Nuclear Physics

Radioactive Decay Chains: What Happens After the First Decay

Definition

A radioactive decay chain is a sequence in which an unstable parent nuclide decays into a daughter that is itself radioactive, and so on until a stable nuclide ends it. The daughter’s activity starts at zero and climbs to a peak at the exact moment it equals the parent’s activity, then falls.

Point a counter at a freshly separated radioactive sample and something odd happens. The count rate does not simply fade the way the textbook curve promises — it climbs for a while first, tops out, and only then settles into the long decline.

Nothing is wrong with the counter. It is hearing a second nuclide: the daughter left behind by the first decay, which turns out to be unstable in its own right. That second nuclide is the whole subject of this page.

What Is a Radioactive Decay Chain?

A radioactive decay chain is a run of successive decays in which each unstable nuclide produces another unstable nuclide, until one of them lands on a stable species and the chain stops there.

Most exam questions quietly stop after one step, as though the daughter were inert. Uranium-238 is the famous counter-example: it runs fourteen steps before finishing as stable lead-206, and every one of those intermediates is made by the step above it and destroyed by the step below.

Three members are enough to show everything that matters, so that is the model used here. A radioactive parent, a radioactive daughter, a stable end product — label them 1, 2 and 3 and you have the whole cast.

One assumption does all the work: the daughter starts at zero. In the laboratory that means somebody separated the parent chemically and started a clock at that moment, which is why every time in this article is measured from a separation rather than from the birth of the sample.

Strontium-90 is the simplest case to picture. It beta-decays to yttrium-90, which beta-decays again to zirconium-90, and zirconium-90 is stable, so there the chain stops after exactly two steps.

Both of those steps turn a neutron into a proton and push the element one place up the periodic table. Our guides to beta decay and to alpha, beta and gamma radiation cover what is physically leaving the nucleus; here the only question is how many decays per second each member is doing.

The Decay Chain Formula for the Daughter

You already know what the parent does. It thins out exponentially, governed by its own decay constant λ1 = ln 2 divided by its half-life — if that sentence needs unpacking, our guide to half-life in physics covers the decay constant, the mean lifetime, activity and the becquerel, and this page takes all five as read.

The daughter is harder, because it is being created and destroyed at the same time. Solving its population equation with none present at the start gives the Bateman result for two members.

N2 = N0 × λ1 / (λ2 – λ1) × (e-λ1t – e-λ2t)

That form is honest but awkward. Divide through to get the ratio of the two activities instead, and almost everything interesting about a chain falls out of a single expression.

A2 / A1 = λ2 / (λ2 – λ1) × (1 – e-(λ2 – λ1)t)

Look at what has vanished. The starting quantity has cancelled, so the ratio depends on the two half-lives and the elapsed time and on nothing else — that is algebra, not a measurement. Feed measured half-lives and one activity into the decay chain calculator and it will return the daughter activity, the parent activity or the elapsed time, whichever you leave blank.

SymbolWhat it isSI unitWorth knowing
N0number of parent atoms at the separation—only sets the vertical scale; it cancels out of every ratio on this page
N2number of daughter atoms at time t—starts at zero by assumption, which is what “since separation” means
λ1decay constant of the parentper secondln 2 divided by the parent half-life T1
λ2decay constant of the daughterper secondln 2 divided by the daughter half-life T2
A1parent activity nowBqλ1N1, and the quantity you can actually measure
A2daughter activity nowBqλ2N2; this is what climbs to a peak and then falls
ttime since the daughter was last separatedsnot the age of the sample and not the date on the label

There is one sharp edge in the first formula: if the two half-lives are equal, λ2 – λ1 is zero and the expression divides by zero. The fix is to factor that difference out, writing A2/A1 = λ2t f(x) with x = (λ2 – λ1)t and f(x) = (1 – e-x)/x.

Since f tends to 1 as x tends to zero, the equal-half-life case is then computed exactly rather than dodged. Worked problem 8 runs that case end to end.

The lab below draws both activities against time. Because the daughter’s activity can never climb above the parent’s starting activity, both curves fit on one shared vertical scale — so set the two half-lives and slide the clock forward, and what you see is a fair comparison rather than two curves stretched to different rulers.

