Stable and unstable nuclei — AQA A-Level Physics
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Stable nuclei have a balance of protons and neutrons that keeps the strong nuclear force dominant over electrostatic repulsion, while unstable nuclei have an excess of protons or neutrons and undergo radioactive decay to reach stability. In AQA A-Level Physics, stability is explained using the N-Z graph, the strong nuclear force, and the ratio of neutrons to protons, with alpha, beta, and gamma emissions changing the nucleus to a more stable configuration.
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Stable and unstable nuclei explained
Stable and unstable nuclei covers 4 specification points, including The strong nuclear force; its role in keeping… and Unstable nuclei; alpha and beta decay..
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The strong nuclear force; its role in keeping the nucleus stable; short-range attraction up to approximately 3 fm, very-short range repulsion closer than approximately 0.5 fm.
The strong nuclear force acts between nucleons and holds the nucleus together despite electrostatic repulsion between protons. It is attractive between nucleons separated by up to about 3 fm, which stabilises nuclei by binding protons and neutrons; at separations below about 0.5 fm it becomes repulsive, preventing nucleons from collapsing into each other and limiting nuclear density. It is much stronger than the electrostatic force at these ranges but acts only over short distances, so it saturates: a nucleon interacts only with near neighbours. For example, in helium-4 the two protons repel electrically, yet the strong force between all nucleon pairs keeps the nucleus bound. In larger nuclei, the short range means the strong force cannot easily stabilise the whole nucleus, contributing to instability.
Unstable nuclei; alpha and beta decay.
Unstable nuclei have an imbalance of protons and neutrons, or too much energy, and decay randomly to become more stable. Alpha decay emits an alpha particle (2 protons and 2 neutrons, a helium nucleus), reducing proton number by 2 and nucleon number by 4; it usually occurs in heavy nuclei where electrostatic repulsion is large. Beta-minus decay converts a neutron into a proton, emitting an electron and an antineutrino, increasing proton number by 1 while nucleon number stays the same; beta-plus decay converts a proton into a neutron, emitting a positron and a neutrino, decreasing proton number by 1. For example, carbon-14 beta-minus decays to nitrogen-14. Each decay is spontaneous and random, and the nucleus often emits gamma radiation afterwards as it loses excess energy.
− Equations for alpha decay, β decay including the need for the neutrino.
Nuclear decay equations track changes in proton and nucleon numbers. In alpha decay, a nucleus emits an alpha particle (a helium nucleus, 2 protons and 2 neutrons), so its proton number decreases by 2 and its nucleon number by 4. For example, uranium-238 becomes thorium-234: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. In beta-minus decay, a neutron converts to a proton, emitting an electron and an antineutrino: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄ₑ. The antineutrino is needed because the electron's kinetic energy varies continuously; without a third particle, energy and momentum would not be conserved. Beta-plus decay emits a positron and neutrino. Balancing proton and nucleon numbers on both sides confirms the equation.
The existence of the neutrino was hypothesised to account for conservation of energy in beta decay.
In beta-minus decay, a neutron becomes a proton and an electron is emitted. If only the electron were emitted, its kinetic energy would have a single fixed value because the energy released is fixed. In practice, beta particles have a continuous range of kinetic energies up to a maximum. This apparent loss of energy suggested a third, unseen particle was emitted. Pauli hypothesised the neutrino to carry away the missing energy and momentum. Later the antineutrino was detected, confirming the idea. The same reasoning applies to beta-plus decay, where a neutrino is emitted. This shows how conservation laws drive the discovery of new particles.
Your focus
- Describe the variation of the strong nuclear force with separation between nucleons, including the approximate distances for attraction and repulsion.
- Explain how the strong nuclear force keeps a nucleus stable despite electrostatic repulsion between protons.
- Apply the short range of the strong nuclear force to account for the stability or instability of nuclei of different sizes.
Show all 12 objectives
- Describe alpha and beta decay in terms of changes to the nucleus and emitted particles.
- Write and balance nuclear decay equations using conservation of nucleon number and charge.
- Explain how alpha or beta decay changes the proton–neutron ratio and moves a nucleus towards stability.
