Radioactive Decay Constant and Activity — Formulas and Worked Examples
Half-life is the famous quantity, but the one that actually appears inside the equations is the decay constant, λ. Activity — the number of nuclei breaking up per second — follows directly from it. Questions in Class 12 nuclear chemistry, IIT-JAM and CSIR-NET regularly give you an activity and ask for a mass, or give a mass and ask for an activity. This guide sets out the formulas, four worked examples with the arithmetic shown, and the unit traps that turn a correct method into a wrong answer.
The formulas
What each symbol means
- λ (decay constant) — the probability per unit time that any one nucleus decays. Unit is the reciprocal of whatever time unit you used: s⁻¹, h⁻¹, y⁻¹.
- N — the number of radioactive nuclei present. Not moles, not grams.
- A (activity) — decays per second. SI unit is the becquerel (Bq), exactly 1 disintegration per second. The older unit is the curie (Ci), defined as exactly 3.7 × 10¹⁰ Bq.
- t½ — half-life, the time for N to fall to half.
- τ (mean life) — average lifetime of a nucleus, always longer than the half-life: τ = t½ ÷ 0.6931 ≈ 1.44 t½.
Note what λ does not depend on: temperature, pressure, chemical form. That independence is what makes radiometric dating possible.
Worked example 1 — decay constant of cobalt-60
Take t½(⁶⁰Co) = 5.27 years, the value printed in most data booklets. Find λ in years⁻¹ and in seconds⁻¹.
In years⁻¹:
λ = 0.6931 ÷ 5.27 = 0.1315 y⁻¹
In seconds⁻¹. First convert the half-life, using 1 year = 365.25 days:
365.25 × 24 × 3600 = 31 557 600 s
t½ = 5.27 × 31 557 600 = 1.663 × 10⁸ s
λ = 0.6931 ÷ (1.663 × 10⁸) = 4.168 × 10⁻⁹ s⁻¹
Mean life τ = 5.27 ÷ 0.6931 = 7.60 years.
Worked example 2 — activity of 1.00 g of cobalt-60
This is the classic "specific activity" question. Molar mass of ⁶⁰Co ≈ 59.93 g/mol.
Step 1 — number of nuclei.
n = 1.00 ÷ 59.93 = 0.016686 mol
N = 0.016686 × 6.022 × 10²³ = 1.005 × 10²² nuclei
Step 2 — activity. λ must be in s⁻¹ to give an answer in Bq:
A = λN = (4.168 × 10⁻⁹) × (1.005 × 10²²) = 4.19 × 10¹³ Bq
Step 3 — in curies.
4.188 × 10¹³ ÷ (3.7 × 10¹⁰) = 1.13 × 10³ Ci, about 1130 Ci per gram.
The lesson buried in that answer: a short half-life means a large λ, which means a large activity for the same mass. Specific activity and half-life are inversely related.
Worked example 3 — activity after a given time (with a second-route check)
A source has an activity of 8000 Bq. Its half-life is 12.0 hours. What is the activity 30.0 hours later?
Route 1 — exponential form.
λ = 0.6931 ÷ 12.0 = 0.05776 h⁻¹ (hours throughout, so the units cancel)
λt = 0.05776 × 30.0 = 1.7329
A = 8000 × e−1.7329 = 8000 × 0.17678 = 1414 Bq
Route 2 — counting half-lives. 30.0 ÷ 12.0 = 2.5 half-lives.
A = 8000 ÷ 22.5 = 8000 ÷ 5.6569 = 1414 Bq ✓
The two routes agree, which is the check you should run whenever the elapsed time is a neat multiple — or half-multiple — of the half-life.
Worked example 4 — from activity back to mass
A phosphorus-32 tracer sample has an activity of 2.50 × 10⁶ Bq. t½(³²P) = 14.3 days. How many nuclei, and what mass, is that?
Step 1 — λ in s⁻¹. 14.3 days = 14.3 × 86 400 = 1 235 520 s
λ = 0.6931 ÷ 1 235 520 = 5.610 × 10⁻⁷ s⁻¹
Step 2 — rearrange A = λN.
