Alpha and Beta Decay Revision Notes | A Level AQA Physics

1. Why Do Nuclei Decay?

Picture a nucleus packed with protons and neutrons. The strong nuclear force holds them together, but it only acts over very short ranges — roughly $10^{-15}$ m. When a nucleus has too many protons, or an imbalance between protons and neutrons, the electrostatic repulsion between protons can overcome the strong force. The nucleus becomes unstable and must shed energy or particles to reach a more stable configuration.

There are two main decay modes at A Level: alpha decay and beta-minus decay. Which one a nucleus undergoes depends on the nature of its instability. Nuclei that are too heavy (high $A$) tend to alpha-decay; nuclei with too many neutrons relative to protons tend to beta-minus decay. Both processes conserve nucleon number and proton number — and that conservation law is the key tool you will use to write every nuclear equation.

Before writing any decay equation, you need two conservation rules locked in: the total nucleon number $A$ must be the same on both sides, and the total proton number $Z$ must be the same on both sides. Think of these as a double-entry bookkeeping system — every particle you emit must be accounted for in both columns.

2. Alpha Decay

An alpha particle is identical to a helium-4 nucleus: two protons and two neutrons, written $^{4}_{2}\text{He}$. When a heavy nucleus emits an alpha particle, it loses 4 from its nucleon number and 2 from its proton number. The daughter nucleus is a completely different element — two places to the left in the periodic table.

Notice that on the left you have $A$ nucleons and $Z$ protons. On the right, the daughter has $A-4$ nucleons and $Z-2$ protons, while the alpha particle contributes 4 nucleons and 2 protons. Adding those up: $(A-4)+4 = A$ and $(Z-2)+2 = Z$. Both conservation laws are satisfied — this is the check you should always perform.

Worked example: Writing an Alpha Decay Equation: Uranium-238

Uranium-238 ($^{238}_{92}\text{U}$) undergoes alpha decay. Write the full nuclear equation and identify the daughter nucleus.

1. $^{238}_{92}\text{U} \rightarrow \text{?} + ^{4}_{2}\text{He}$ — Start with the parent on the left and the alpha particle on the right. Leave the daughter as unknown.
2. $A_{\text{daughter}} = 238 - 4 = 234$ — Nucleon number is conserved: subtract 4 (the alpha particle's nucleon number) from 238.
3. $Z_{\text{daughter}} = 92 - 2 = 90$ — Proton number is conserved: subtract 2 (the alpha particle's proton number) from 92.
4. $Z = 90 \Rightarrow \text{Thorium (Th)}$ — Use the periodic table. Proton number 90 is thorium.
5. $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$ — Check: nucleon numbers 234 + 4 = 238 ✓. Proton numbers 90 + 2 = 92 ✓. Both conservation laws satisfied.

Answer: $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$

Alpha decay is most common in heavy nuclei (typically $A > 200$) where the electrostatic repulsion between the large number of protons is hardest for the strong force to overcome. Emitting an alpha particle reduces both the charge and the mass of the nucleus significantly, moving it towards a more stable region of the nuclear chart.

3. Beta-Minus Decay and the Neutrino

Beta-minus decay occurs in nuclei that have too many neutrons relative to protons. Here is what happens at the level of the nucleus: a neutron converts into a proton, and in doing so emits a fast electron — the beta-minus particle ($\beta^{-}$) — along with an electron antineutrino ($\bar{ u}_{e}$). The proton stays in the nucleus, so the proton number increases by 1 while the nucleon number stays the same.

Check the conservation laws here. Nucleon number: $A + 0 + 0 = A$ ✓. Proton number: $(Z+1) + (-1) + 0 = Z$ ✓. The antineutrino carries zero nucleon number and zero charge, so it does not disturb either conservation law — but its role in energy conservation is the most important part of this story.