Decay Chain Lab

Why the Daughter Climbs, Peaks, Then Follows

Two things happen to the daughter every second: some is made by the parent, and some of what exists decays away. Write that as a rate and the daughter’s population grows at λ1N1 – λ2N2, which is simply A1 – A2.

That single line explains the shape. Early on there is no daughter to speak of, so A2 is tiny, the difference is large and positive, and the stock of daughter atoms builds fast. As the stock grows, A2 grows with it and the gap narrows.

When the gap closes to nothing the growth stops. The daughter’s activity is at its maximum at precisely the instant the two activities are equal — not near it, exactly at it, which makes the peak a genuinely checkable prediction rather than a feature of a drawing.

A line graph of activity in megabecquerels against time since separation in days, for barium-140 decaying to lanthanum-140 and on to stable cerium-140, starting from 500 megabecquerels of barium-140 and no lanthanum-140. The gold parent curve falls smoothly from 500 megabecquerels at day zero to about 98 megabecquerels at day 30. The wine daughter curve starts at zero, rises steeply, meets the parent curve at day 5.6548 where both read 367.7 megabecquerels, reaches its own highest value at that same instant, and then falls, staying a little above the parent curve for the rest of the graph. A dashed guide line runs from the vertical axis to the meeting point. The vertical axis is activity in megabecquerels, ticked 0, 100, 200, 300, 400 and 500; the horizontal axis is time since the barium-140 was separated in days, ticked 0, 5, 10, 15, 20, 25 and 30.
Barium-140 feeding lanthanum-140, from 500 MBq of barium and no lanthanum at all. The curves meet at day 5.6548, where both read 367.7 MBq, and that crossing is the daughter peak. After it the daughter runs a little above the parent and the two fall together.

Past the peak, provided the parent is the longer-lived of the two, the daughter cannot decay on its own schedule, because it is being topped up by a parent that is itself shrinking. The pair lock into a fixed ratio, and from then on both curves fall at the parent’s rate — a short-lived daughter, decaying on a clock its own half-life has nothing to do with.

Watch the right-hand end of the graph above. The daughter holds a constant factor of 1.1516 above the parent from there on, so the visible gap between the curves keeps narrowing as both of them fall — a fixed ratio does not mean a fixed distance.

The Three Equilibrium Regimes

How the story ends depends entirely on which nuclide outlives the other. Three cases exhaust the possibilities, and the deciding quantity is the limiting ratio the pair settle onto, which works out to T1/(T1 – T2) whenever the parent is the longer-lived of the two.

RegimeWhen it appliesLimiting ratio A2/A1What the daughter ends up doing
Secular equilibriumparent at least 100 times longer-lived than the daughter, by conventionjust above 1 — 1.000253 for strontium-90 feeding yttrium-90settles onto a near-constant activity and stays there for as long as the parent lasts
Transient equilibriumparent longer-lived than the daughter, but by less than thatnoticeably above 1 — 1.1516 for barium-140 feeding lanthanum-140peaks, then falls at the parent rate while staying a fixed factor above it
No equilibriumdaughter longer-lived than the parentnone at allkeeps climbing relative to the parent, and is still there long after the parent has gone
A line graph of daughter activity divided by parent activity against time since separation in days, for three decay chains. The gold strontium-90 to yttrium-90 curve rises quickly and flattens onto a dashed guide line labelled 1.000253. The wine barium-140 to lanthanum-140 curve rises past one and flattens onto a second, higher dashed guide line labelled 1.151600. The dark bismuth-210 to polonium-210 curve rises slowly at first, then steepens, passes 1.25 and runs off the top of the plot before day 30 without ever flattening. The vertical axis is daughter activity divided by parent activity, ticked 0.00, 0.25, 0.50, 0.75, 1.00 and 1.25; the horizontal axis is time since separation in days, ticked 0, 5, 10, 15, 20, 25 and 30. A note states that strontium-90 settles on 1.000253 rather than on 1, and that at this scale the gap is far too small to see.
The same quantity for all three chains: daughter activity divided by parent activity. Two of them flatten onto a ceiling; the third has no ceiling to flatten onto and leaves the top of the plot before day 30.

That factor of a hundred deserves a warning. It is a naming convention and nothing else — no physical boundary sits there, and a chain does not change behaviour as it crosses the line. The limiting ratio moves smoothly from one chain to the next, and only the label attached to it jumps.

