18.3.2 Nuclear Fusion

Nuclear fusion occurs when two smaller atomic nuclei combine to form a larger nucleus.

For example, a deuteron (H-2) can combine with a triton (H-3) to create a helium nucleus (He-4), releasing one neutron in the process.

$\displaystyle \displaystyle {}_{1}^{2}H+{}_{1}^{3}H\to {}_{2}^{4}He+{}_{0}^{1}n$

Let calculate the energy released during this fusion.

BE of $\displaystyle {}_{1}^{2}H=2.224573\text{ MeV}$ BE of $\displaystyle {}_{2}^{4}He=28.295673\text{ MeV}$

BE of $\displaystyle {}_{1}^{3}H=8.481821\text{ MeV}$ BE of $\displaystyle {}_{0}^{1}n=0$

Total BE before fusion $\displaystyle \sum B{{E}_{i}}=2.224573+8.481821=10.706394\text{ MeV}$

Total BE after fusion $\displaystyle \sum B{{E}_{f}}=28.295673+0=28.295673\text{ MeV}$

Increase in BE,

$\displaystyle \begin{aligned}\Delta BE&=\sum B{{E}_{f}}-\sum B{{E}_{i}}\\&=28.295673-10.706394\\&=\text{ }17.6\text{ MeV}\end{aligned}$

An increase in total BE implies lower total rest energy. Why? Because an increase in total BE implies an increase in total mass defect (since $\displaystyle BE=\Delta m.{{c}^{2}}$ ), which implies a decrease in total mass. So an increase in total BE of 17.6 MeV implies that 17.6 MeV of energy is released during a deuterium-tritium fusion. This energy is mainly carried by the KE of the He-4 and the neutron.

Our Sun (and all stars) generates its energy from the hydrogen fusion series: the larger nuclei formed from each stage serve as the inputs to the next stage. So hydrogen combine to form deuterium to form tritium to form helium, and tremendous amounts of energy is released at each stage.

You may wonder why hydrogen fusion does not occur among the hydrogen atoms in our atmosphere. Well, in order for fusion to occur, the nuclei have to be brought within “touching distance” of each other. Unfortunately, the electrical repulsion between the two protons in the hydrogen nuclei is formidable.

Assuming the radius of a proton to be $\displaystyle {{10}^{{-15}}}\text{ m}$ , the electrical potential energy when two protons are at “touching distance” is about

$\displaystyle \frac{1}{{4\pi {{\varepsilon }_{0}}}}\frac{{(1.60\times {{{10}}^{{-19}}})(1.60\times {{{10}}^{{-19}}})}}{{1\times {{{10}}^{{-15}}}}}=1.44\text{ MeV}$

1.44 MeV represents a lot of KE for a hydrogen nucleus. Even the hydrogen nuclei in the Sun’s core, at temperature of $\displaystyle 15\times {{10}^{6}}\text{ K}$ , need a little help from quantum tunneling to achieve fusion. On Earth, we have yet to develop a physically and economically viable process to generate power using fusion. Which is a pity, because nuclear fusion promises an even cleaner and safer source of energy compared to nuclear fission.