Beta decay and the neutrino
Key idea: Beta-minus electrons have a continuous energy spectrum because an antineutrino shares the released energy and momentum with the electron and recoiling daughter nucleus.
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Guided lesson
Guided steps
- Read why one decay branch has fixed energy but variable electron energies.
- Predict whether two products could give a continuous energy spectrum.
- Follow how energy and momentum fix a two-product decay.
- Read how a third particle allows variable energy sharing.
- Find the energy carried away by the antineutrino.
- Explain why the energy-sharing graph is not a beta spectrum.
- Explain the third-particle argument using both conservation laws.
- Find electron energy and decide whether recoil momentum can be ignored.
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ContinueIn the 1920s, measurements of beta decay seemed to show energy disappearing. Every decay along a given branch releases the same energy, yet the emitted electrons came out with a whole range of energies. In 1930 Pauli proposed that an undetected particle was carrying off the rest; it was finally detected in 1956. This lesson follows that argument: how conservation of energy and momentum predicted the (anti)neutrino.
The puzzle: electrons with a range of energies
For a particular decay branch, the parent and daughter nuclear states are fixed, so the energy released, Q, is fixed too. A nuclide can have several branches leading to different daughter energy levels; compare electrons within one branch. It comes from the decrease in rest mass.
Alpha decay behaves as you would expect from that: alpha particles from one two-body decay branch have the same kinetic energy. Beta-minus decay does not. If you plot the number of electrons against their kinetic energy, you get a continuous spectrum. It starts at zero, rises to a peak at a fairly low energy and falls to zero at a maximum kinetic energy that is practically equal to Q. Most electrons have much less kinetic energy than Q.
Check your understanding 1
A nucleus at rest decays along one branch. Suppose the only products were an electron and the daughter nucleus. Could the daughter take a different share of Q in each decay, giving a continuous range of electron energies? Commit to a prediction before reading the two-product argument.
Show answer
No. The momenta have equal magnitudes and opposite directions, and the kinetic energies must add to the same Q. For these two fixed masses, there is only one energy-sharing result. Allowing daughter recoil does not create a range of electron energies.
Two products would share the energy in only one way
Suppose a nucleus at rest decayed into just two products: the daughter nucleus and the electron.
- Conservation of momentum: the total momentum stays zero, so the two products move off in opposite directions with equal momenta.
- Conservation of energy: their kinetic energies add up to Q.
These two conditions fix both kinetic energies, so every electron from that decay would have the same kinetic energy, just below Q. The daughter’s recoil is already part of this calculation; it cannot absorb a different amount of energy each time. That is what happens in a two-body alpha decay branch, and it is why that branch gives one alpha energy.
The beta spectrum shows that this two-product picture is wrong. A momentum imbalance between the electron and daughter would also require another particle: the two visible momenta alone would not add to zero.
A third particle carries away the difference
Rather than give up conservation of energy and momentum, Pauli proposed that a third particle is emitted. In beta-minus decay it is the antineutrino, ν bar. It has no charge and a rest mass so small that it can be neglected here, and it interacts so weakly with matter that early detectors could not register it.
With three products, energy and momentum can be shared in many different ways:
where E_(k,e) is the electron’s kinetic energy, E_(ν bar) the antineutrino’s energy and E_(k,D) the daughter’s recoil kinetic energy. The three momenta add up to zero, but they need not lie along one line. When recoil energy is neglected, the electron energy approaches Q as the antineutrino takes almost no energy; in most decays the antineutrino takes a large share.
The daughter nucleus is thousands of times heavier than the electron, so its recoil kinetic energy is tiny and can usually be neglected in the energy account. Its momentum cannot be neglected: a heavy particle can carry a large momentum with very little kinetic energy.
With the recoil energy neglected, the electron and antineutrino energies add up to Q. For an illustrative decay with Q = 0.80 MeV:
Fixed total energy, variable electron energy
A straight line falls from an antineutrino energy of 0.80 MeV at zero electron kinetic energy to zero at an electron kinetic energy of 0.80 MeV. The two energies have an approximately fixed sum.
Scroll across the figure to read all labels.
View figure data
| Electron kinetic energy (MeV) | Antineutrino energy (MeV) |
|---|---|
| 0.0 | 0.8 |
| 0.1 | 0.7 |
| 0.2 | 0.6 |
| 0.3 | 0.5 |
| 0.4 | 0.4 |
| 0.5 | 0.3 |
| 0.6 | 0.2 |
| 0.7 | 0.1 |
| 0.8 | 0.0 |
This graph shows how the energy is shared in any one decay: an electron with 0.20 MeV leaves 0.60 MeV for the antineutrino. It is not the beta spectrum, which shows how many electrons have each energy.
Check your understanding 2
In a beta-minus decay with Q = 0.60 MeV, an electron is emitted with 0.17 MeV of kinetic energy. Neglecting the recoil kinetic energy of the daughter, find the energy of the antineutrino.
Show answer
E_(ν bar) = 0.60-0.17 = 0.43 MeV. The energy has not disappeared: the antineutrino carries it away undetected.
Check your understanding 3
The graph above slopes down. A learner says this means fewer electrons are emitted at higher energies. What is wrong with that reading? Refer to both axes and explain what would be plotted on a beta spectrum.
Show answer
This graph plots antineutrino energy against electron kinetic energy for a fixed Q, neglecting recoil energy. Its downward slope means that more energy for the electron leaves less for the antineutrino in that decay. A beta spectrum instead plots the number of electrons in each energy interval against electron kinetic energy. Check that your answer names the vertical quantity in each graph; an energy-sharing line is not a count distribution.
Exam-style question 1
Why the beta spectrum needed another particle
Examination question
The kinetic energies of beta-minus particles emitted along one decay branch form a continuous spectrum up to a maximum value. Explain how this observation, together with the conservation laws, led to the prediction of an additional particle emitted in beta decay. [4 marks]
Write your answer before viewing the mark scheme
Show solution step by step
State the fixed energy
1 markMethod
Each decay along the same branch releases the same energy.Reason
The parent and daughter nuclear states are fixed for this branch.Working
Answer
The energy released, Q, is the same for every decay along this branch.
Apply both conservation laws to two products
1 markMethod
Show that two products give one electron energy.Reason
Momentum and energy conservation together fix how two products share Q.Working
Answer
If only the daughter and the electron were emitted, conservation of momentum and energy would give every electron the same kinetic energy.
Compare with the observation
1 markMethod
The observed electron energies vary.Reason
A prediction that disagrees with measurement means the model is incomplete.Working
Answer
The continuous spectrum shows that electrons have a range of energies, so energy appears to be missing in most decays.
Conclude
1 markWorking
Answer
A third, uncharged particle, the (anti)neutrino, must be emitted. It shares the energy and momentum with the electron and daughter in varying proportions, so energy and momentum are conserved in every decay.
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Self-mark one point for each statement.
Check your understanding 4
In one beta-minus decay branch, Q = 0.70 MeV. An antineutrino carries away 0.48 MeV. Find the electron’s kinetic energy, neglecting the daughter’s recoil kinetic energy. Does that small recoil kinetic energy let you omit the daughter’s momentum from the momentum balance?
Show answer
E_(k,e) = 0.70-0.48 = 0.22 MeV. Neglecting a small recoil energy is an approximation to the energy sum. It does not remove a product from the vector momentum sum: vector pₑ + vector p_(ν bar) + vector pD = 0. A large daughter mass makes its recoil energy small for a given momentum.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
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