3a. Why the Neutrino Had to Exist

Here is where the physics gets genuinely interesting. In alpha decay, every alpha particle emitted from a given isotope has the same kinetic energy. That is exactly what you would expect if only two particles are produced — conservation of energy and momentum fix the energy of each particle precisely. But in beta decay, the emitted electrons have a continuous range of kinetic energies, from near zero up to a maximum value.

This was deeply troubling to physicists in the 1920s and 1930s. If only the daughter nucleus and the beta particle were produced, energy conservation demands a single fixed electron energy — yet experiments consistently showed a spectrum. Either energy was not conserved in beta decay (which would overturn a fundamental law of physics), or something else was carrying away the missing energy unseen.

In 1930, Wolfgang Pauli proposed a bold solution: a third particle must be emitted alongside the beta particle, one that is electrically neutral and has negligible mass, making it almost impossible to detect directly. This particle — later named the neutrino by Fermi — carries away the remaining energy and momentum, restoring conservation laws. The total energy released in each decay is fixed, but it is shared between the beta particle and the antineutrino in varying proportions, producing the observed continuous spectrum.

The chart above shows the characteristic shape of a beta energy spectrum. The x-axis represents the kinetic energy of the emitted beta particle as a fraction of the maximum possible energy $E_{\text{max}}$. At $E_{\text{max}}$, the antineutrino carries away zero kinetic energy; at lower beta energies, the antineutrino carries the remainder. The continuous, peaked distribution — rather than a sharp line — is the experimental signature that demanded the neutrino's existence.

Worked example: Writing a Beta-Minus Decay Equation: Carbon-14

Carbon-14 ($^{14}_{6}\text{C}$) undergoes beta-minus decay. Write the full nuclear equation, including the antineutrino.

1. $^{14}_{6}\text{C} \rightarrow \text{?} + ^{0}_{-1}e + \bar{ u}_{e}$ — In beta-minus decay, a beta particle and an electron antineutrino are always emitted. Leave the daughter nucleus unknown.
2. $A_{\text{daughter}} = 14 - 0 = 14$ — The beta particle has nucleon number 0 and the antineutrino has nucleon number 0. Nucleon number is conserved, so the daughter also has $A = 14$.
3. $Z_{\text{daughter}} = 6 - (-1) = 6 + 1 = 7$ — Proton number is conserved. The beta particle has charge number $-1$, so the daughter must have $Z = 7$ to balance: $7 + (-1) = 6$.
4. $Z = 7 \Rightarrow \text{Nitrogen (N)}$ — Proton number 7 is nitrogen.
5. $^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}e + \bar{ u}_{e}$ — Check nucleon numbers: 14 = 14 + 0 + 0 ✓. Check proton numbers: 6 = 7 + (−1) + 0 ✓. Both conservation laws satisfied.

Answer: $^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}e + \bar{ u}_{e}$

4. Comparing Alpha and Beta-Minus Decay

Now that both decay modes are established, it is worth placing them side by side. The differences in nuclear effect, emitted particles, and the role of conservation laws are all examinable — and examiners frequently ask you to compare rather than describe one in isolation.

One pattern worth noting: in alpha decay the neutron number also decreases by 2, because the alpha particle contains two neutrons. In beta-minus decay, a neutron converts into a proton, so the neutron number decreases by 1 while the proton number increases by 1 — the nucleon number stays fixed throughout. This is a clean way to remember why the two decays affect the periodic table position of the daughter differently.

5. Exam Technique and Common Mistakes

Nuclear equations and the neutrino argument are both high-frequency exam targets. The marks are very accessible if you apply the conservation laws methodically — but they are easy to drop if you rush. The guidance below consolidates the most common errors seen in mark schemes and examiners' reports.

Across this lesson, every spec requirement connects back to the same two pillars: conservation of nucleon number and conservation of proton number govern the nuclear equations for both alpha and beta-minus decay, while conservation of energy — applied to the continuous beta spectrum — is the physical argument that demanded the neutrino's existence. Bring both pillars into any extended-answer response and you will cover the marks that matter.