- Write balanced nuclear equations for alpha, beta-minus and beta-plus decay.
- Explain why a neutrino or antineutrino must be emitted in beta decay.
- Apply conservation of nucleon number and charge to check decay equations.
- Explain how the continuous beta energy spectrum provides evidence for the neutrino.
- Describe how conservation of energy and momentum led to the neutrino hypothesis.
- Distinguish between neutrino and antineutrino emission in beta decay.
Which command word is this?
In AQA A-Level Physics, 'explain' requires a detailed justification with causal links. You must state the relevant physics principle (e.g., balance of strong nuclear force and electrostatic force) and apply it to the specific nucleus to show why it is unstable or why a particular decay occurs. Simply stating a fact without reasoning will not gain full marks.
Stable and unstable nuclei exam tips
Topic Overview
This topic explores why some nuclei are stable while others undergo radioactive decay. You will learn how the interplay between the strong nuclear force and electrostatic repulsion determines stability, and how the neutron-to-proton ratio changes across the N-Z graph. Understanding stability is crucial for explaining alpha, beta, and gamma emissions and for predicting the products of decay.
Stable and unstable nuclei form the foundation for later topics such as radioactive decay, nuclear energy, and particle physics. It also connects to the fundamental forces and conservation laws in AQA A-Level Physics. Mastery of this topic is essential for tackling exam questions on decay equations, the N-Z graph, and the evidence for neutrinos.
Key Concepts
- →The strong nuclear force is attractive between all nucleons, acts over a very short range (about 3 fm), and is much stronger than the electrostatic force at short distances.
- →Electrostatic repulsion acts between protons and has a longer range than the strong force, so in large nuclei the repulsion can overcome the strong force, leading to instability.
- →The N-Z graph shows stable nuclei clustered around a line where N ≈ Z for light nuclei and N > Z for heavier nuclei; unstable nuclei lie above or below this band.
- →Alpha decay reduces both proton and neutron numbers by 2, moving the nucleus diagonally towards the stability band; beta-minus decay converts a neutron to a proton, moving it up and left; beta-plus decay converts a proton to a neutron, moving it down and right.
- →Gamma emission occurs when a nucleus is in an excited state and releases energy without changing its proton or neutron numbers.
Marking Points
- States that the strong nuclear force acts between nucleons (protons and neutrons) and is attractive at separations up to approximately 3 fm.
- Explains that this attraction counteracts the electrostatic repulsion between protons and binds the nucleus, keeping it stable.
- States that at separations closer than approximately 0.5 fm the strong force becomes repulsive, preventing nucleon overlap and limiting nuclear density.
- Notes that the strong force is short-range and much stronger than the electrostatic force at nucleon separations, but does not act beyond a few femtometres.
- Applies the range idea to explain why the strong force saturates and why very large nuclei can be unstable.
- Defines an unstable nucleus as one with an imbalance of protons and neutrons or excess energy that leads to spontaneous decay.
- Describes alpha decay as emission of an alpha particle (2 protons and 2 neutrons), decreasing proton number by 2 and nucleon number by 4.
- Describes beta-minus decay as a neutron changing into a proton with emission of an electron and an antineutrino, increasing proton number by 1.
- Describes beta-plus decay as a proton changing into a neutron with emission of a positron and a neutrino, decreasing proton number by 1.
- Applies conservation of nucleon number and charge to write or check nuclear decay equations.
- Alpha decay: nucleon number decreases by 4 and proton number decreases by 2, with an alpha particle (⁴₂He) emitted.
- Beta-minus decay: a neutron changes into a proton, so proton number increases by 1 while nucleon number stays the same, with an electron (⁰₋₁e) emitted.
- Beta-plus decay: proton number decreases by 1 and a positron (⁰₊₁e) is emitted.
- The neutrino or antineutrino carries away energy and momentum, explaining the continuous beta energy spectrum.
- Equations must balance nucleon number and proton number (or charge) on both sides.
- Correct symbols and notation, such as ⁴₂He, ⁰₋₁e, ν̄ₑ and νₑ, should be used.