So read the number, not the name. And be careful with one sentence you will meet almost everywhere: in secular equilibrium the two activities are routinely said to become equal, and they do not. Strontium-90 feeding yttrium-90 settles onto 1.000253, a hair above 1, and that tiny surplus is exactly what keeps the daughter supplied.

How long does settling take? Because the ratio approaches its ceiling as 1 – e-(λ2 – λ1)t, reaching 99 per cent of the way there takes ln(100) divided by that same difference — usually quoted as “about seven daughter half-lives”, and worked out properly in problem 5.

Three Real Decay Chains Side by Side

Nothing beats real nuclides for making the three regimes concrete. The half-lives below were pulled from the ground-state data behind the IAEA Live Chart of Nuclides, and every derived column is computed from them rather than copied from a textbook.

A block diagram of a three-member decay chain. A gold-edged box labelled Sr-90, strontium-90, half-life 28.91 years, radioactive parent, is joined by an arrow labelled beta-minus to a wine-edged box labelled Y-90, yttrium-90, half-life 64.05 hours, radioactive daughter, which is joined by a second beta-minus arrow to a dark-edged box labelled Zr-90, zirconium-90, stable, end of the chain. Notes beneath read that each arrow is one beta-minus decay in which a neutron becomes a proton and the atomic number rises by one, and that the daughter is made at the parent rate and destroyed at its own rate, so its activity climbs from zero, peaks, and then falls in step with the parent that feeds it. A final line gives the quoted uncertainties, 0.03 years on the strontium-90 half-life and 0.05 hours on the yttrium-90 half-life.
Row one of the table, drawn out: strontium-90 to yttrium-90 to stable zirconium-90, two beta-minus decays and then nothing. Half-lives are the IAEA ground-state values, with the quoted uncertainties printed beneath.
ParentT1DaughterT2Stable endT1/T2RegimeLimiting ratioDaughter peaks at
Sr-9028.91 yY-9064.05 hZr-903957Secular1.00025331.90 d
Ba-14012.751 dLa-1401.67858 dCe-1407.5963Transient1.15165.6548 d
Bi-2105.012 dPo-210138.376 dPb-2060.03622Nonenone24.8944 d

Read the precision honestly. The quoted uncertainties are 0.03 y on Sr-90, 0.05 h on Y-90, 0.004 d on Ba-140, 0.00021 d on La-140, 0.005 d on Bi-210 and 0.002 d on Po-210, and every derived column inherits them. Those columns are quoted to the precision the worked problems use, which in two or three cells runs a figure or two past what the uncertainties strictly support, so read the trailing digits as bookkeeping rather than as measurement.

Two footnotes on that table. Bismuth-210 also has an alpha branch, but at 0.000132 per cent of its decays it changes nothing here. And strontium-90’s peak time shifts in the sixth significant figure depending on whether you convert 28.91 years using a 365.25-day year or a 365.2422-day one, which is why it is quoted to four.

The third row is the interesting one. Polonium-210 outlives bismuth-210 by a factor of nearly twenty-eight, so the daughter never catches up in the ordinary sense — it simply accumulates, peaks almost five parent half-lives in, and is still sitting there when the parent has effectively vanished.

Common Misconceptions About Decay Chains

Every one of the five below comes from reading a number too generously, and a separate point about scope follows them. None is an exotic mistake.

  • “In secular equilibrium the activities become equal.” They become constant in ratio, which is a different claim. Strontium-90 to yttrium-90 settles on 1.000253, not 1.000000, and that gap of one part in about four thousand is what feeds the daughter.
  • “A hundred to one is where secular equilibrium starts.” It is where the word starts, not the physics. The limiting ratio is continuous across that line, so a chain at 99 to 1 and a chain at 101 to 1 behave almost identically while collecting different labels.
  • “A stronger source changes the shape of the chain.” It cannot. The starting quantity cancels out of the ratio algebraically, so the regime, the limiting ratio and the peak time are fixed by the two half-lives alone and the source strength only sets the vertical scale.
  • “The daughter always builds up towards equilibrium.” Only when the parent outlives it. Bismuth-210 feeding polonium-210 has no limiting ratio to build towards, and the ratio of the two activities climbs without ever flattening.
  • “Total activity only ever falls.” Not at the start of a fresh chain. Take the barium-140 shipment of problem 3: the parent alone stood at 688.88 MBq at separation, and ten days later the pair together are putting out 847.87 MBq, because the daughter has arrived.

