PhysicsParticles › Stable and unstable nuclei

Stable and unstable nuclei

A nucleus packs mutually repelling protons into a tiny volume, and the strong nuclear force is what holds it together. That force acts only over a few femtometres, and where it cannot balance the repulsion the nucleus rearranges itself by alpha or beta decay.

Builds on Constituents of the atom.

IN THIS TOPIC

  • Explain the role of the strong nuclear force, including its attractive and repulsive ranges.
  • Write balanced equations for alpha and beta-minus decay.
  • Explain why the neutrino was hypothesised from the beta-decay energy spectrum.

COMMON MISCONCEPTION

Positive charges can never stick together.

The force that shouldn't need to exist

Every proton in a nucleus electrostatically repels every other, at separations where that repulsion is enormous. For nuclei to exist at all, something stronger has to be gluing the nucleons together. That something is the strong nuclear force, an attraction between nucleons, protons and neutrons alike, which comfortably beats the electrostatic repulsion at nuclear range.

The strong nuclear force between nucleons: repulsive inside half a femtometre, attractive out to about three, gone beyondseparationforce0.5 fm3 fmrepulsionattractionbeyond 3 fm:negligible
FIG. 1The strong force between nucleons, with fierce repulsion closer than about 0.5 fm, attraction from there out to about 3 fm, and negligible beyond about 3 fm.

Its defining feature is its reach. The attraction operates only out to about 3 fm and is negligible beyond, so the strong force plays no part in anything you can see. Closer than about 0.5 fm it turns fiercely repulsive, and that repulsion is what stops a nucleus collapsing to a point. Nucleons settle at a comfortable spacing between the two limits, held in a pocket where the forces balance.

Alpha decay

Some nuclei, mostly very large ones, are unstable and rearrange themselves. In alpha decay the nucleus emits an alpha particle, which is a helium-4 nucleus, two protons and two neutrons in one tightly bound package. A falls by 4, Z falls by 2, and a new element results. Uranium-238 gives the standard example.

238U → 234Th + 4He, with the proton numbers balancing as 92 = 90 + 2.

Both numbers must balance across the arrow, always, because nucleons are conserved and so is charge.

Beta decay, and the particle it demanded

In beta-minus decay a neutron inside the nucleus becomes a proton, emitting a fast electron, the beta particle, along with a second and almost undetectable particle, the electron antineutrino, written as ν with a bar over it.

Beta-minus decay: inside the nucleus a neutron becomes a proton, emitting an electron and an electron antineutrinoneutronprotonelectronνelectron antineutrinoA is unchanged; Z rises by one
FIG. 2Beta-minus decay, in which a neutron becomes a proton and an electron leaves with an electron antineutrino. A is unchanged and Z rises by one.

So A stays fixed while Z rises by one. Carbon-14 dating runs on exactly this decay, 14C → 14N + e + an electron antineutrino, with 6 = 7 + (−1) + 0 balancing the charge.

The beta-decay energy spectrum: electrons emerge with a continuous range of energies, so something invisible carries the restelectron energycountexpected: all of the energyobserved: a continuous spreadthe antineutrino carries the missing energy
FIG. 3The beta energy spectrum, a continuous spread of electron energies always falling short of the total available. The shortfall is the antineutrino's share.

The antineutrino was hypothesised before it was detected, to rescue conservation of energy. Each decay releases a fixed total energy, so an electron leaving alone would always carry the same amount. Measured beta electrons do nothing of the kind. They show a continuous spectrum, from almost nothing up to a maximum. Either energy conservation fails, or an unseen partner takes the variable remainder. Pauli backed the partner, and decades later the neutrino turned up, exactly as billed.

WORKED EXAMPLE

Writing a beta-minus equation

Carbon-14 (Z = 6) undergoes β decay. Write the full equation and audit it.

β turns a neutron into a proton, so A stays put, Z rises by one, and the electron leaves with an electron antineutrino.

14C → 14N + e + νe, with nitrogen at Z = 7.

Now audit it. Nucleons, 14 = 14 + 0 + 0. Charge, 6 = 7 + (−1) + 0. Both books balance, and the antineutrino has earned its place by carrying off the energy the electron is missing. That shortfall is how it was predicted in the first place.

GUIDED PRACTICE

Thorium takes its turn

Thorium-234 (Z = 90) undergoes β decay to protactinium (Pa). Write the equation and run both audits.

Show the working

234Th → 234Pa + e + νe, with Pa at Z = 91.

Nucleons, 234 = 234 + 0 + 0. Charge, 90 = 91 − 1 + 0. The daughter has moved one place right in the periodic table while its mass number is unchanged, and that pairing identifies β.

INDEPENDENT PRACTICE

Predicting the decay mode

A nuclide has significantly more neutrons than its stable neighbours. Predict its likely decay mode, and what the decay does to its neutron and proton counts.

Show the working

Neutron-rich nuclei correct themselves by β decay, turning one neutron into a proton.

N falls by one, Z rises by one, and the nuclide steps diagonally towards the stable band. The N against Z map in the nuclear unit draws that step in exactly those terms.

ASSESSMENT FOCUS

  • The strong-force numbers are quotable content. Attraction out to about 3 fm, repulsion inside about 0.5 fm. Range questions want both, and they want the word “nucleons”, because the force acts on protons and neutrons alike.
  • Why does the nucleus not collapse? The very-short-range repulsion. Why does it not fly apart? The strong attraction beats the electrostatic repulsion. Two questions, two different halves of the same graph.
  • Decay equations are marked on the balancing. A and Z must each sum equally across the arrow, with the beta electron counted as A = 0 and Z = −1.
  • Alpha takes A down by 4 and Z down by 2. Beta-minus leaves A alone and takes Z up by 1. Naming the new element usually carries a mark of its own.
  • The neutrino argument is an energy-conservation argument, and the order matters. Continuous electron spectrum, fixed decay energy, therefore an undetected particle takes the rest.

CHECK YOURSELF

Write the decay equation for radium-226 (Z = 88) undergoing alpha decay to radon (Rn), and check both balances.

Show a hint

The alpha particle takes four nucleons, two of them protons.

Show the answer

226Ra → 222Rn + 4He.

Nucleons balance, 226 = 222 + 4. Protons balance, 88 = 86 + 2, so radon sits at Z = 86.

Both books hold, and the element changed because Z changed. Changing Z is what makes a decay a change of element.

The strong force attracts out to 3 fm and repels inside 0.5 fm.

Every decay balances A and Z exactly.

The antineutrino exists because the beta spectrum is continuous.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the stable and unstable nuclei questions page.

5 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

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  • Explain the role of the strong nuclear force, including its attractive and repulsive ranges.
  • Write balanced equations for alpha and beta-minus decay.
  • Explain why the neutrino was hypothesised from the beta-decay energy spectrum.

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