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Bounded modern-physics survey · Interactive lesson

Q values, two-body kinematics, and conservation ledgers

This is the audit every nuclear energy claim has to pass. Read the mass table and count its electrons, let momentum decide who is paid, charge the beam for the centre of mass it cannot shed, and run the ledgers that rule a process out. None of it says how fast anything happens, and knowing that is part of the skill.

UPIII-11.04 · Lesson 4 of 9 in this unit. Position is not a completion score.
Starting model: Q values, two-body kinematics, and conservation ledgersEnergies against the projectile's lab kinetic energy Kₐ, target at rest. Dashed diagonal: Kₐ itself. Solid line: the share available in the centre-of-mass frame, Kₐ m_X/(mₐ + m_X). It meets the reaction's need, −Q = 1.2 MeV, at the threshold Kₜₕ = 1.54 MeV; the shaded gap above, 0.34 MeV, is centre-of-mass kinetic energy that momentum conservation keeps out of reach.−Q = 1.2 MeV, so threshold Kₜₕ = 1.54 MeVshaded: locked in CM motion, 0.34 MeVKₐKcm−Q01010 MeVprojectile lab kinetic energy Kₐ / MeV
Start with a picture. Change the model in the lab below.

Bounded modern-physics survey. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment.

1

Build the model

The mass table sets the budget, momentum sets the split, the ledgers set what is allowed.

The key idea

The Q value of a decay or reaction is the rest energy that disappears, Q = (Σmᵢₙᵢₜᵢₐₗ − Σmfinal)c²: the kinetic energy released when Q is positive, and the least energy the collision must supply in its centre-of-mass frame when Q is negative.

Example. ²¹⁰Po → ²⁰⁶Pb + α: atomic masses 209.9828737 u, 205.9744653 u and 4.0026033 u leave 0.0058051 u, so Q = 0.0058051 × 931.494 MeV = 5.41 MeV, of which the α carries 5.30 MeV and the recoiling lead 0.10 MeV.

The complete physical model and assumptions

A decay or a reaction is treated here as a relativistic collision with only three kinds of bookkeeping: energy, momentum, and a short list of conserved counts. Rest energies come from a mass table, so Q = (Σmᵢₙᵢₜᵢₐₗ − Σmfinal)c² is the kinetic energy the process releases, the same in every frame; the table lists neutral atoms, so the electrons must be audited before Q means anything. Momentum then rations Q. A parent at rest that splits in two has no freedom left, so each fragment's energy is fixed and the spectrum is a set of lines. A reaction driven by a beam must leave the centre of mass moving, so its threshold always exceeds −Q. On top sit the ledgers of charge, nucleon number and lepton number: a process that breaks one cannot occur at any energy. That is where the model stops. It says which doors are open and how much energy passes through them, not how often: decay constants, branching ratios and cross sections need dynamics, tunnelling and the weak interaction, and a three-body final state leaves the split undetermined, giving an endpoint rather than a line.

1Read the mass table, then count its electrons

Q = (Σmᵢₙᵢₜᵢₐₗ − Σmfinal)c², with 1 u c² = 931.494 MeV. Mass tables list neutral atoms, so every entry carries Z electrons, and a Q value is right only once they balance. In α decay they do: Z on the left, Z − 2 in the daughter atom plus 2 in the helium atom. In β⁻ decay the daughter atom's extra electron is the one emitted, so Q = [M(Z) − M(Z + 1)]c² as it stands. Electron capture takes its electron from the parent's own cloud, so again the bare difference is Q, less the binding energy of that electron, which returns as X-rays or Auger electrons. β⁺ decay is the exception: the daughter atom has one electron too many and a positron is created as well, so 2mₑc² = 1.022 MeV comes off. For ¹⁸F → ¹⁸O the atomic difference is 0.0017777 u, or 1.656 MeV: that is Q for capture, while β⁺ gets only 0.634 MeV. Below 1.022 MeV, β⁺ is closed altogether: ⁷Be → ⁷Li has 0.862 MeV and decays by capture alone.

2Momentum decides who is paid

A parent at rest has zero momentum, so two fragments leave back to back with equal momenta. Nonrelativistically K = p²/2m, so K₁/K₂ = m₂/m₁, and with K₁ + K₂ = Q the light fragment takes K₁ = Q m₂/(m₁ + m₂). For ²²⁶Ra → ²²²Rn + α, Q = 4.8706 MeV; the α leaves with 4.8706 × 222.018/226.020 = 4.784 MeV and the radon recoils with 0.086 MeV. Two things follow. The α energy is fixed, not merely likely: every decay to the same final state emits the same energy, so α spectra are lines, and a line below the ground-state value labels an excited level of the daughter. And the recoil is small only beside Q: 86 keV is some 10⁴ times a chemical bond energy, so the daughter atom is torn out of whatever molecule held its parent.