The scope point is simpler. This three-member model assumes one decay path, one radioactive daughter and no daughter present at the start; a branching parent, a longer natural series or an incomplete separation each break one of those assumptions and need more than the formula above.

Where Decay Chains Turn Up in Everyday Life

The places where a chain matters share a signature: a daughter whose activity is not what somebody assumed, because the clock started at a separation rather than at a birth.

A domestic radon test kit, the everyday end of a radioactive decay chain
Radon is a daughter product, not a primordial gas: it is made underground by the chain above it and arrives through the floor.
  • Radon in buildings. Radon-222 is not a primordial gas hanging around since the Earth formed — at 3.8235 days it would have gone long ago. It is manufactured continuously underground by radium-226, whose 1600-year half-life makes that pair as deeply secular as any pair in this article. The EPA calls radon a naturally occurring radioactive gas that can cause lung cancer and notes that testing is the only way to know your exposure.
  • Radionuclide generators. Hold a long-lived parent on a column, wash the short-lived daughter off when you need it, and the daughter rebuilds from zero on a schedule the formula above predicts exactly. The strontium-90 row of the table is that cycle.
  • Fresh fission products in transit. Barium-140 and lanthanum-140 come out of a reactor together, and what a consignment is actually emitting on arrival depends on how many days ago it left the separation plant, not on the number printed on the drum.
  • Dating from two lines in one spectrum. Since the activity ratio depends on elapsed time and nothing else, measuring both lines gives the time since separation directly. Problem 7 works it through, including the case where the answer is honestly “no solution”.
  • Shielding a fresh source. Radiation leaves a young chain from both members at once, so a shield designed around the parent figure alone will be undersized whenever the daughter has had time to grow.

How Decay Chains Fit With Half-Life, Dating and Reactors

This page sits one layer above the single-nuclide picture and one layer below the full natural series. If your daughter is stable, or you simply do not care what it does, you do not need any of this — the ordinary exponential and the half-life calculator are enough.

Radiometric dating is the same mathematics read backwards. Our article on carbon dating uses a parent whose daughter is stable nitrogen-14, which is why one measurement suffices there; on a chain with a radioactive daughter you get a second, independent clock in the ratio of the two activities.

Reactors are where most short chains on Earth are made. Fission fragments are neutron-rich, so they beta-decay in short runs of two or three steps before reaching stability — the barium-to-lanthanum pair in the table is one such run, and how a nuclear reactor works covers where those fragments come from.

To change the half-lives and watch the regime change with them, the decay chain simulator puts the parent curve, the daughter curve and the ratio on screen together.

