
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.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.
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.
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 Jβ⁺ 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 QTwo-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 m₂c²Two-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 speedLab 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; exactElectron 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 AInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
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.
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
This experiment is kept only while this page is open. It does not change saved course progress.
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.
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.
- Balance the three. β⁻: ⁶⁴Cu → ⁶⁴Zn + e⁻ + ν̄ₑ, charge 29 = 30 − 1, Lₑ 0 = 1 − 1. Capture: ⁶⁴Cu + e⁻ → ⁶⁴Ni + νₑ, charge 29 − 1 = 28, Lₑ 1 = 1. β⁺: ⁶⁴Cu → ⁶⁴Ni + e⁺ + νₑ, charge 29 = 28 + 1, Lₑ 0 = −1 + 1. A = 64 throughout.
- β⁻: the zinc atom's 30th electron is the one emitted, so atomic masses need no correction. Qβ⁻ = (63.9297642 − 63.9291420) u × 931.494 MeV u⁻¹ = 0.0006222 × 931.494 MeV = 0.5796 MeV.
- Capture: the electron came from the copper atom's own 29, leaving the 28 a nickel atom needs, so again no correction. QEC = (63.9297642 − 63.9279660) × 931.494 MeV = 0.0017982 × 931.494 MeV = 1.6750 MeV.
- β⁺: the nickel atom has one electron to spare and a positron is created too, so subtract 2mₑc² = 1.0220 MeV. Qβ⁺ = 1.6750 − 1.0220 = 0.6530 MeV.
Answer. Qβ⁻ = 0.580 MeV, QEC = 1.675 MeV and Qβ⁺ = 0.653 MeV. All three are positive and all three occur; how often each happens is a rate question no ledger answers.
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.
- Electrons balance, 95 = 93 + 2, so atomic masses give Q directly: Δm = 241.0568293 − 237.0481736 − 4.0026033 = 0.0060524 u, and Q = 0.0060524 × 931.494 MeV = 5.63777 MeV.
- The parent is at rest, so the α takes the fraction M(²³⁷Np)/[M(²³⁷Np) + M(⁴He)] = 237.0481736/241.0507769 = 0.983395 of whatever energy is released.
- To the ground state: Kα = 0.983395 × 5.63777 MeV = 5.5442 MeV, and the neptunium recoils with 5.6378 − 5.5442 = 0.0936 MeV.
- The 5.4856 MeV line is lower, so that branch releases less: invert the split, Q* = 5.4856/0.983395 = 5.57823 MeV. The daughter is left in a level E* = Q − Q* = 5.63777 − 5.57823 = 0.05954 MeV above its ground state.
- That level decays by the γ-ray, which carries E* less a recoil of about 0.01 eV, so Eγ = 59.5 keV.
Answer. Q = 5.638 MeV; Kα = 5.544 MeV to the ground state; Eγ ≈ 59.5 keV, the line every ²⁴¹Am source is known for, recovered from masses and one α energy.
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.
- Ledger: charge 1 + 6 = 0 + 7, A 1 + 13 = 1 + 13, and 7 electrons on each side, so atomic masses give Q with no correction. Δm = 1.0078250 + 13.0033548 − 1.0086649 − 13.0057386 = −0.0032237 u, so Q = −0.0032237 × 931.494 MeV = −3.0029 MeV.
- Nonrelativistic threshold: −Q(1 + mₐ/mX). The atomic masses of ¹H and ¹³C stand in for the bare proton and nucleus that collide; the true masses would lower the result by only about 0.1 keV. So −Q(1 + mₐ/mX) = 3.0029 × (1 + 1.0078250/13.0033548) = 3.0029 × 1.077505 = 3.2356 MeV.
- Relativistic term: Q²/(2mXc²) = 3.0029²/(2 × 13.0033548 × 931.494) MeV = 9.0174/24225 MeV = 3.72 × 10⁻⁴ MeV. So the exact threshold is Kₜₕ = 3.2360 MeV; the correction is 1 part in 8700, and correspondence holds.
- At threshold the products are created at rest in the centre-of-mass frame, so both move at its velocity. The energy locked in that motion is Kₜₕ mₐ/(mₐ + mX) = 3.2360 × 0.071930 = 0.2328 MeV, and 3.2360 − 0.2328 = 3.0032 MeV ≈ −Q, as it must be to the nonrelativistic accuracy used here.
- Moving together, the products share 0.2328 MeV in proportion to mass: Kₙ = 0.2328 × 1.0086649/14.0144035 = 0.0168 MeV.
Answer. Q = −3.003 MeV; Kₜₕ = 3.236 MeV, with the relativistic term only 0.37 keV; the threshold neutrons carry about 16.8 keV, all moving forward with the centre of mass.
Quick check
Select an answer and explain it before checking.
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.