Types Of Radioactive Decay: Alpha, Beta, Gamma

​Radioactive​ is a word that's not that well understood. Swamped in fear and inherently seeming alien and dangerous, the nature of radioactive decay is something that is worth learning about whether you're a physics student or just an interested layman.

The reality is that radioactivity essentially describes nuclear reactions that lead to a change in the atomic number of an element and/or a release of gamma radiation. It is dangerous in large amounts because the radiation released is "ionizing" (i.e., it has enough energy to strip electrons from atoms) but it's an interesting physical phenomenon and in practice, most people will never be around radioactive materials enough to be at risk.

Nuclei can achieve a lower energy state by fusion – which is when two nuclei fuse together to create a heavier nucleus, releasing energy in the process – or by fission, which is the splitting of heavy elements into lighter ones. Fission is the source of the energy in nuclear reactors, and also in nuclear weapons, and this in particular is what most people picture when they think about radioactivity. But most of the time, when nuclei change to a lower energy state in nature, it's down to radioactive decay.

There are three types of radioactive decay: alpha decay, beta decay and gamma decay, although beta decay in itself comes in three different types. Learning about these forms of nuclear decay is a crucial part of any nuclear physics course.

Alpha decay occurs when a nucleus emits what's called an "alpha particle" (α-particle). An alpha particle is a combination of two protons and two neutrons, which if you know your periodic table you'll recognize as a helium nucleus.

The process is fairly easy to understand in terms of the mass and properties of the resulting atom: It loses four from its mass number (two from the protons and two from the electrons) and two from its atomic number (from the two protons lost). This means that the original atom (i.e., the "parent" nucleus) becomes a different element (based on the "daughter" nucleus) after undergoing alpha decay.

When computing the energy released in alpha decay, you need to subtract the mass of the helium nucleus and the daughter atom from the mass of the parent atom, and convert this into a value of energy using Einstein's famous equation ​E​ = ​mc​². It's usually easier to perform this calculation if you work in atomic mass units (amu) and multiply the missing mass by the factor ​c​² = 931.494 MeV / amu. This returns a value of energy in MeV (i.e., mega electronvolts), with an electronvolt being equal to 1.602 × 10⁻⁹ joules and generally a more convenient unit for working in energies at the atomic scale.

Since beta decay has three different varieties, it's helpful to learn about each one in turn, although there are a lot of similarities between them. Beta-plus decay is when a proton turns into a neutron, with the release of a beta-plus particle (i.e., a β+ particle) along with an uncharged, near-massless particle called a neutrino. As a result of this process, the daughter atom will have one less proton and one more neutron than the parent atom, but the same overall mass number.

The beta-plus particle is actually called a positron, which is the antimatter particle corresponding the electron. It has a positive charge of the same size as the negative charge on the electron, and the same mass as an electron. The neutrino released is technically called an electron neutrino. Notice that one particle of regular matter and one particle of antimatter are released in this process.

Calculating the energy released in this decay process is a little more complicated than for other forms of decay, because the mass of the parent atom will include the mass of one more electron than the daughter atom's mass. On top of this, you also have to subtract the mass of the β+ particle that is emitted in the process. Essentially, you have to subtract the mass of the daughter particle and ​two​ electrons from the mass of the parent particle, and then convert to energy as before. The neutrino is so tiny that it can be safely neglected.

Beta-minus decay is essentially the opposite process of beta-plus decay, where a neutron turns into a proton, releasing a beta-minus particle (a β− particle) and an electron antineutrino in the process. Because of this process, the daughter atom will have one less neutron and one more proton than the parent atom.

The β− particle is actually an electron, but it has a different name in this context because when the beta emission for the decay was first discovered, nobody knew what the particle actually was. Additionally, calling them beta particles is useful because it reminds you that it comes from the beta decay process, and it can be useful when you're trying to remember what happens in each – the positive beta particle is released in beta-plus decay and the negative beta particle is released in beta-minus decay. In this case, though, the neutrino is an antimatter particle, but again, one antimatter and one regular matter particle are released in the process.

Calculating the energy released in this type of beta decay is a little simpler, because the extra electron possessed by the daughter atom cancels out with the electron lost in the beta emission. This means that to calculate ∆​m​, you simply subtract the mass of the daughter atom from that of the parent atom and then multiply by the speed of light squared (​c​²), as before, expressed in mega electronvolts per atomic mass unit.

The last type of beta decay is quite different from the first two. In electron capture, a proton "absorbs" an electron and turns into a neutron, with the release of an electron neutrino. This therefore reduces the atomic number (i.e., the number of protons) by one and increases the number of neutrons by one.

This might seem like it violates the pattern so far, with one matter and one antimatter particle being emitted, but it gives a hint at the actual reason for this balance. The "lepton number" (which you can think of as an "electron family" number) is conserved, and an electron or electron neutrino has a lepton number of 1, while the positron or electron antineutrino has a lepton number of −1.