Worked Problems

Problem 1
Barium-140 has a half-life of 12.751 d and its daughter lanthanum-140 has a half-life of 1.67858 d. Find both decay constants in units of per day.
Show Solution
Solution: Step 1: The decay constant is λ = ln 2 divided by the half-life. Carrying ln 2 as 0.6931472 keeps seven figures, which is what the later problems need. Step 2: Parent. λ1 = 0.6931472 / 12.751 = 0.0543602 per day. Step 3: Daughter. λ2 = 0.6931472 / 1.67858 = 0.4129366 per day. Answer: λ1 = 0.0543602 per day and λ2 = 0.4129366 per day Divide one by the other and you get 0.4129366 / 0.0543602 = 7.5963, which is the same figure as T1/T2 in the table further up. A decay constant runs inversely with its half-life, so the two ratios come out equal with the order flipped.
Problem 2
Strontium-90 has a half-life of 28.91 y and yttrium-90 one of 64.05 h. Which equilibrium regime is this chain in, and what limiting activity ratio does it settle onto?
Show Solution
Solution: Step 1: Put both half-lives in the same unit. The IAEA table gives 28.91 y as 912310729.9 s, which is 10559.1520 d, and 64.05 h is 64.05 / 24 = 2.66875 d. Step 2: Compare them. T1 / T2 = 10559.1520 / 2.66875 = 3956.5909, which the table above rounds to 3957 because 28.91 y carries only four significant figures. Either way it is far above the conventional 100, so the chain is called secular. Step 3: The limiting ratio is T1 / (T1 – T2). Bottom first: 10559.1520 – 2.66875 = 10556.48325. Then 10559.1520 / 10556.48325 = 1.000253. Answer: secular equilibrium, with a limiting activity ratio of 1.000253 Note what that answer is not. It is not 1.000000, and the difference is not rounding noise — it is the surplus of parent decays over daughter decays that keeps the daughter stocked.
Problem 3
A barium-140 source was separated 10 d ago and its parent activity now reads 400 MBq. What is the lanthanum-140 activity?
Show Solution
Solution: Step 1: Use the factored ratio, A2/A1 = λ2t f(x) with x = (λ2 – λ1)t and f(x) = (1 – e-x)/x, taking the decay constants from problem 1. Step 2: The exponent. λ2 – λ1 = 0.4129366 – 0.0543602 = 0.3585764 per day, so x = 0.3585764 × 10 = 3.585764. Step 3: Evaluate f, then the ratio. e-3.585764 = 0.0277155, so 1 – 0.0277155 = 0.9722845 and f = 0.9722845 / 3.585764 = 0.2711513. Then A2/A1 = 0.4129366 × 10 × 0.2711513 = 1.119683, and A2 = 1.119683 × 400 = 447.8732 MBq. Answer: A2 = 447.8732 MBq, or 447.9 MBq to four significant figures The daughter is out-radiating its own parent by twelve per cent, on a sample that contained none of it ten days earlier.
Problem 4
For the same barium-140 to lanthanum-140 chain, at what time after separation does the daughter activity reach its maximum?
Show Solution
Solution: Step 1: The peak sits where the growth rate A1 – A2 hits zero. Solving that gives t = ln(1 + d/λ1) / d, with d = λ2 – λ1 = 0.3585764 per day. Step 2: The bracket. d / λ1 = 0.3585764 / 0.0543602 = 6.596304, so 1 + 6.596304 = 7.596304 and ln(7.596304) = 2.027662. Step 3: Divide. t = 2.027662 / 0.3585764 = 5.6548 d. Answer: the daughter peaks 5.6548 d after separation Check it against problem 3: at 10 d the ratio was already 1.119683, comfortably past 1, so the crossing must lie earlier — and 5.6548 d is exactly where the two activities are equal.
Problem 5
How long does strontium-90 feeding yttrium-90 take to come within 1 per cent of its limiting ratio, and how many yttrium-90 half-lives is that?
Show Solution
Solution: Step 1: The ratio approaches its ceiling as 1 – e-dt, so 99 per cent of the way there means e-dt = 0.01 and t = ln(100) / d. The starting quantity and the ceiling both drop out. Step 2: Get d. λ2 = 0.6931472 / 2.66875 = 0.2597273 per day and λ1 = 0.6931472 / 10559.1520 = 0.0000656 per day, so d = 0.2597273 – 0.0000656 = 0.2596617 per day. Step 3: Divide, then convert. t = 4.605170 / 0.2596617 = 17.7353 d, and 17.7353 / 2.66875 = 6.6455 yttrium-90 half-lives. Answer: 17.7353 d, which is 6.6455 daughter half-lives That is where the “about seven half-lives” rule of thumb comes from: log2(100) = 4.605170 / 0.6931472 = 6.6439, and the small surplus over it is the parent’s own decay during the wait.
Problem 6
Bismuth-210 has a half-life of 5.012 d and decays to polonium-210, half-life 138.376 d. A sample separated 20 d ago reads 250 MBq of bismuth-210. Find the polonium-210 activity and the time of its peak.
Show Solution
Solution: Step 1: Decay constants first. λ1 = 0.6931472 / 5.012 = 0.1382975 per day and λ2 = 0.6931472 / 138.376 = 0.005009158 per day, so d = 0.005009158 – 0.1382975 = -0.133288342 per day. The difference is negative, which is the whole character of this chain. Step 2: With x = -0.133288342 × 20 = -2.6657668, the function f(x) = (1 – e-x)/x is cleanest read as (e2.6657668 – 1) / 2.6657668 = (14.378971 – 1) / 2.6657668 = 13.378971 / 2.6657668 = 5.018808. Step 3: Ratio and activity. A2/A1 = 0.005009158 × 20 × 5.018808 = 0.502800, so A2 = 0.502800 × 250 = 125.7000 MBq. For the peak, d / λ1 = -0.133288342 / 0.1382975 = -0.9637798, then 1 – 0.9637798 = 0.0362202 and ln(0.0362202) = -3.318138, giving t = -3.318138 / -0.133288342 = 24.8944 d. Answer: A2 = 125.7000 MBq, and the daughter peaks at 24.8944 d That peak arrives 4.97 parent half-lives after separation, by which point the bismuth is down to about three per cent of what it started with. There is no equilibrium here and never will be.
Problem 7
A barium-140 sample reads 400 MBq of barium-140 and 447.8732 MBq of lanthanum-140. How long ago was it separated, and what would a reading of 480 MBq of lanthanum-140 imply?
Show Solution
Solution: Step 1: Take the ratio and check it against the ceiling. R = 447.8732 / 400 = 1.119683, and the ceiling is T1/(T1 – T2) = 12.751 / 11.07242 = 1.151600. The measurement sits below it, so an elapsed time exists. Step 2: Invert the ratio formula: t = -ln(1 – R d / λ2) / d. First the bracket, 1.119683 × 0.3585764 / 0.4129366 = 0.9722846, so 1 – 0.9722846 = 0.0277154. Step 3: Take the logarithm and divide. -ln(0.0277154) = 3.585767, and t = 3.585767 / 0.3585764 = 10.0000 d. Answer: 10.0000 d since separation; a reading of 480 MBq has no solution at all 480 / 400 is 1.2000, which is above the 1.151600 ceiling this chain can ever reach. No elapsed time produces it, so “no solution” is the correct answer and it tells you something real — the separation was incomplete, a second parent is feeding the daughter, or one of the two readings is wrong.
Problem 8
Suppose a parent and its daughter happened to share exactly the same half-life T. What is the activity ratio after one half-life, and what daughter activity does a 500 MBq parent imply?
Show Solution
Solution: Step 1: The textbook formula divides by λ2 – λ1, which is now zero. Use the factored form instead: A2/A1 = λ2t f(x), and with x = 0 the function f(0) = 1 exactly. Step 2: So A2/A1 = λt with no approximation. At t = T that is (ln 2 / T) × T = ln 2 = 0.693147, whatever the half-life happens to be. Step 3: Scale it. Carrying ln 2 to nine figures, A2 = 0.693147181 × 500 = 346.5736 MBq. Answer: A2/A1 = 0.693147 and A2 = 346.5736 MBq Those extra digits earn their keep. Rounding ln 2 to 0.693147 before multiplying prints 346.5735, one unit out in the last place — a small illustration of why working values are carried further than answers are.

