University Physics V · Radioactive Decay · 15.2
Half-Life, Mean Life, Width & Branching
One unstable state, five currencies: a mean life in seconds, a width in electronvolts, a half-life on the chart, an activity in becquerel, a branching ratio with no units at all. This is the exchange desk between them, and the place to learn that when several decay routes compete it is the widths that add, never the half-lives.
Build the model
Connect the measurement to the mechanism.
An unstable state is not a stationary state. Couple a bound configuration to a continuum of decay products and the discrete level becomes a pole of the resolvent at complex energy E₀ − iΓ/2, so the survival probability is e(−Γt/ħ) and everything in this topic is bookkeeping on that single imaginary part. Because the hazard rate λ = Γ/ħ does not depend on how long the nucleus has already waited, the waiting time is exponentially distributed, p(t) = λe(−λt), whose mean τ = 1/λ and median T½ = τ ln 2 differ by 31% and are routinely confused; an ensemble of N such nuclei then disintegrates at A = λN becquerel.
When several final states are open, each contributes its own −iΓᵢ/2, imaginary parts add, and so Γ = ΣΓᵢ, λ = Σλᵢ, bᵢ = Γᵢ/Γ, with the partial half-lives combining in parallel as 1/T½ = Σ1/Tᵢ — which is why a nuclide always lives shorter than its fastest single channel. The costs are worth naming out loud. This is an ensemble statement about a mean, silent on which nucleus goes next; the exponential is exact only in the intermediate-time window, failing quadratically at very short times and to a power law at very long ones; and a becquerel counts disintegrations, so it fixes no dose until energy per decay, geometry, radiation weighting and biology are supplied.
- Simple definition
- The decay constant λ is an unstable state's hazard rate — the probability per unit time that it decays — equal to Γ/ħ, to 1/τ, and to ln 2/T½, so a width, a mean life and a half-life are three ways of writing one number.
- Example
- Tc-99m has T½ = 6.01 h = 2.16×10⁴ s, so λ = ln 2/T½ = 3.21×10⁻⁵ s⁻¹, τ = 3.12×10⁴ s, and Γ = ħ/τ = 2.11×10⁻²⁰ eV — a linewidth no spectrometer will ever resolve, so this one is timed rather than measured.
One complex pole E₀ − iΓ/2 is the whole input; every other number on this page is arithmetic on its imaginary part.
Γ in eV, ħ = 6.582×10⁻¹⁶ eV s, τ and T½ in s, λ in s⁻¹
Mean and median differ by 31%, so the mean life is 44% longer than the half-life; swapping them corrupts every inventory.
p in s⁻¹; τ is the mean of the density, T½ its median; T½ = 0.693 τ
Turns a quantity of material into a counting rate: 1 g of Co-60 (T½ = 5.27 y) runs at 4.2×10¹³ Bq, about 1130 curie.
N nuclei, A in becquerel = 1 decay s⁻¹; 1 Ci = 3.7×10¹⁰ Bq exactly
Rates add, lifetimes do not: two equal channels halve the lifetime rather than averaging it.
Γᵢ is the partial width of channel i; bᵢ is dimensionless with Σbᵢ = 1
Tᵢ is what a golden-rule or WKB calculation predicts, one channel at a time; the chart quotes their parallel sum.
Tᵢ is the half-life channel i would have alone, so Tᵢ ≥ T½ always
Short-lived means hot: F-18 reaches 3.5×10¹⁸ Bq g⁻¹, while U-238 manages only 1.2×10⁴ Bq g⁻¹.
