Types of Fusion Reactions

Fusion offers several usable reactions, and the choice of reaction greatly affects the required temperature, how the energy is extracted, and how easily the fuel can be obtained. This page covers four representative reactions (D-T, D-D, D-He3, and p-B11), looking in turn at each reaction equation and its energy, how the likelihood of a reaction changes with temperature, and the strengths and challenges of each fuel, from simple intuition all the way to the research frontier.

Start with Intuition (High School)

Fusion is a reaction in which light atomic nuclei stick together to form a heavier nucleus, releasing a large amount of energy in the process. However, every atomic nucleus carries a positive electric charge, so when you try to bring two of them close together they repel each other strongly, like the same poles of two magnets. To overcome this wall of repulsion, the nuclei must collide at tremendous speed, which requires extremely high temperatures.

The height of the repulsion wall is set by the amount of charge each nucleus carries. Light nuclei with little charge have a low wall and can be joined under relatively mild conditions. That is why the fuel for fusion is best suited to the lightest members of the hydrogen family (deuterium and tritium). Conversely, reactions that use nuclei with a lot of charge, such as boron, have a much higher wall and require enormously higher temperatures.

Think of the type of reaction as essentially a difference in “which combination of fuels you use.” Let us line up the representative combinations using familiar images.

The D-T reaction combines deuterium (D) and tritium (T), and it is like the “well-burning firewood” that catches fire most easily. It is the star of current fusion research, and ITER uses this reaction too. When it burns, though, it releases many particles called neutrons, and these neutrons can damage the walls of the device and make them radioactive.

The D-D reaction can be burned using deuterium alone, so the fuel is easy to obtain, but it is firewood that is harder to ignite. The D-He3 and p-B11 reactions are “clean firewood that releases almost no neutrons,” but they burn very reluctantly and require even higher temperatures to ignite. Choosing among them is a matter of balancing how easily the fuel burns against how easily it can be handled.

Understand the Physics (Undergraduate)

First, let us estimate the repulsion wall, the Coulomb barrier. The electrostatic energy when two nuclei with charges $Z_1 e$ and $Z_2 e$ approach to a distance $r$ can be written as follows.

$$U(r) = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r}$$

Here $\varepsilon_0$ is the permittivity of vacuum and $e$ is the elementary charge. Substituting the distance at which the nuclei touch (about a few femtometers), the wall for D-T comes out to roughly a few hundred keV. Converting this to temperature corresponds to several billion kelvin, which classically is far out of reach. Yet in practice the reaction proceeds at 100-200 million degrees. The reason is quantum tunneling, in which a small probability remains for a particle to “slip through” the wall.

Let us organize the reaction equations and energies of the main reactions. The released energy $Q$ comes from the mass $\Delta m$ lost across the reaction being converted to energy according to $E=\Delta m c^2$.

D-T reaction:

$$\mathrm{D} + \mathrm{T} \rightarrow {}^4\mathrm{He}\,(3.5\ \mathrm{MeV}) + n\,(14.1\ \mathrm{MeV})$$

It releases a total of 17.6 MeV. About 80% of the energy is carried by the neutron $n$, and about 20% by the alpha particle (${}^4\mathrm{He}$).

The D-D reaction splits into two branches with nearly equal probability.

$$\mathrm{D} + \mathrm{D} \rightarrow \mathrm{T}\,(1.01\ \mathrm{MeV}) + p\,(3.02\ \mathrm{MeV})$$ $$\mathrm{D} + \mathrm{D} \rightarrow {}^3\mathrm{He}\,(0.82\ \mathrm{MeV}) + n\,(2.45\ \mathrm{MeV})$$

D-He3 reaction:

$$\mathrm{D} + {}^3\mathrm{He} \rightarrow {}^4\mathrm{He}\,(3.6\ \mathrm{MeV}) + p\,(14.7\ \mathrm{MeV})$$

Both products are charged particles carrying electric charge, and it does not directly produce a neutron.

p-B11 reaction:

$$p + {}^{11}\mathrm{B} \rightarrow 3\,{}^4\mathrm{He} + 8.7\ \mathrm{MeV}$$

It splits into three alpha particles and, again, produces no neutron.