N = A ÷ λ = (2.50 × 10⁶) ÷ (5.610 × 10⁻⁷) = 4.46 × 10¹² nuclei
Step 3 — convert to mass (M ≈ 31.97 g/mol):
moles = 4.456 × 10¹² ÷ (6.022 × 10²³) = 7.40 × 10⁻¹² mol
mass = 7.40 × 10⁻¹² × 31.97 = 2.37 × 10⁻¹⁰ g, about 0.24 nanograms.
A perfectly usable laboratory source weighs a fraction of a nanogram. That is why radioactive tracers can be used at concentrations far too small to disturb the chemistry they are tracing.
Worked example 5 — finding the elapsed time
The same equation run backwards answers dating and shelf-life questions. Rearranging A = A₀e−λt gives:
A source with a half-life of 8.00 days is measured at 2000 Bq, having been 6000 Bq when it was prepared. How old is it?
A₀ ÷ A = 6000 ÷ 2000 = 3
ln 3 = 1.0986, ln 2 = 0.6931, so ln 3 ÷ ln 2 = 1.5850
t = 8.00 × 1.5850 = 12.7 days
Sanity check. One half-life (8 days) would take 6000 to 3000, and two half-lives (16 days) would take it to 1500. Our answer of 12.7 days sits between the two, exactly as it must.
Notice that only the ratio A₀ ÷ A matters, never the absolute activities. This is why the method works even when the detector is not perfectly calibrated — a constant efficiency factor cancels top and bottom.
Fraction remaining after n half-lives
| Half-lives elapsed | Fraction of nuclei left | Percentage left |
|---|---|---|
| 1 | 1/2 | 50% |
| 2 | 1/4 | 25% |
| 3 | 1/8 | 12.5% |
| 4 | 1/16 | 6.25% |
| 5 | 1/32 | 3.125% |
| 10 | 1/1024 | 0.098% |
Two things follow from this table. First, activity never reaches zero — it only halves again, which is why storage times for radioactive waste are quoted as multiples of the half-life. Second, the "ten half-lives" rule of thumb used in laboratories comes from the last row: after ten half-lives less than 0.1% remains, which for most short-lived tracers is indistinguishable from background.
Units at a glance
| Quantity | Symbol | SI unit | Note |
|---|---|---|---|
| Decay constant | λ | s⁻¹ | Any reciprocal time unit is valid if used consistently |
| Activity | A | becquerel, Bq | 1 Bq = 1 decay per second |
| Activity (old unit) | A | curie, Ci | 1 Ci = 3.7 × 10¹⁰ Bq exactly |
| Absorbed dose | D | gray, Gy | Energy absorbed per kg — a different quantity from activity |
| Equivalent dose | H | sievert, Sv | Dose weighted for biological effect |
Common mistakes that cost marks
- Mixing time units. λ in s⁻¹ with t in hours is the single most common error. Decide on one unit and convert the half-life first.
- Using log instead of ln. λ = ln 2 ÷ t½. If you use the base-10 form, it is λ = 2.303 log 2 ÷ t½ — the 2.303 is not optional.
- Putting moles or grams into A = λN. N is a count of nuclei; multiply moles by 6.022 × 10²³ first.
- Confusing activity with dose. Becquerels describe the source; grays and sieverts describe what a person receives. Exam questions mix these deliberately.
- Treating decay as linear. "Half gone in one half-life, so all gone in two" is wrong — after two half-lives a quarter remains.
- Forgetting that A ∝ N. Because activity and number of nuclei fall in exactly the same proportion, you can put activities straight into the decay equation with no conversion at all.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 12 | Nuclear chemistry numericals; λ from half-life; fraction remaining |
| JEE/NEET | Activity after n half-lives; mean life vs half-life comparisons |
| IIT-JAM / CUET-PG | Specific activity; tracer-sample mass calculations |
| GATE / CSIR-NET | Decay chains, secular equilibrium, radiochemical dating |
Check your λ and your remaining activity in seconds. The suite's radioactivity tool works with half-life, elapsed time and the fraction remaining, so you can confirm every step of the examples above.
Open the Radioactivity / Half-Life Calculator →Nuclear chemistry numericals are where a lot of Class 12 marks are won and lost. ABC Chemistry runs Class 11–12 chemistry coaching at its Gurugram centre and online classes across India — details at abcchemistry.in.