N-Z Graph and Decay Modes Revision Notes | A Level AQA Physics

1. Why Some Nuclei Are Stable and Others Are Not

Picture a nucleus as a crowd of protons and neutrons packed into an incredibly small space. Protons repel each other electrostatically, so something must hold them together — that something is the strong nuclear force. The balance between these two forces determines whether a nucleus is stable or whether it will decay.

The key insight is this: stability is not just about having the right number of protons. It is about having the right ratio of neutrons to protons. Too many or too few neutrons relative to protons, and the nucleus becomes unstable. The N-Z graph makes this relationship visible at a glance.

2. The N-Z Graph for Stable Nuclei

The N-Z graph plots neutron number N on the vertical axis against proton number Z on the horizontal axis. Every known stable nucleus appears as a point on this graph, and together those points form a narrow band called the stability band (sometimes called the belt of stability).

For light nuclei (low Z, roughly up to Z = 20), the stability band follows the line N = Z very closely. Helium-4 has 2 protons and 2 neutrons; carbon-12 has 6 of each. Equal numbers of protons and neutrons is sufficient for stability when the nucleus is small.

For heavier nuclei (Z above about 20), the stability band curves above the N = Z line. Lead-208, for example, has 82 protons but 126 neutrons — far more neutrons than protons. The extra neutrons dilute the proton-proton repulsion without adding to it, effectively acting as a nuclear glue. This is why the N/Z ratio increases with Z for stable nuclei.

There is an upper limit to stability: no nucleus with Z greater than 82 (lead) is permanently stable. Beyond this point, the electrostatic repulsion between the large number of protons becomes too great for the strong nuclear force to overcome, regardless of how many neutrons are present.

3. Unstable Nuclei: Position Relative to the Stability Band

Once you know where the stability band sits, you can immediately read off what is wrong with any unstable nucleus. A nucleus that plots above the stability band has too many neutrons for its number of protons — it is neutron-rich. A nucleus that plots below the band has too few neutrons — it is proton-rich. Each situation demands a different fix, and nature provides one through radioactive decay.

This table is the core logic of the whole topic. Every decay mode is essentially the nucleus correcting its N/Z ratio. Keep this framing in mind as you work through each mode — it turns memorisation into understanding.

4. Alpha Decay

Alpha decay is the characteristic decay of very heavy nuclei — those with Z greater than 82. The nucleus ejects an alpha particle, which is identical to a helium-4 nucleus: 2 protons and 2 neutrons, written $^{4}_{2}\text{He}$ or $^{4}_{2}\alpha$.

In one alpha decay, Z decreases by 2 and A decreases by 4. The daughter nucleus therefore moves down and to the left on the N-Z graph, closer to the stability band. For very heavy nuclei, this is the most effective way to shed both protons and neutrons simultaneously and reduce the overall size of the nucleus.

Worked example: Writing an Alpha Decay Equation

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

1. $^{238}_{92}\text{U} \rightarrow ^{A}_{Z}Y + ^{4}_{2}\alpha$ — Start with the parent nucleus on the left and the alpha particle on the right. Leave the daughter nucleus unknown for now.
2. $238 = A + 4 \Rightarrow A = 238 - 4 = 234$ — The total mass number must be the same on both sides. Subtract 4 from 238 to find the daughter's mass number.
3. $92 = Z + 2 \Rightarrow Z = 92 - 2 = 90$ — The total proton number must also be conserved. Subtract 2 from 92 to find the daughter's proton number.
4. $Z = 90$ corresponds to Thorium (Th) — Use the periodic table to match proton number 90 to the element name.
5. $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\alpha$ — Check: top numbers 238 = 234 + 4 ✓ and bottom numbers 92 = 90 + 2 ✓. Both conservation laws satisfied.

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

5. Beta-Minus Decay

Beta-minus ($\beta^-$) decay is the correction mechanism for neutron-rich nuclei — those sitting above the stability band. Inside the nucleus, a neutron transforms into a proton. The nucleus simultaneously emits a fast-moving electron (the beta-minus particle) and an antineutrino ($\bar{ u}_e$).