- Beta decay emits electrons with a continuous range of kinetic energies, not a single fixed value.
- Conservation of energy requires the missing energy to be carried by another particle.
- The neutrino was hypothesised as a neutral, nearly massless particle to account for this missing energy.
- Conservation of momentum also requires a third particle in beta decay.
- The antineutrino is emitted in beta-minus decay and the neutrino in beta-plus decay.
- Experimental detection of the neutrino later supported the hypothesis.
Examiner Tips
- 💡Sketch a force–separation graph with repulsion below 0.5 fm, attraction up to about 3 fm and zero beyond, labelling the axes with force and separation in fm.
- 💡Use the phrase 'short-range attraction up to approximately 3 fm' and 'very-short range repulsion closer than approximately 0.5 fm' when explaining stability.
- 💡Link the range to nuclear size: compare the diameter of a nucleus with 3 fm to justify why the strong force acts only between neighbouring nucleons.
- 💡Balance nuclear equations by checking that the total nucleon number and total charge are the same on both sides.
- 💡Use the notation for alpha, beta-minus and beta-plus particles correctly, including the antineutrino or neutrino, and state that decay is spontaneous and random.
- 💡When explaining stability, link the type of decay to whether the nucleus has too many neutrons, too many protons or is simply too heavy.
- 💡Write the full nuclear equation and check both nucleon and proton numbers balance before moving on.
- 💡State clearly that the neutrino is needed to account for conservation of energy and momentum in beta decay.
- 💡Use correct symbols for alpha, beta-minus, beta-plus, neutrino and antineutrino to avoid losing communication marks.
- 💡Link the continuous beta energy spectrum directly to the need for the neutrino.
- 💡Mention both energy and momentum conservation when explaining the hypothesis.
- 💡Use the correct particle names and symbols for neutrino and antineutrino in your answer.
- 💡When explaining stability, always refer to both the strong nuclear force and the electrostatic force, and mention their relative strengths and ranges. Vague statements like 'the nucleus is too big' will not gain marks.
- 💡For decay equations, check that both nucleon number and proton number are conserved. Write the element symbol and atomic numbers clearly to avoid careless errors.
- 💡In questions about the N-Z graph, sketch or describe the position of the nucleus relative to the stability band and state the direction of movement after each type of decay. This shows clear understanding.
Common Mistakes
- Treating the strong force as purely attractive at all separations; the correction is that it is repulsive below about 0.5 fm.
- Confusing the strong force with the electrostatic force or claiming it acts on electrons; the correction is that it acts between nucleons only.
- Assuming the strong force has infinite range like gravity or electrostatics; the correction is that it is short-range, with attraction up to about 3 fm.
- Writing alpha decay as changing proton number by 4 and nucleon number by 2; the correction is proton number decreases by 2 and nucleon number decreases by 4.
- Forgetting the antineutrino in beta-minus decay or the neutrino in beta-plus decay; the correction is to include the appropriate neutrino or antineutrino to conserve energy and momentum.
- Treating beta particles as orbiting electrons; the correction is that beta particles are emitted from the nucleus when a neutron or proton changes type.
- Writing beta decay as if the electron came from the nucleus without changing a neutron; the correction is that a neutron transforms into a proton, increasing proton number by 1.
- Omitting the antineutrino in beta-minus decay; the correction is to include it to conserve energy and momentum.
- Forgetting to balance nucleon numbers in alpha decay; the correction is to check that the total nucleon number is the same on both sides.
- Thinking the neutrino is charged; the correction is that it is neutral, which is why it is hard to detect.
- Believing energy is not conserved in beta decay; the correction is that the neutrino carries away the missing energy.
- Confusing neutrino and antineutrino; the correction is that beta-minus decay emits an antineutrino while beta-plus decay emits a neutrino.
- Students often think that unstable nuclei are 'radioactive' because they have too much energy, but the real reason is an imbalance of forces within the nucleus. The decay process is a way to rearrange nucleons to achieve a more stable force balance.
- Many students believe that beta particles come from the electron shells. In fact, beta-minus particles are created in the nucleus when a neutron decays into a proton and an electron (plus an antineutrino).
