We have shown that transmuting hydrogen into helium would release a large amount of energy. Let’s be more specific. The first step is to use the easiest reaction possible, which is the following:

D + T ^ a + n + 17.6 MeV.

Remember that D stands for deuterium, a hydrogen isotope containing one proton and one neutron, and T stands for tritium, containing one proton and two neutrons. Alpha (a) stands for a helium nucleus (He4), containing two protons and two neu­trons. There is one neutron (n) left over, which flies off carrying most of the energy released in the fusion, which is 17.6 million electron volts. This reaction is depicted in Fig. 4.3. The intermediate state shown there with five nucleons in it is not stable and immediately breaks up into an a particle and a neutron. The a particle, being ordinary helium, is very stable; and its energy will be used to keep the reaction going. The neutron carries 80% of the energy released (about 14 MeV), and it has to be captured and its energy transformed into heat, replacing the heat we now get from burning fossil fuels to run a power plant. Although neither reaction product is itself radioactive, the neutron can induce radioactivity in the walls of the reactor, and this material has to be buried. We shall see in Chap. 9 that the amount of long — lived radioactive waste is about 1,000 times smaller than for fission reactors. The D-T reaction is the worst of the fusion reactions in this regard, but it is the easiest to start with. Chapter 10 will show advanced reactions which have less radioactivity or none at all.

Fig. 4.3 The D-T reaction

As pointed out before, deuterium is easy to separate out from water, but tritium has to be made in a nuclear reaction. In a fusion reactor, tritium is regenerated in a “blanket” containing lithium. Leaving this Chap. 9 topic aside for the moment, let us see how we can make this reaction go, because it’s not easy. Since D and T have one proton apiece, they each have a positive charge. Like charges repel, so if we fire a beam of deuterons into a tritium target, the D’s will most likely bounce off the T’s without ever getting close enough to combine. Only a head-on collision with energy larger than 280 keV can overcome the electric repulsion (the so-called Coulomb barrier). Once inside this barrier, the nuclear force takes over, and the force becomes attractive instead of repulsive. Most of the time, however, the D’s will bounce off without penetrating the barrier and lose most of the energy used to accelerate them. It is possible to use beams of around 60 keV energy and get net energy out, but not enough to justify the large number accelerators needed to make a dent on the power grid. There is a better solution. And that is not to use beams of particles at all but to heat a hydrogen gas, half in the form of deuterons and half in the form of tritons (tritium nuclei), to such a high temperature that there are always some high-energy collisions that result in fusion. The energy in failed collisions is not lost; it returns to the gas to keep it hot. This hot gas, called a plasma, perks away steadily, releasing enough fusion energy to keep itself hot and generate power besides. That’s what happens in the interior of the sun. The fusion power generated comes out as solar radiation, of which the earth receives a small portion.

It is hard for people to understand why a hot plasma is necessary when you can simply shoot a beam of deuterons from a particle accelerator and hit a solid tritium target with enough energy to penetrate the electric barrier and get the D and T close enough to fuse. Or, one might circulate a beam of deuterons in one direction and a beam of tritons in the opposite direction in a round accelerator. Once in a while there will be a head-on collision and a fusion. But not often enough to pay for the energy used in accelerating the beams! Believe me, many proposals for using beams for fusion have been tried and have failed. Here is an analogy to illustrate how plasma fusion works. Imagine a friction-free pool table which has no pockets at the edge. However, there are pockets all over the middle of the table, and each pocket is surrounded by a hill, like a deep crater at the center of a volcano. The hills represent Coulomb repulsion. A pool player then adds billiard balls randomly, shooting them with insufficient accuracy and speed to climb the hills and get into the hole. Since there is no friction, the numerous balls keep bouncing around at random until one is lost by chance by jumping off the table, whereupon it is replaced by another shot. Since the balls bounce against one another, once in a while, one will undergo several favorable bounces in a row and end up with more

than the average energy. If it has enough energy and is going right toward a crater, it will be able to climb the hill and get into the pocket at the top. This represents a fusion reaction yielding 17.6 MeV of energy. You might have to wait a long time for this to happen, but after the initial energy used to shoot the balls, no more energy is needed other than to replace those lost over the edge. This is the idea of plasma-induced fusion. A small amount of energy is invested in shooting the balls in, and then one waits for a long time before a ball by chance climbs a hill and gets into the pocket. But the payoff in energy is so huge that there is a large energy gain even if it takes many collisions to get one fusion.