The likelihood of a reaction is expressed by the cross section $\sigma$. The quantity $\sigma$ expresses, in units of area, “what fraction of encounters between particles result in a reaction,” and it depends strongly on the relative velocity of the particles (that is, on temperature). In a real plasma, particles have a velocity distribution (the Maxwell distribution), so we use the reaction rate coefficient $\langle \sigma v \rangle$, obtained by multiplying the cross section $\sigma$ by the relative velocity $v$ and averaging over the velocity distribution. The reaction rate per unit volume can be written using the number densities $n_1$ and $n_2$ of the fuel as follows.

$$R = n_1 n_2 \langle \sigma v \rangle$$

The $\langle \sigma v \rangle$ for D-T is one to two orders of magnitude larger than for the other reactions in the temperature range of 100-200 million degrees (10-20 keV). At the same temperature D-D is roughly 1/100 of D-T, and its required temperature is about 5 times higher. This “low temperature at which $\langle \sigma v \rangle$ is large” is precisely the physical reason D-T is chosen as the first fuel.

Deepen the Theory (Graduate)

The reaction cross section, mediated by tunneling, can be decomposed into the following form.

$$\sigma(E) = \frac{S(E)}{E}\exp\!\left(-\sqrt{\frac{E_G}{E}}\right)$$

Here $E$ is the center-of-mass energy, and $S(E)$ is called the astrophysical S-factor, which contains the properties of the nuclear reaction itself in a slowly varying form. The exponential part represents the tunneling probability, and $E_G$ is the Gamow energy, defined as follows.

$$E_G = (\pi \alpha Z_1 Z_2)^2 \, 2 m_r c^2$$

Here $\alpha$ is the fine-structure constant and $m_r$ is the reduced mass. Because $E_G$ is proportional to the square of $Z_1 Z_2$, we can quantitatively understand that nuclei with larger charge have more difficulty crossing the barrier.

When calculating $\langle \sigma v \rangle$ with the Maxwell distribution, the integrand becomes the product of two competing factors. On the low-energy side, the distribution function $\exp(-E/k_B T)$ is large while the tunneling probability is small, and on the high-energy side the tunneling probability is large while the number of particles in the distribution drops off sharply. The product of these two forms a sharp peak at a specific energy. This is the Gamow peak. The peak position $E_0$ is considerably higher than the thermal energy $k_B T$ and is approximated by the following.

$$E_0 = \left(\frac{\sqrt{E_G}\,k_B T}{2}\right)^{2/3}$$

In other words, the reactions are actually carried out by a subset of particles moving faster than average within the distribution, and this population in the tail governs the fusion output. Understanding the Gamow peak explains why the temperature dependence of $\langle \sigma v \rangle$ is so steep, and why raising the temperature only slightly greatly increases the output.

The handling of neutrons is also an important point connecting theory and engineering. The 14.1 MeV neutrons released by D-T are electrically neutral and cannot be confined by a magnetic field, so they fly straight into the reactor wall. These neutrons carry two meanings. One is energy recovery, in which the neutron energy is converted to heat in the surrounding blanket and used for power generation. The other is fuel breeding, which handles the tritium production described later. On the other hand, fast neutrons knock material atoms out of place, creating lattice defects (neutron irradiation damage) and activating the structural material. The neutron wall loading is a fundamental metric of reactor design, governing material lifetime and maintenance intervals.

Advanced fuels (D-He3, p-B11) fundamentally reduce this neutron problem. However, in D-He3 the side reaction D-D produces 2.45 MeV neutrons, so it is not completely aneutronic. The main reaction of p-B11 is aneutronic, but because of its large charge ($Z=5$) the Gamow energy is extremely large, requiring temperatures on the order of about 3 billion degrees for ignition. Furthermore, in high-temperature, high-$Z$ plasmas, electromagnetic-wave losses from bremsstrahlung increase in proportion to $Z^2$, making it hard for the fusion output to keep up with the losses. This balance of output versus loss is the heart of the difficulty of treating p-B11 with a simple Lawson criterion.