- Some students think that gamma decay changes the element. Gamma emission only releases excess energy from an excited nucleus; it does not change the proton or neutron numbers, so the element remains the same.
Revision Plan
- 1Day 1-2: Review atomic structure and the fundamental forces. Learn the definitions of atomic number, mass number, and isotopes, and understand the range and strength of the strong nuclear force.
- 2Day 3-4: Study the N-Z graph. Plot stable nuclei and identify the stability band. Learn how alpha, beta-minus, beta-plus, and gamma decay move a nucleus on the graph.
- 3Day 5-6: Practice writing nuclear decay equations. Ensure you can balance both nucleon number and proton number, and identify the daughter nuclide.
- 4Day 7-8: Learn the evidence for the neutrino from the beta decay energy spectrum. Understand why the continuous spectrum implies a third particle.
- 5Day 9-10: Complete past paper questions on stable and unstable nuclei. Focus on explanations of stability, decay equations, and graph interpretation. Review mark schemes to refine your answers.
Exam Question Types
- 📋Multiple choice questions on identifying the correct decay equation or the change in N and Z after a given decay. Advice: check conservation of A and Z carefully, and remember that gamma decay does not change A or Z.
- 📋Structured questions asking you to explain why a particular nucleus is unstable in terms of forces. Advice: use precise terminology about the strong nuclear force and electrostatic repulsion, and refer to the N:Z ratio.
- 📋Graph interpretation questions where you must plot or identify a nucleus on the N-Z graph and describe its decay path. Advice: remember that alpha decay moves diagonally down-left, beta-minus moves up-left, and beta-plus moves down-right.
- 📋Questions on the beta decay energy spectrum and the neutrino. Advice: explain that the continuous spectrum is due to energy sharing with the antineutrino, and that conservation of energy and momentum require a third particle.
Command Word Expectations (AQA)
In AQA A-Level Physics, 'explain' requires a detailed justification with causal links. You must state the relevant physics principle (e.g., balance of strong nuclear force and electrostatic force) and apply it to the specific nucleus to show why it is unstable or why a particular decay occurs. Simply stating a fact without reasoning will not gain full marks.
'Describe' asks for a detailed account of the process or pattern without necessarily giving reasons. For example, describing the changes in proton and neutron numbers during alpha decay, or describing the shape of the N-Z graph. Be precise with numbers and directions.
'Calculate' requires you to use given data and equations to find a numerical answer. Show your working clearly, include units, and round appropriately. In nuclear decay, you may need to calculate the number of neutrons or the energy released using E=mc^2.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A nucleus of uranium-238 (Z=92) undergoes alpha decay. Write the nuclear equation and state the changes in proton and neutron numbers.
- 1.Step 1: Identify the parent nucleus: U-238 has 92 protons and 238 - 92 = 146 neutrons.
- 2.Step 2: An alpha particle is a helium nucleus, 4He2, so it has 2 protons and 2 neutrons.
- 3.Step 3: Apply conservation of nucleon number and proton number: the daughter nucleus will have A = 238 - 4 = 234 and Z = 92 - 2 = 90.
- 4.Step 4: Identify the daughter element: Z = 90 is thorium (Th). Write the equation: 238U92 -> 234Th90 + 4He2.
- 5.Step 5: State changes: the proton number decreases by 2 and the neutron number decreases by 2 (since 146 - 2 = 144).
Question: Explain why beta-minus decay produces a continuous energy spectrum for the emitted beta particles, and how this led to the prediction of the antineutrino.
- 1.Step 1: In beta-minus decay, a neutron transforms into a proton, an electron, and an antineutrino: n -> p + e- + antineutrino.
- 2.Step 2: The energy released in the decay is shared between the electron and the antineutrino.
- 3.Step 3: Because the share of energy between the two particles is variable, the beta particle can have any kinetic energy from zero up to a maximum value (the Q value).
- 4.Step 4: If only the electron were emitted, conservation of energy and momentum would require it to have a single, discrete energy. The observed continuous spectrum therefore implied another particle was emitted.
- 5.Step 5: This particle must be neutral and have negligible mass, and was named the antineutrino.