{\varepsilon_0}{2} \int (\vec E \cdot \vec E) dV \\ &= \frac{\varepsilon_0}{2} \left[ \int_0^{R} \left(\dfrac{Q r}{4 \pi \varepsilon_0 R^3}\right)^2 4 \pi r^2 dr + \int_R^{\infty} \left(\dfrac{Q}{4 \pi \varepsilon_0 r^2}\right)^2 4 \pi r^2 dr \right] \\ &= \frac{Q^2}{4 \pi \varepsilon_0} \left[ \int_0^{R} \dfrac{r^4}{R^6} dr + \int_R^{\infty} \frac{1}{r^2} dr \right] \\ &= \frac{Q^2}{4 \pi \varepsilon_0} \left[ \frac{1}{5 R} + \frac{1}{R} \right] \\ &= \frac{3 Q^2}{10 \pi \varepsilon_0 R} \end{aligned}$ The energy released during fusion thus results $\begin{aligned} \Delta E &= 2 \cdot (W_\text{surface}^\text{C} + W_\text{Coulomb}^\text{C}) - (W_\text{surface}^\text{Mg} + W_\text{Coulomb}^\text{Mg}) \\ &= 4 \pi \sigma (2 \cdot R_\text{C}^2 - R_\text{Mg}^2) + \frac{3 e^2}{10 \pi \varepsilon_0} \left(2 \frac{6^2}{R_\text{C}} - \frac{12^2}{R_\text{Mg}} \right) \\ & \approx 1.5\cdot 10^{-12} \,\text{J} \\ & \approx 9.6 \, \text{MeV} \end{aligned}$