Research Frontier (PhD)

Research on advanced fuels has become active in recent years, motivated by the goal of neutron-free power generation. Regarding p-B11 (proton-boron 11), there are reports of high reaction yields under laser drive, and there is discussion of the idea of actively using non-thermal velocity distributions to circumvent the limits of thermal-equilibrium plasmas. Because bremsstrahlung losses tend to dominate in a thermal-equilibrium Maxwell distribution, non-thermal approaches, such as giving particles specific velocity components through beam injection, are being examined.

Refinement of the reaction rate coefficient $\langle \sigma v \rangle$ itself also continues. Cross section data are compiled mainly in evaluated nuclear data libraries, and especially for p-B11 the contributions of low-energy resonances and how electron screening lowers the effective barrier are the focus of quantitative evaluation.

Coupling with the fuel cycle is also a research theme. In D-T, tritium self-sufficiency is a requirement, and tritium is bred through the reaction of lithium and neutrons in the blanket,

$${}^6\mathrm{Li} + n \rightarrow {}^4\mathrm{He} + \mathrm{T} + 4.8\ \mathrm{MeV}$$

It must produce more than it consumes, and whether the tritium breeding ratio (TBR) can be made greater than 1 is a condition for the reactor to be viable. The arrangement of neutron multipliers and breeding materials, and the experimental verification of the TBR, are challenges.

The resource question for He3 is also a long-term theme. He3 is scarce on Earth, but it is said that He3 of solar-wind origin has accumulated in the lunar regolith, and it is mentioned as a future resource. That said, the cost of extraction and transport is unestablished, and at present it is a concept still at the research stage. Keeping in mind the keywords that appear frequently in papers, such as S-factor, Gamow peak, reactivity, aneutronic fusion, bremsstrahlung loss, and ignition condition, will make the reading easier to follow.

Check Your Understanding

Q1. Why are lighter atomic nuclei better suited for fusion?

A Because lighter nuclei have a larger mass defect, so the extractable energy always increases B Because a smaller charge means a lower Coulomb barrier and a higher reaction probability through tunneling, allowing the reaction to occur at a lower temperature C Because light nuclei have no neutrons, so no repulsion occurs D Because for lighter nuclei the electrons detach on their own without needing high temperatures

Q2. What is the total released energy of the D-T reaction in MeV, and how much energy do the neutron and the alpha particle each carry?

A A total of 3.5 MeV, with the neutron and alpha particle each carrying half B A total of 17.6 MeV, with the alpha particle carrying 14.1 MeV and the neutron carrying 3.5 MeV C A total of 17.6 MeV, with the neutron carrying 14.1 MeV (about 80 percent) and the alpha particle carrying 3.5 MeV (about 20 percent) D A total of 8.7 MeV, all carried away by the neutron

Q3. How does the reaction rate coefficient (the average of sigma v) differ from the reaction cross section sigma?

A The cross section sigma does not depend on temperature, whereas the reaction rate coefficient is a different quantity determined by temperature alone B The reaction rate coefficient multiplies the cross section sigma by the relative velocity v and averages over the plasma velocity distribution, giving the actual reaction rate R = n1 n2 (the average of sigma v) C The two are exactly the same quantity, just called by different names D The cross section sigma has the dimension of volume, and the reaction rate coefficient has the dimension of length

Q4. What does the Gamow peak refer to?

A A phenomenon in which reactions concentrate at exactly the energy that coincides with the thermal energy kB T B A sharp peak formed by the product of the Maxwell distribution (large at low energy) and the tunneling probability (large at high energy), representing the energy band of the particles that actually carry out the reactions C The position where the cross section is maximized, at an energy equal to the height of the barrier D A peak on the low-energy side whose position drops as the temperature rises

Q5. Why is the p-B11 reaction, despite being aneutronic, difficult to realize?

A Because the reaction produces large amounts of neutrons that strongly activate the reactor wall B Because boron's large charge makes the Gamow energy extremely large, requiring temperatures on the order of about 3 billion degrees, and furthermore bremsstrahlung losses increase in proportion to the square of Z, making it hard for the output to keep up with the losses C Because boron barely exists on Earth, so the fuel cannot be procured D Because the products carry no charge, so no energy can be extracted

Related Topics

• What Is Fusion
• Lawson Criterion
• Fuel Cycle