You should be able to see that all of the other processes fulfill this easily. For electron capture, the lepton number decreases by 1 when the electron is captured, so to balance this, a particle with a lepton number of 1 has to be emitted.

Calculating the energy released in electron capture is pretty simple: Because the electron comes from the parent atom, you don't need to worry about accounting for the difference in the number of electrons between the parent and daughter atoms. You find ∆​m​ by simply subtracting the mass of the daughter atom from that of the parent atom. The expression for the process will generally be written with the electron on the left hand side, but the simple rule reminds you that this is actually part of the parent atom in terms of the mass.

Gamma decay involves the emission of a high-energy photon (electromagnetic radiation), but the number of protons and neutrons in the atom doesn't change as a result of the process. It's analogous to the emission of a photon when an electron transitions from a higher energy state to a lower energy state, but the transition in this case takes place in the nucleus of the atom.

Just like in the analogous situation, the transition from a higher energy state to a lower energy state is balanced out by the emission of a photon. These have energies over 10 keV and are generally called gamma rays, although the definition isn't really strict (the energy range overlaps with X-rays, for instance).

Alpha or beta emission can leave a nucleus in a higher-energy, excited state, and the energy released as a result of these processes is done in the form of gamma rays. However, the nucleus can also end up in a higher-energy state after colliding with another nucleus or being struck by a neutron. The result in all cases is the same: The nucleus drops from its excited state into a lower energy state and releases gamma rays in the process.

Uranium-238 decays into thorium-234 with the release of an alpha particle (i.e., a helium nucleus), and this is one of the most well-known examples of radioactive decay. The process can be represented as:

\(^{238}\text{U} \to \;^{234}\text{Th} + \;^4\text{He}\)

In order to calculate how much energy is released in this process, you'll need the atomic masses: ²³⁸U = 238.05079 amu, ²³⁴Th = 234.04363 amu and ⁴He = 4.00260 amu, with all of the masses expressed in atomic mass units. Now to work out how much energy is released in the process, all you need to do is find ∆​m​ by subtracting the masses of the products from the mass of the original parent atom, and then calculate the amount of energy this represents.

\(\begin{aligned}
∆m &= \text{(mass of parent)}- \text{(mass of products)} \
&= 238.05079 \text{ amu} – 234.04363 \text{ amu} – 4.00260 \text{ amu} \
&= 0.00456 \text{ amu}\)
\(E &= ∆mc^2 \
&= 0.00456 \text{ amu} × 931.494 \text{ MeV / amu} \
&= 4.25 \text{ MeV}
\end{aligned}\)

Radioactive decay often happens in chains, with multiple steps between the starting point and the final point. These decay chains are long and would require many steps to calculate how much energy is released in the whole process, but taking a piece of one such chain illustrates the approach.

If you look at the decay chain of thorium-232, close to the end of the chain, an unstable nucleus (i.e., an atom of an unstable isotope, with a short half-life) of bismuth-212 undergoes beta-minus decay into polonium-212, which then undergoes alpha decay into lead-208, a stable isotope. You can calculate the energy released in this process by taking it step by step.

First, the beta-minus decay from bismuth-212 (​m​ = 211.99129 amu) into polonium-212 (​m​ = 211.98887 amu) gives:

\(\begin{aligned}
∆m &= \text{(mass of parent)} -\text{(mass of daughter)} \
&= 211.99129 \text{ amu} – 211.98887 \text{ amu} \
&= 0.00242 \text{ amu}
\end{aligned}\)

Remembering that the change in electron numbers cancels out in beta-minus decay. That releases:

\(\begin{aligned}
E &= ∆mc^2 \
&= 0.00242 \text{ amu} × 931.494 \text{ MeV / amu} \
&= 2.25 \text{ MeV}
\end{aligned}\)

The next stage is the alpha decay from polonium-212 to lead-208 (​m​ = 207.97665 amu) and one helium nucleus.

\(\begin{aligned}
∆m &= \text{(mass of parent)} -\text{(mass of products)} \
&= 211.98887\text{ amu} – 207.97665\text{ amu}- 4.00260\text{ amu} \
&= 0.00962\text{ amu}
\end{aligned}\)

And the energy is:

\(\begin{aligned}
E &= ∆mc^2 \
&= 0.00962 \text{ amu} × 931.494 \text{ MeV / amu} \
&= 8.96 \text{ MeV}
\end{aligned}\)

In total then, there is 2.25 MeV + 8.96 MeV = 11.21 MeV of energy released in the process. Of course, if you're careful (including the alpha particle, and additional electrons if your process includes a beta-plus decay) you can calculate the difference in mass in a single step and then convert, but this approach tells you the energy released at each stage.