M in g mol⁻¹, NA = 6.022×10²³ mol⁻¹, a in Bq g⁻¹
The decay constant is an imaginary energy
Solve the full problem — a bound configuration coupled to the continuum of its decay products — and the discrete level stops being an eigenvalue of a Hermitian operator. It becomes a pole of the resolvent (E − H)⁻¹ at complex energy E₀ − iΓ/2, equivalently an eigenvalue of the non-Hermitian effective Hamiltonian left behind once the continuum is projected out. The amplitude to still be in that state is then a(t) = e(−iE₀t/ħ) e(−Γt/2ħ), so |a(t)|² = e(−Γt/ħ). Compare with e(−λt) and the identification is forced: λ = Γ/ħ. The decay constant is nothing more than the imaginary part of an energy divided by ħ, and the rest of this topic is unit conversion on that one number. With ħ = 6.582×10⁻¹⁶ eV s, a width of 1 eV is a mean life of 6.58×10⁻¹⁶ s; the ρ meson's Γ = 149 MeV is τ = 4.4×10⁻²⁴ s; Tc-99m's τ = 3.12×10⁴ s is Γ = 2.1×10⁻²⁰ eV. One relation, twenty-eight orders of magnitude.
A constant hazard rate forces the exponential
Write S(t) for the probability that one given nucleus is still there at time t. Constancy of λ is the physical claim, and it is a strong one: the chance of decaying in the next dt is λ dt whatever t is, because the nucleus carries no record of how long it has already waited. Then S(t + dt) = S(t)(1 − λ dt), which is S′ = −λS and S = e(−λt). The waiting-time density is p(t) = −S′ = λe(−λt), and every summary number falls out of it. The mean is τ = ∫₀^∞ t λe(−λt) dt = 1/λ. The median solves e(−λT½) = ½, giving T½ = ln 2/λ = 0.693 τ. These are different numbers and neither is a sloppy name for the other: at t = τ the surviving fraction is e(−1) = 36.8%, not 50%. The density is also violently skewed — ⟨t²⟩ = 2/λ² makes the standard deviation exactly equal to the mean — so a lifetime quoted as one number is the summary of a spread running from zero to many τ.
Activity is a rate of disintegrations, not of energy
N nuclei each carrying hazard λ give a mean disintegration rate A = λN, the activity, in becquerel — one decay per second — with the historical curie fixed at exactly 1 Ci = 3.7×10¹⁰ Bq. Two consequences should be kept apart. First, A obeys the same exponential as N, so a ratio of two activities dates a sample without anyone ever counting nuclei, and A = ln 2 · N/T½ says that of two samples holding equal numbers of nuclei the shorter-lived one is the fiercer. Second, a becquerel is a rate of events, not a delivery of energy. A routine 37 MBq injection of Tc-99m emits 140.5 keV per decay, so the entire source radiates 3.7×10⁷ × 1.405×10⁵ × 1.602×10⁻¹⁹ = 8.3×10⁻⁷ W — under a microwatt. Absorbed dose in gray needs that power, the fraction stopping in the organ of interest, and the mass it stops in; equivalent dose then multiplies by wR, which is 1 for this photon and 20 for an alpha. Equal activities of an alpha and a gamma emitter are not equal hazards, and that is precisely where the decay law stops helping.
Channels add as widths; half-lives combine in parallel
Second-order perturbation theory hands the level one −iΓᵢ/2 from each open decay channel, and imaginary parts simply add: Γ = ΣᵢΓᵢ. Divide by ħ and λ = Σᵢλᵢ. Nothing about that algebra is optional. The branching ratio is a channel's share of the total, bᵢ = Γᵢ/Γ = λᵢ/λ with Σbᵢ = 1, and the partial half-life is the half-life the channel would have if it acted alone, Tᵢ = ln 2/λᵢ = T½/bᵢ. So the measured half-life is a parallel combination, 1/T½ = Σᵢ1/Tᵢ, structurally identical to conductances in parallel, and it is always shorter than the shortest partial half-life. K-40 makes it concrete: T½ = 1.248×10⁹ y split 89.28% β⁻ against 10.72% electron capture plus β⁺, so the partial half-lives are 1.248×10⁹/0.8928 = 1.398×10⁹ y and 1.248×10⁹/0.1072 = 1.164×10¹⁰ y. It is the partial half-life a nuclear-structure calculation predicts; the chart quotes their parallel sum.
