Secular and Transient Radioactive Equilibrium: Parent and Daughter Activity
When a radioactive parent decays into a radioactive daughter, the daughter activity does not simply fall with its own half-life. It rises first, as the parent keeps supplying new atoms, and later follows the parent. This gives two named situations, secular equilibrium and transient equilibrium, which appear in nuclear and radiochemistry questions in IIT-JAM, GATE, CSIR-NET and CUET-PG.
The Bateman relation for a parent–daughter pair
Let the parent P (decay constant λ1) decay to the daughter D (λ2), with only the parent present at t = 0. Then the number of daughter atoms is:
λ = ln 2 ÷ t½ A = λN
N1(0) is the initial number of parent atoms. The activities are A1 = λ1N1 and A2 = λ2N2. If the parent is the only source of the daughter, and every parent decay gives one daughter atom, the ratio of activities after the parent’s exponential term dominates is given below.
A2 ÷ A1 = λ2 ÷ (λ2 − λ1) = t½,1 ÷ (t½,1 − t½,2)
Time of maximum daughter activity: tmax = ln(λ2 ÷ λ1) ÷ (λ2 − λ1)
Secular equilibrium (t½,1 » t½,2): A2 ≈ A1 and N2 ÷ N1 = t½,2 ÷ t½,1
Secular equilibrium: the parent is very long-lived
If the parent half-life is much longer than the daughter’s, then λ1 is tiny compared with λ2, and the denominator λ2 − λ1 is almost λ2. The parent hardly changes during the experiment, so the daughter builds up until it decays as fast as it is formed. At that stage the two activities are equal.
Question. Take the half-life of Ra-226 as 1600 years and that of its daughter Rn-222 as 3.82 days (textbook values; use the values given in your own question). Find the ratio of Rn atoms to Ra atoms in secular equilibrium and the fraction of equilibrium reached after three daughter half-lives.
Step 1. Convert 1600 years to days: 1600 × 365.25 = 584 400 days.
Step 2. N2 ÷ N1 = t½,2 ÷ t½,1 = 3.82 ÷ 584 400 = 6.54 × 10−6.
Step 3. The daughter grows as 1 − 2−n of its final value after n daughter half-lives. For n = 3: 1 − 1/8 = 0.875, or 87.5%.
Answer. At equilibrium there are only about 6.5 Rn atoms for every million Ra atoms, but their activities are equal. After 7 daughter half-lives the fraction is 1 − 1/128 = 99.2%, so equilibrium is practically complete.
Transient equilibrium: parent is longer-lived, but only modestly
When the parent half-life is longer than the daughter’s but not enormously so, the parent decays noticeably during the observation. After a long enough time, the daughter activity stays in a constant ratio to the parent activity, but A2 is larger than A1, and both fall together with the parent’s half-life. Do not call this secular.
Question. Treat a parent with half-life 66 h and a daughter with half-life 6.0 h (about the values used for a Mo-99 / Tc-99m generator; this simplified question assumes every parent decay gives one daughter atom). Find (a) the activity ratio A2/A1 in transient equilibrium, (b) the time of maximum daughter activity.
Step 1. λ1 = 0.6931 ÷ 66 = 0.010502 h−1. λ2 = 0.6931 ÷ 6.0 = 0.11552 h−1.
Step 2 (a). A2/A1 = λ2 ÷ (λ2 − λ1) = 0.11552 ÷ 0.10502 = 1.100. Check with half-lives: 66 ÷ (66 − 6) = 66 ÷ 60 = 1.100.
Step 3 (b). λ2 ÷ λ1 = 66 ÷ 6 = 11, so ln 11 = 2.3979. tmax = 2.3979 ÷ 0.10502 = 22.8 h.
Answer. In transient equilibrium the daughter activity is about 10% higher than the parent’s. The daughter activity is greatest after about 22.8 h, so the timing of collection matters when such a generator is used.
Check on tmax: at tmax the daughter activity equals the parent activity (A2 = A1), because the daughter is momentarily neither growing nor shrinking. Putting t = 22.8 h into the full relation gives A2/A1 = 1.000. The ratio then climbs to 1.10 over the following days, for example 1.09 at 48 h.
Comparing the cases
| Case | Condition | Activity ratio A2/A1 |
|---|---|---|
| Secular | Parent half-life » daughter | About 1 |
| Transient | Parent half-life > daughter, comparable order | λ2 ÷ (λ2 − λ1), greater than 1 |
| No equilibrium | Parent half-life < daughter | Ratio keeps growing; daughter outlives parent |
Common mistakes
- Equal atoms instead of equal activities. In secular equilibrium it is the activities that are equal, not the numbers of atoms. The atom ratio is t½,2 ÷ t½,1.
- Mixing time units. Convert both half-lives to the same unit, such as days, before dividing.
- Using the secular result for a modest half-life ratio. With 66 h and 6 h the ratio is 1.10, not 1.
- Thinking the daughter decays at the parent’s rate in a physical sense. It still decays with its own λ; the observed fall is the parent’s supply.
- Ignoring branching. If only a fraction of parent decays reach the daughter, multiply the ratio by that fraction. Use the data your question gives.
Exam relevance
Questions usually ask you to name the type of equilibrium from the two half-lives, to find the ratio of activities or atoms, or to find the time of maximum daughter activity. Do the arithmetic with half-lives in one unit and with λ = ln 2 ÷ t½. Check your own syllabus for the exact form of the Bateman relation expected. Exact half-life values differ slightly between data tables, so use the numbers the question provides.
Convert half-lives to decay constants and activities with the Half-Life tool.
Open the Half-Life Calculator →Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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