Frequently Asked Questions

What is a radioactive decay chain in simple terms?
A radioactive decay chain is what happens when the product of a decay is itself radioactive. The parent decays into a daughter, the daughter decays into something else, and the sequence continues until a stable nuclide ends it. Uranium-238 runs fourteen steps before it reaches stable lead-206; many chains stop after one or two.
Do parent and daughter activities become equal in secular equilibrium?
No, they settle onto a constant ratio slightly above 1 rather than becoming equal. For strontium-90 feeding yttrium-90 that ratio is 1.000253, and the small surplus is what keeps supplying the daughter. The useful statement is that the ratio stops changing, not that the two numbers match.
When does the daughter activity reach its maximum?
At the exact instant the two activities are equal. The daughter’s stock grows at the parent activity minus the daughter activity, so growth stops when that difference reaches zero. For barium-140 feeding lanthanum-140 the peak falls 5.6548 days after separation, and the formula is t = ln(1 + d divided by lambda-1) divided by d.
What is the difference between secular and transient equilibrium?
Only how far the limiting ratio sits above 1. A parent at least 100 times longer-lived than its daughter is called secular by convention and settles very close to 1; a shorter-lived parent that still outlives its daughter is called transient and settles noticeably higher. That factor of 100 is a naming rule, not a physical boundary.
Can a decay chain reach no equilibrium at all?
Yes, whenever the daughter outlives the parent. Bismuth-210, at 5.012 days, feeds polonium-210 at 138.376 days, so there is no limiting ratio for the pair to approach. The daughter accumulates, peaks almost five parent half-lives in, and is still radiating long after the parent has effectively disappeared.
How can two activities date a sample?
Because the ratio of the two depends on the elapsed time and on nothing else, not even on how strong the source is. Measure both lines in one spectrum, form the ratio, and invert the formula for time. The method stops working once the ratio reaches its ceiling, which takes about thirty to thirty-five daughter half-lives depending on the chain.
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