Whether Γ can be measured depends on how big it is
λ, τ, T½ and Γ carry identical information, but experiments do not reach them equally, and choosing the currency is really choosing the apparatus. A width is read straight off a line shape only when Γ is comparable to the resolution of whatever spreads the energy out. Hadronic resonances live at that end: the ρ at Γ = 149 MeV is a bump 149 MeV wide in an invariant-mass spectrum, and no stopwatch is involved anywhere. Nuclear levels sit far below any spectrometer — the 14.4 keV state of Fe-57 has τ = 142 ns, hence Γ = 4.6×10⁻⁹ eV and a relative width Γ/E = 3.2×10⁻¹³ — which only recoil-free Mössbauer absorption resolves, and it succeeds because the absorber's own line is equally narrow. Below that, timing is the only route: a six-hour isomer's Γ = 2×10⁻²⁰ eV will never appear as a line, so you count decays against a clock and quote λ.
One λ with many channels is still one exponential
Branching and mixing look alike on the page and behave differently on a semi-log plot, and telling them apart is the commonest reason to draw one. A nuclide with three open channels still has a single λ: every gated channel, every partial activity, the total count rate, all fall with the same T½, and ln A against t is exactly straight. A sample holding two different nuclides has two decay constants, and a sum of two exponentials is not an exponential — ln A against t is concave upward, steep early and shallow late, so a straight-line fit returns an apparent half-life that drifts upward through the run. That drift is the diagnostic. Peel the tail: fit the late, straight portion for the long-lived λ, extrapolate it back, subtract it from the early data, and the residual is the short-lived component, which should now plot straight. If it does not, there is a third nuclide, or the background subtraction is wrong.
Change one variable at a time
Make the relationship visible.
Set both partial half-lives to 20 h: the combined half-life is 10 h, not 20, and the solid curve sits below both dashed ones because opening a channel can only add width. Now drag channel 2 out to 30 h and watch the branch swing to 0.60 while the half-life creeps only from 10 h to 12 h.
HALF-LIFE T½6.67 h
MEAN LIFE τ9.62 h
BRANCH b₁ = Γ₁/Γ0.667
TOTAL WIDTH Γ1.90 ×10⁻²⁰ eV
Live interpretationHALF-LIFE T½: 6.67 h. MEAN LIFE τ: 9.62 h. BRANCH b₁ = Γ₁/Γ: 0.667. TOTAL WIDTH Γ: 1.90 ×10⁻²⁰ eV
Catch the common trap
Explain before calculating.
A nuclide decays by two channels with partial half-lives 10.0 h and 20.0 h. What is its half-life, and what share of decays uses the 10.0 h channel?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyBa-137m, the 662 keV emitter eluted from a caesium generator, has T½ = 153 s. Find its decay constant, mean lifetime and natural linewidth Γ, then the activity of a freshly eluted sample holding 2.5×10¹² of these nuclei.
- λ = ln 2/T½ = 0.693147/153 s = 4.53×10⁻³ s⁻¹.
- τ = 1/λ = 221 s, which is also T½/ln 2 = 153/0.693 — the mean life runs 44% longer than the half-life, so state which one you are quoting.
- Γ = ħ/τ = 6.582×10⁻¹⁶ eV s ÷ 220.7 s = 2.98×10⁻¹⁸ eV. Against the 662 keV transition energy that is Γ/E = 4.5×10⁻²⁴, so no detector will ever see this as a line width: the half-life here is measured by timing decays, not by resolving a line shape.
- A = λN = 4.53×10⁻³ s⁻¹ × 2.5×10¹² = 1.13×10¹⁰ Bq.
Answerλ = 4.53×10⁻³ s⁻¹, τ = 221 s, Γ = 2.98×10⁻¹⁸ eV, and A = 1.13×10¹⁰ Bq (11.3 GBq) at the moment of elution.