The net effect on the nucleus: Z increases by 1, A stays the same (because a neutron has become a proton — the total nucleon count is unchanged), and N decreases by 1. On the N-Z graph, the daughter nucleus moves one step down and one step to the right, shifting towards the stability band.

The antineutrino carries away some of the energy released, which is why beta particles are emitted with a continuous range of kinetic energies rather than a fixed value — the energy is shared between the electron and the antineutrino in varying proportions.

6. Beta-Plus Decay and Electron Capture

Both beta-plus decay and electron capture are the corrections for proton-rich nuclei — those sitting below the stability band. They achieve the same nuclear outcome by different routes: converting a proton into a neutron, which increases N and decreases Z by 1.

In beta-plus ($\beta^+$) decay, a proton inside the nucleus transforms into a neutron, emitting a positron ($^{0}_{+1}e$, the antiparticle of the electron) and a neutrino ($ u_e$). The positron is the beta-plus particle. On the N-Z graph, the daughter nucleus moves one step up and one step to the left — towards the stability band from below.

Electron capture is an alternative route to the same nuclear outcome. Here, the proton-rich nucleus absorbs one of its own inner-shell orbital electrons. That electron combines with a proton to produce a neutron and a neutrino. The nuclear equation looks like this:

The nuclear result of electron capture is identical to beta-plus decay: Z decreases by 1, N increases by 1, A stays the same. The daughter nucleus is the same in both cases. The difference is in what is emitted: beta-plus decay produces a positron, while electron capture produces only a neutrino (plus characteristic X-rays as outer electrons fill the vacancy left by the captured electron).

7. Comparing All Four Decay Modes

Now that all four modes are in place, it is worth comparing them side by side. This is exactly the kind of comparison an examiner might ask you to make, so having a clear mental map of the changes in Z, N, and A for each mode is essential.

Notice that beta-plus decay and electron capture produce identical changes to Z, N, and A. The distinction lies in the mechanism and the emitted particles, not the nuclear outcome. Both move the daughter nucleus one step up and to the left on the N-Z graph.

Worked example: Predicting Decay Mode from N-Z Position

Carbon-14 has Z = 6 and A = 14. Stable carbon isotopes have approximately N = 6 to N = 7 for Z = 6. Predict the decay mode of carbon-14 and write the nuclear equation.

1. $N = A - Z = 14 - 6 = 8$ — Carbon-14 has 8 neutrons and 6 protons.
2. Stable carbon has N ≈ 6. Carbon-14 has N = 8, which is above the stable value for Z = 6. — The nucleus sits above the stability band — it is neutron-rich.
3. Neutron-rich nucleus → beta-minus decay — Beta-minus decay converts a neutron into a proton, reducing N and increasing Z, moving the daughter towards the stability band.
4. $Z: 6 + 1 = 7$ (nitrogen); $A: 14$ (unchanged) — Z increases by 1 to give Z = 7 (nitrogen). A stays at 14.
5. $^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}e + \bar{ u}_e$ — Check top numbers: 14 = 14 + 0 ✓. Check bottom numbers: 6 = 7 + (-1) ✓. Both conservation laws satisfied.

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

8. Exam Technique and Common Mistakes

The N-Z graph and decay modes topic rewards students who think in terms of conservation laws and N/Z ratios rather than trying to memorise isolated facts. The following guidance targets the specific places where marks are most commonly dropped.

For a 'describe the N-Z graph' question, structure your answer in three parts: describe the axes, describe the position of the stability band for low-Z nuclei (N ≈ Z), and then describe how the band curves above the N = Z line for high-Z nuclei. Three clear statements typically secure three marks.

Every decay mode covered in this lesson — alpha, beta-minus, beta-plus, and electron capture — is the nucleus finding the most direct route back to the stability band. Keep that unifying idea at the centre of your revision, and the specific details of each mode will stay anchored to something meaningful rather than floating as isolated facts.