Alpha, beta and gamma decay

A rock that fogs a photographic plate

In 1896 Henri Becquerel left a uranium salt on a wrapped photographic plate in a drawer. When he developed the plate, it was dark. The salt had sent out rays that went through the paper. Marie and Pierre Curie then found two new elements, polonium and radium, that did the same thing much more strongly. We call the effect radioactivity. A nucleus that changes by itself and sends out a particle or a ray is a radioactive nucleus, and the change is a decay.

Rutherford named the three kinds of rays by the first letters of the Greek alphabet.

• Alpha particle ($\alpha$): a helium-4 nucleus, ${}^{4}_{2}\text{He}$, with two protons, two neutrons and charge $+2e$. It makes the most ionisation along its path, and it is stopped by a sheet of paper or a few centimetres of air.
• Beta particle ($\beta$): an electron ($\beta^-$) or a positron ($\beta^+$), which is a particle like an electron with the opposite charge. It is stopped by a few millimetres of aluminium.
• Gamma ray ($\gamma$): a photon of very high energy, with no charge and no mass. It goes through a lot of material, and thick lead is needed to stop it.

In an electric field the alpha particle bends towards the negative plate, but only a little, since it is heavy. The beta-minus particle bends the other way, much more, since it is light. The gamma ray does not bend at all.

Two rules for every decay

Write a nucleus as ${}^{A}_{Z}\text{X}$. In every nuclear change two things stay the same:

1. The total mass number. The number of nucleons is conserved.
2. The total charge. The sum of the atomic numbers is conserved, with the charge of an electron counted as $-1$ and of a positron as $+1$.

The nucleus that decays is the parent, and the nucleus it makes is the daughter. Rule 1 and rule 2 let you write any decay equation. NCERT, A-Level and IB all use this idea. A-Level and IB often write the electron as ${}^{0}_{-1}e$ or $\beta^-$, with the mass number $0$ and the charge $-1$.

Alpha decay

A heavy nucleus sends out an alpha particle:

The daughter has $4$ fewer nucleons and $2$ fewer protons. Radium-226 decays like this:

The mass numbers add up ($226 = 222 + 4$) and so do the atomic numbers ($88 = 86 + 2$). The energy released is called the Q value. It is the mass that disappears, times $c^2$. With atomic masses, $M$,

For radium-226 the masses are $226.025410\ \text{u}$, $222.017578\ \text{u}$ and $4.002603\ \text{u}$. The lost mass is $0.005229$ u, so $Q = 0.005229\ \text{u} \times 931.494\ \text{MeV/u} = 4.871\ \text{MeV}$.

Who gets this energy? The parent was at rest, so the two products fly apart with equal and opposite momenta. The lighter alpha particle takes the larger share of the kinetic energy: $K_\alpha = Q \times \dfrac{M_d}{M_d + m_\alpha}$, close to $Q\,\dfrac{A - 4}{A}$. For radium, $K_\alpha = 4.785$ MeV and the radon nucleus recoils with $0.086$ MeV. These are the numbers of the real decay, and the alpha particles that come from one kind of nucleus all have the same energy.

Beta decay

Inside the nucleus a neutron can change into a proton, or the other way round.

The particle $\nu$ is the neutrino and $\bar{\nu}$ is the antineutrino. They have no charge and almost no mass, and they hardly ever react with matter. The electron does not live inside the nucleus before the decay. It is made at the moment of the decay.

For the whole nucleus:

The mass number stays the same. The atomic number goes up by one in $\beta^-$ and down by one in $\beta^+$. A third kind, electron capture, has the nucleus take in one of its own atomic electrons: $p + e^- \to n + \nu$. Its result on $A$ and $Z$ is the same as $\beta^+$.

Why do we need the neutrino? If the nucleus sent out only an electron, the electron would always take the same energy, as the alpha particle does. Instead the electrons come with every energy from zero up to a largest value. Wolfgang Pauli suggested in 1930 that a third, unseen particle takes the rest. Energy, momentum and angular momentum all balance only with it.

