Beta Decay Explained: Nuclear Equations and Emitted Particles
Learn how beta-minus decay changes a neutron into a proton, how to write nuclear equations, and why electrons and antineutrinos are emitted.
A strange thing happens when you look at some radioactive nuclei. A nucleus can contain too many neutrons, yet it does not simply throw out a neutron. Instead, one neutron can transform into a proton, creating a completely different element.
Imagine predicting what comes out of the nucleus. If a neutron disappears and a proton appears, what particle should be emitted to keep charge balanced? The nucleus started with no charge from that neutron, but it finishes with an extra positive charge from the proton. Something with a negative charge must leave.
That prediction leads us to beta-minus decay.
01The problem beta decay solves: an unstable neutron-to-proton balance
Inside a nucleus, protons and neutrons are held together by the strong nuclear force. A nucleus is more likely to be stable when it has a suitable balance between these two particles.
For light nuclei, the number of neutrons is often close to the number of protons. For heavier nuclei, extra neutrons help reduce the effect of proton-proton repulsion because neutrons add strong nuclear force attraction without adding electrical repulsion.
However, some nuclei have too many neutrons. A simple way to picture this is a crowded team where too many players are doing the same job. The team can become more balanced if one neutron changes into a proton.
The neutron transformation is:
\[ \text{neutron} \rightarrow \text{proton} + \text{electron} + \text{antineutrino} \]
The emitted electron is called a beta-minus particle, written as \( \beta^- \).
A common prediction is:
“An electron is tiny and exists outside atoms, so it probably cannot come from inside the nucleus.”
That feels reasonable because electrons are usually shown orbiting atoms. However, in beta decay, the electron is not an old atomic electron being released. It is created during the neutron-to-proton transformation inside the nucleus.
02Building the nuclear equation
A nuclear equation must conserve two important quantities:
- mass number, \(A\): the total number of protons and neutrons
- atomic number, \(Z\): the number of protons
The beta-minus decay equation is:
\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}\beta + {}^{0}_{0}\bar{\nu}_e \]
where:
- \(X\) is the original element
- \(Y\) is the new element
- \(A\) is the mass number
- \(Z\) is the atomic number
- \( \beta^- \) is the emitted electron
- \( \bar{\nu}_e \) is the electron antineutrino
Notice what changes:
- The mass number stays the same because one neutron becomes one proton. The total number of nucleons is unchanged.
- The atomic number increases by 1 because the nucleus gains a proton.
This means beta-minus decay creates a new element.
For example, carbon-14 undergoes beta decay:
\[ {}^{14}_{6}\ce{C} \rightarrow {}^{14}_{7}\ce{N} + {}^{0}_{-1}\beta + {}^{0}_{0}\bar{\nu}_e \]
Carbon has 6 protons. After decay, the nucleus has 7 protons, so it becomes nitrogen.
03Why is the antineutrino needed?
At first, the decay seems simple:
\[ \text{neutron} \rightarrow \text{proton} + \text{electron} \]
But experiments showed that the emitted electrons did not always have the same energy. Some energy and momentum appeared to be missing.
The solution was another particle: the electron antineutrino.
The antineutrino has:
- almost no mass,
- no electrical charge,
- energy and momentum.
It carries away the missing energy so that conservation laws still work.
This is a good example of physics working backwards. Scientists did not directly see the antineutrino at first. They inferred that something else had to exist because the measurements did not match the simpler prediction.
04A useful comparison: alpha decay versus beta decay
Beta decay is often confused with alpha decay. The key question is:
“Is the nucleus trying to reduce its size, or is it trying to change its neutron-proton balance?”
| Decay type | What leaves | Effect on mass number | Effect on atomic number |
|---|---|---|---|
| Alpha decay | \( {}^{4}_{2}\ce{He} \) nucleus | Decreases by 4 | Decreases by 2 |
| Beta-minus decay | Electron and antineutrino | No change | Increases by 1 |
You can compare beta decay with alpha decay in more detail here: Alpha decay: nuclear equations and radiation properties.
05The particle is not the same as an atomic electron
A common HSC mistake is writing beta-minus decay as if an electron was already inside the nucleus.
A student might write:
\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z}Y + e^- \]
and think, “The nucleus just released an electron.”
The problem is that this does not explain how the proton number changes.
The actual process is:
\[ n \rightarrow p + e^- + \bar{\nu}_e \]
The neutron changes identity. The electron is created in that process.
A useful analogy is a person changing teams. Imagine a person leaves the “neutron team” and joins the “proton team”. The team change creates a “receipt” showing the change happened. The electron is like that receipt: it leaves the system, but it was created during the change.
The analogy breaks down because particles are not people changing jobs. The neutron is not made of a proton, electron, and antineutrino sitting together waiting to separate. The transformation happens through the weak nuclear interaction.
06Worked example: identify the daughter nucleus
A nucleus of phosphorus-32 undergoes beta-minus decay. Write the nuclear equation.
Step 1
Beta-minus decay increases the atomic number by 1 but keeps the mass number unchanged.
Phosphorus has:
\[ A = 32, \qquad Z = 15 \]
Step 2
The new atomic number is:
\[ Z = 15 + 1 = 16 \]
Element 16 is sulfur.
Step 3
\[ {}^{32}_{15}\ce{P} \rightarrow {}^{32}_{16}\ce{S} + {}^{0}_{-1}\beta + {}^{0}_{0}\bar{\nu}_e \]
The result means phosphorus-32 becomes sulfur-32 because one neutron has changed into a proton.