3Exact kinematics and the correspondence check

Relativity keeps the logic and changes only the energy–momentum relation. With E₁ + E₂ = Mc², equal momenta, and E² − p²c² = m²c⁴ for each fragment, E₁ = (M² + m₁² − m₂²)c²/2M, so K₁ = Q(Q + 2m₂c²)/(2Mc²). When Q ≪ m₂c² this tends to Q m₂/(m₁ + m₂), the nonrelativistic split, as correspondence demands; for ²²⁶Ra the two differ by about one part in 10⁵. Where a fragment is massless the old formula fails outright. In π⁺ → μ⁺ + νμ, Q = (139.570 − 105.658) MeV = 33.912 MeV, and with m₂ = 0 the exact result is Kμ = Q²/(2Mc²) = 4.12 MeV, while Q m₂/(m₁ + m₂) would give zero. A third body breaks the argument differently: three momenta can sum to zero in endlessly many ways, so β decay gives a continuous electron spectrum, and kinematics fixes only its endpoint.

4A beam must pay for the centre of mass

Fire projectile a at target X at rest. The invariant s = Eₜₒₜ² − pₜₒₜ²c² is the squared energy in the centre-of-mass frame, and here s = (mₐc² + mXc²)² + 2mXc² Kₐ. The reaction is just possible when √s equals the products' rest energy, all of them created at rest in that frame. Solving gives Kₜₕ = −Q(1 + mₐ/mX) + Q²/(2mXc²), exactly. The first term is the familiar result: the extra fraction mₐ/mX pays for the kinetic energy of the centre of mass, which momentum conservation will not let the reaction spend. For ¹⁴N(α, p)¹⁷O, Q = −1.192 MeV, so Kₜₕ = 1.192 × (1 + 4.003/14.003) = 1.533 MeV, and the relativistic term adds only 54 eV. In particle physics it matters: p + p → p + p + π⁰ has Q = −135.0 MeV and a threshold of 279.7 MeV, 3.6% above the nonrelativistic 270.0 MeV. An exothermic reaction has no kinematic threshold at all, though a Coulomb barrier can still bar the way.

5Ledgers forbid, but never set a rate

Three counts must balance in every process here: charge, nucleon number A, and electron lepton number Lₑ, which is +1 for e⁻ and νₑ and −1 for e⁺ and ν̄ₑ. The ledger names the neutrino: n → p + e⁻ + ν̄ₑ balances Lₑ as 0 = 1 − 1, with Q = [M(n) − M(¹H)]c² = 0.782 MeV. It also separates forbidden from merely uphill. A free proton cannot decay as p → n + e⁺ + νₑ, yet every ledger balances; the obstacle is Q = −1.804 MeV, and inside ¹⁸F the nucleus pays it. By contrast p → e⁺ + π⁰ would release 803 MeV and balances charge, but breaks nucleon number and lepton number, and the measured lower limit on its lifetime is above 10³⁴ years. What no ledger gives is a rate: ²¹²Po (Q = 8.95 MeV) and ²³²Th (Q = 4.08 MeV) pass every ledger, yet their α half-lives, 0.3 μs and 1.4 × 10¹⁰ years, differ by a factor of 10²⁴.

2

Key equations

Read the units, assumptions and limits before substituting.

Q value from rest energies

Q = (Σmᵢₙᵢₜᵢₐₗ − Σmfinal)c²

Q is the change in total kinetic energy, the same in every frame. Positive Q is released, negative Q must be supplied, and its sign alone says whether a nucleus at rest can decay.

masses in kg or u; Q in J or MeV; 1 u c² = 931.494 MeV and 1 MeV = 1.602 × 10⁻¹³ J

β⁺ decay from atomic masses

Qβ⁺ = [M(Z, A) − M(Z − 1, A)]c² − 2mₑc²

The daughter atom has an electron to spare and a positron is made as well. Below 1.022 MeV of atomic difference only capture is open: ⁷Be, at 0.862 MeV, decays by capture alone.

M: neutral-atom masses in u, each counting Z electrons; mₑc² = 0.511 MeV. For β⁻ and electron capture the bare difference is Q

Two-body split, parent at rest

K₁ = Q m₂/(m₁ + m₂)

Equal and opposite momenta with K = p²/2m hand the light fragment the large share: ²²⁶Ra gives its α 4.784 of 4.871 MeV, and the radon recoils with the other 86 keV.

K₁ the kinetic energy of fragment 1 in J or MeV; m₁, m₂ the fragment masses in kg or u; valid for Q ≪ m₁c² and m₂c²

Two-body split, exact

K₁ = Q(Q + 2m₂c²)/(2Mc²)

For Q ≪ m₂c² it returns Q m₂/(m₁ + m₂), as correspondence requires. With a massless partner it still works: in π⁺ → μ⁺ + νμ the muon gets Q²/(2Mc²) = 4.12 MeV.

M the parent's mass, m₂ the partner fragment's, in kg or MeV/c²; from E₁ = (M² + m₁² − m₂²)c²/2M, valid at any speed

Lab threshold, target at rest

Kₜₕ = −Q(1 + mₐ/mX) + Q²/(2mXc²)

The first term is the nonrelativistic threshold; the second is relativity's correction, 54 eV for ¹⁴N(α, p)¹⁷O but 9.7 MeV for p + p → p + p + π⁰.