The Q value in beta decay. Use atomic masses and the electrons look after themselves:

In $\beta^+$ decay there is an extra $2m_e$, which is $1.022$ MeV, because the daughter atom has one electron too many and a positron is made. Cobalt-60 turns into nickel-60 by $\beta^-$. The masses are $59.933817$ u and $59.930786$ u, so $Q = 0.003031\ \text{u} \times 931.494\ \text{MeV/u} = 2.823$ MeV. This energy is shared among the electron, the antineutrino and the gamma rays that follow. Nitrogen-13 ($13.005739$ u) turns into carbon-13 ($13.003355$ u) by $\beta^+$, and $Q = 0.002384\ \text{u} \times 931.494\ \text{MeV/u} - 1.022\ \text{MeV} = 1.199\ \text{MeV}$.

Gamma decay

After an alpha or a beta decay the daughter is often left in an excited state, a state with extra energy, just as an atom can be excited. It gets rid of the energy by sending out a photon. The nucleus changes in energy but not in $A$ or $Z$. We write an excited nucleus with a star, ${}^{60}_{28}\text{Ni}^{*}$. The photon energies are from a few keV to some MeV. Cobalt-60 gives the two gamma rays of $1.173$ and $1.332$ MeV, which are used in hospitals. The largest energy of the beta particle in this decay is $2.823 - 1.173 - 1.332 = 0.318$ MeV.

Which nuclei decay, and how?

Plot the neutron number $N$ against the proton number $Z$. The stable nuclei lie in a narrow band. A nucleus above the band has too many neutrons, and a $\beta^-$ decay turns one into a proton. A nucleus below it has too many protons, and $\beta^+$ decay or electron capture helps. Very heavy nuclei, with $Z$ above $82$, are too big, and they shed an alpha particle to become smaller.

Set $n_\alpha = 8$ and $n_\beta = 6$. Which nucleus do you reach? Now try $n_\alpha = 6$ and $n_\beta = 4$. What mass number do you reach?

Practise with new numbers

Try it in Python

The program follows the main path of the uranium series, one decay at a time. An alpha decay changes $(A, Z)$ by $(-4, -2)$. A beta-minus decay changes it by $(0, +1)$.

Common mistakes

• Changing the mass number in beta decay. An electron is not a nucleon. $A$ stays the same.
• Mixing up the sign of the atomic number change. $\beta^-$ raises $Z$ by $1$. $\beta^+$ lowers it by $1$.
• Saying the electron was in the nucleus. It is made in the decay, together with the antineutrino.
• Forgetting the neutrino. Without it the beta energy, the momentum and the spin do not balance.
• Leaving $2m_e$ out of the $\beta^+$ Q value. With atomic masses, a positron decay needs $1.022$ MeV more than the mass difference.
• Giving the alpha particle all the energy. The recoiling daughter takes a small share, and it is not zero.

Remember these

• Alpha: $A$ falls by $4$ and $Z$ by $2$. Beta-minus: $Z$ rises by $1$. Beta-plus: $Z$ falls by $1$. Gamma: $A$ and $Z$ stay.
• Conserve the total mass number and the total charge in every decay.
• $Q = \Delta m\,c^2$, and with atomic masses, $\beta^-$ needs $M_P - M_D$, and $\beta^+$ needs $M_P - M_D - 2m_e$.
• Alpha particles from one decay have one energy. Beta particles have a range of energies, because of the neutrino.
• The lighter product of a two-body decay takes the larger share of the kinetic energy.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Complete the equation: ${}^{234}_{90}\text{Th} \to {}^{A}_{Z}\text{Pa} + e^- + \bar{\nu}$. What are $A$ and $Z$?

2. A nucleus ${}^{A}_{Z}\text{X}$ emits one alpha particle and then two beta-minus particles. What nucleus is left, compared with X?

3. Why are the energies of the beta particles from one kind of nucleus spread over a range, but the energies of the alpha particles are not?

4. Is the decay ${}^{226}_{88}\text{Ra} \to {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}$ possible only if the mass of radium is larger than the two products together? Why?

References

• NCERT Physics Part II (Class 12, rationalised edition), Chapter 13 Nuclei, on radioactivity.
• OpenStax University Physics Volume 3, Chapter 10 Nuclear Physics, on radioactive decay.
• AME2020 atomic mass evaluation (Wang and others, Chinese Physics C, 2021), for the atomic masses.