07Worked example: finding the original nucleus
A beta-minus decay produces nitrogen-16:
\[ {}^{16}_{7}\ce{N} \rightarrow {}^{16}_{8}\ce{O} + {}^{0}_{-1}\beta + {}^{0}_{0}\bar{\nu}_e \]
A student detects the daughter nucleus as oxygen-16. What was the original nucleus?
Step 1
Beta-minus decay increases the atomic number by 1.
So the parent nucleus must have had:
\[ Z = 8 - 1 = 7 \]
Step 2
Atomic number 7 is nitrogen.
The mass number does not change, so:
\[ A = 16 \]
Step 3
The original nucleus was:
\[ {}^{16}_{7}\ce{N} \]
This means the decay process can be reversed in reasoning even though radioactive decay itself is not reversible.
08How beta decay connects to energy
The energy released in radioactive decay comes from the difference between the mass of the original nucleus and the masses of the products.
A small amount of mass is converted into energy:
\[ E = mc^2 \]
where:
- \(E\) is energy released in joules,
- \(m\) is the missing mass in kilograms,
- \(c\) is the speed of light in metres per second.
For nuclear reactions, the mass difference is tiny, but \(c^2\) is enormous, so the energy can be significant.
You can explore the connection between missing mass and nuclear energy here: Mass defect: how missing mass becomes nuclear binding energy.
The rate at which radioactive nuclei decay is a separate idea from the type of decay. Once you know what beta decay does to a nucleus, the next step is understanding how quickly a sample changes, which is described using decay constants and half-life: Decay constant and half-life: HSC Physics explained.
09Questions and solutions
Question 1
A nucleus of iodine-131 undergoes beta-minus decay. Determine the daughter nucleus.
Solution 1
The daughter nucleus is xenon-131.
Beta-minus decay keeps the mass number constant and increases the atomic number by 1.
Iodine has atomic number 53, so:
\[ Z = 53 + 1 = 54 \]
Atomic number 54 is xenon.
The nuclear equation is:
\[ {}^{131}_{53}\ce{I} \rightarrow {}^{131}_{54}\ce{Xe} + {}^{0}_{-1}\beta + {}^{0}_{0}\bar{\nu}_e \]
The common trap is changing the mass number. A beta particle has a mass number of zero because it is not a nucleon, and the total number of nucleons in the nucleus has not changed.
Question 2
A student claims:
“Because beta particles are electrons, beta-minus decay happens when an electron escapes from the atom.”
Explain why this statement is incorrect.
Solution 2
The statement is incorrect because the beta particle is created during a neutron-to-proton transformation inside the nucleus.
The process is:
\[ n \rightarrow p + e^- + \bar{\nu}_e \]
An atomic electron already exists outside the nucleus, but the emitted beta particle is produced by the weak interaction inside the nucleus.
The tempting idea is reasonable because beta particles have the same charge and mass properties as electrons. However, the key clue is that the atomic number increases by one. That can only happen if a neutron becomes a proton.
Question 3
A radioactive sample initially contains a nucleus that decays by beta-minus emission to form silicon-31.
Find the original nucleus and write the decay equation.
Solution 3
The original nucleus was aluminium-31.
Beta-minus decay increases atomic number by 1, so the parent must have one fewer proton.
Silicon has atomic number 14:
\[ Z_{\text{parent}} = 14 - 1 = 13 \]
Atomic number 13 is aluminium.
The mass number stays 31:
\[ {}^{31}_{13}\ce{Al} \rightarrow {}^{31}_{14}\ce{Si} + {}^{0}_{-1}\beta + {}^{0}_{0}\bar{\nu}_e \]
The trap is looking only at the emitted beta particle. The important information is how the nucleus changes before and after the emission.
Question 4
Two radioactive nuclei have the same mass number. Nucleus A has one more neutron than nucleus B. Nucleus A undergoes beta-minus decay.
Explain what happens to the neutron and proton numbers after the decay, and compare the new nucleus with nucleus B.
Solution 4
After beta-minus decay, nucleus A has one fewer neutron and one more proton, so it has the same neutron-proton balance as nucleus B.
A beta-minus decay changes:
\[ n \rightarrow p + e^- + \bar{\nu}_e \]
For nucleus A:
- neutron number decreases by 1,
- proton number increases by 1,
- total nucleon number stays constant.
Because A started with one extra neutron compared with B, the decay can make the daughter nucleus have the same neutron and proton counts as B.
The trap is thinking the nucleus simply loses a neutron. If that happened, the atomic number would not increase. The neutron changes identity instead.
Question 5
A scientist observes an unknown nucleus undergoing beta-minus decay. The daughter nucleus is stable. The scientist notices that the daughter has fewer neutrons than the original nucleus, but the same mass number.
Another scientist argues:
“The original nucleus must have lost a neutron, so the daughter should be a different isotope of the same element.”
Decide which explanation is correct and justify your answer using conservation laws.
Solution 5
The first explanation is correct: the neutron changed into a proton, not simply left the nucleus.
In beta-minus decay:
\[ n \rightarrow p + e^- + \bar{\nu}_e \]
The mass number remains unchanged because the total number of nucleons is unchanged. However, the neutron count decreases by one while the proton count increases by one.
The second scientist’s explanation is tempting because the daughter does have fewer neutrons. If a neutron were simply emitted, the daughter would keep the same atomic number and become a different isotope. However, beta-minus decay is identified by the emitted electron and the increase in proton number.
The decisive constraint is charge conservation. A neutron becoming a proton creates one extra positive charge in the nucleus, so the emitted electron carries away one negative charge. The element must change.
This is why beta decay is not a process of the nucleus getting smaller. It is a process of the nucleus changing its identity. Understanding that idea makes nuclear equations much easier because you are tracking what changes, rather than memorising a symbol.