MediumCu-64 has T½ = 12.70 h and decays 38.5% by β⁻, 17.6% by β⁺ and 43.9% by electron capture. Find the total decay constant and the three partial half-lives, verify that they combine to the measured half-life, then find how many positrons a 500 MBq source emits each second.
- Total: λ = ln 2/T½ = 0.693147/(12.70 × 3600 s) = 0.693147/45720 s = 1.516×10⁻⁵ s⁻¹.
- Partial constants are shares of that total, λᵢ = bᵢλ: λ(β⁻) = 5.84×10⁻⁶ s⁻¹, λ(β⁺) = 2.67×10⁻⁶ s⁻¹, λ(EC) = 6.66×10⁻⁶ s⁻¹. They sum back to λ, as they must.
- Partial half-lives Tᵢ = T½/bᵢ: 12.70/0.385 = 33.0 h for β⁻, 12.70/0.176 = 72.2 h for β⁺, 12.70/0.439 = 28.9 h for EC. Every one exceeds 12.70 h, which is the check that they are partial and not total.
- Combine in parallel: 1/33.0 + 1/72.2 + 1/28.9 = 0.03031 + 0.01386 + 0.03457 = 0.07874 h⁻¹, and 1/0.07874 = 12.70 h ✓. The reciprocals add; the times themselves never do.
- Positron rate = b(β⁺) × A = 0.176 × 5.00×10⁸ Bq = 8.80×10⁷ s⁻¹. Activity counts every disintegration; only this fraction produces the annihilation pair a PET scanner needs.
Answerλ = 1.52×10⁻⁵ s⁻¹; partial half-lives 33.0 h (β⁻), 72.2 h (β⁺) and 28.9 h (EC), whose reciprocals sum to 0.0787 h⁻¹ = 1/12.70 h; the 500 MBq source emits 8.80×10⁷ positrons per second.
HardA source holds two independent nuclides: 8.0×10¹⁰ nuclei with T½ = 10.0 min and 2.0×10⁹ nuclei with T½ = 300 min. Find the total activity at t = 0 and at t = 60 min, then find the apparent half-life that a straight-line semi-log fit would return over 0–20 min and over 100–120 min, and say what the difference between them proves.
- Decay constants: λ₁ = ln 2/600 s = 1.1552×10⁻³ s⁻¹ and λ₂ = ln 2/18000 s = 3.8508×10⁻⁵ s⁻¹.
- Activities at t = 0 from A = λN: A₁ = 9.242×10⁷ Bq and A₂ = 7.70×10⁴ Bq, total 9.250×10⁷ Bq. The short-lived nuclide dominates 1200 to 1 even though there are only 40 times as many of its nuclei — activity weights each nucleus by its own λ.
- At 60 min the first has run six half-lives (÷64) and the second only 0.2 of one: A₁ = 1.444×10⁶ Bq, A₂ = 7.70×10⁴ × 0.8706 = 6.70×10⁴ Bq, total 1.511×10⁶ Bq.
- Early interval: A(20)/A(0) = 2.318×10⁷/9.250×10⁷ = 0.2506, so Tₐₚₚ = 20 ln 2/ln(1/0.2506) = 13.86/1.384 = 10.0 min. The fit sees the fast component and nothing else.
- Late interval: A(100) = 9.03×10⁴ + 6.11×10⁴ = 1.514×10⁵ Bq and A(120) = 2.26×10⁴ + 5.84×10⁴ = 8.09×10⁴ Bq, ratio 0.5346, so Tₐₚₚ = 13.86/0.6262 = 22.1 min.
- A single decay constant gives one slope at all times. A drifting apparent half-life therefore proves the sample is a mixture of nuclides, not a nuclide with two channels: branching still gives one λ and a semi-log plot that is exactly straight.
AnswerA(0) = 9.25×10⁷ Bq and A(60 min) = 1.51×10⁶ Bq; the apparent half-life rises from 10.0 min over 0–20 min to 22.1 min over 100–120 min and keeps climbing toward 300 min. Curvature on a semi-log plot means two nuclides, never two channels.