Q negative; mₐ, mX the projectile and target masses in kg or u; Kₜₕ the projectile's lab kinetic energy in J or MeV; exact

Electron lepton-number ledger

ΣLₑ before = ΣLₑ after

It names the neutrino: β⁻ decay must emit ν̄ₑ, while β⁺ decay and electron capture emit νₑ. A process that breaks it is forbidden at every energy.

Lₑ = +1 for e⁻ and νₑ, −1 for e⁺ and ν̄ₑ, 0 for nucleons and nuclei; a pure number, like charge number and nucleon number A
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1.2 MeV
0.5 to 2.5 MeV; steps of 0.1.
4 u
1 to 12 u; steps of 1.
14 u
4 to 60 u; steps of 1.

Try this. Set mₐ = 1 u and mX = 60 u and the threshold sits almost on −Q; then fire a heavy beam at a light target, mₐ = 12 u on mX = 4 u, and it climbs to four times −Q, because three quarters of the beam energy rides along with the centre of mass.

Live results

Kₜₕ NONRELATIVISTIC
1.543 MeV
LOCKED IN CM MOTION
0.343 MeV
Kₜₕ / (−Q)
1.286
RELATIVISTIC Q²/2mXc²
55.2 eV

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics III modelEnergies against the projectile's lab kinetic energy Kₐ, target at rest. Dashed diagonal: Kₐ itself. Solid line: the share available in the centre-of-mass frame, Kₐ m_X/(mₐ + m_X). It meets the reaction's need, −Q = 1.2 MeV, at the threshold Kₜₕ = 1.54 MeV; the shaded gap above, 0.34 MeV, is centre-of-mass kinetic energy that momentum conservation keeps out of reach.−Q = 1.2 MeV, so threshold Kₜₕ = 1.54 MeVshaded: locked in CM motion, 0.34 MeVKₐKcm−Q01010 MeVprojectile lab kinetic energy Kₐ / MeV
Energies against the projectile's lab kinetic energy Kₐ, target at rest. Dashed diagonal: Kₐ itself. Solid line: the share available in the centre-of-mass frame, Kₐ mX/(mₐ + mX). It meets the reaction's need, −Q = 1.2 MeV, at the threshold Kₜₕ = 1.54 MeV; the shaded gap above, 0.34 MeV, is centre-of-mass kinetic energy that momentum conservation keeps out of reach.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Energy deficit −Q: 1.2 MeV; Projectile mass mₐ: 4 u; Target mass mX: 14 u

Baseline 1.2; choose 0.5 to 2.5, in steps of 0.1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The tempting rule is that an endothermic reaction starts once the beam brings −Q of kinetic energy. It cannot. Before the collision the system carries the projectile's momentum, so afterwards the products must carry the same momentum, and with it at least the kinetic energy of the centre of mass, a fraction mₐ/(mₐ + mX) of the beam energy. Only the remainder, the energy in the centre-of-mass frame, can become rest energy. For α particles on ¹⁴N, with Q = −1.19 MeV, that locks away 4/18 of the beam energy, and the threshold is 1.53 MeV, not 1.19 MeV. Swap the roles and fire ¹⁴N at helium: now 14/18 is locked away, and the same reaction with the same Q needs a 5.36 MeV beam. −Q is the threshold only in the centre-of-mass frame, or for an infinitely heavy target.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

The PET isotope ⁶⁴Cu can decay three ways: β⁻ to ⁶⁴Zn, and electron capture or β⁺ to ⁶⁴Ni. With atomic masses M(⁶⁴Cu) = 63.9297642 u, M(⁶⁴Zn) = 63.9291420 u and M(⁶⁴Ni) = 63.9279660 u, write each process with its ledger and find its Q value.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

²⁴¹Am (241.0568293 u) α-decays to ²³⁷Np (237.0481736 u); M(⁴He) = 4.0026033 u. Its strongest α line is at 5.4856 MeV and is often followed by a γ-ray. Find Q, the α energy for a decay straight to the ground state of ²³⁷Np, and the energy of the γ-ray.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

A cyclotron can make the PET isotope ¹³N by ¹³C(p, n)¹³N. Atomic masses: ¹H 1.0078250 u, ¹³C 13.0033548 u, n 1.0086649 u, ¹³N 13.0057386 u. Find Q, the lab threshold proton energy, exactly and in the nonrelativistic limit, and the kinetic energy of the neutrons emitted at threshold.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

Which of these proposed processes is forbidden by a conservation ledger, so that no amount of available energy could make it happen?
7

Review & continue

The Q value of a decay or reaction is the rest energy that disappears, Q = (Σmᵢₙᵢₜᵢₐₗ − Σmfinal)c²: the kinetic energy released when Q is positive, and the least energy the collision must supply in its centre-of-mass frame when Q is negative.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.