The halving rule is great for whole half-lives, but real problems rarely land on neat numbers. The fix is the decay constantλ — the probability that any one nucleus decays each second. It links directly to the half-life through one clean equation, and once you have λ you can find the activity, the number of nuclei, or the half-life for any time at all. A big λ means a jumpy, fast-decaying source; a small one means a slow, long-lived isotope.
📚 What you need to know
The decay constantλ is the probability an individual nucleus decays per unit time (unit: s−1)
A larger λ means a higher activity and faster decay
Activity: A = −ΔN/Δt = λN — activity depends on how many undecayed nuclei remain
The number of nuclei falls exponentially: N = N0e−λt
Half-life and decay constant are linked by t½ = ln 2 / λ
Half-life and decay constant are inversely proportional: short half-life → large λ
A log graph of ln N against t gives a straight line with gradient −λ
The decay constant
Since decay is random, we describe it with a probability. The decay constantλ is the chance that any one nucleus decays in a given time. A source with a high λ has lots of nuclei decaying each second, so a high activity.
Activity and the decay constantA = −ΔN/Δt = λNA = activity (Bq) • λ = decay constant (s−1) • N = number of undecayed nuclei
This tells you three things at once: a bigger λ gives a bigger activity; the activity depends on how many undecayed nuclei are left; and the minus sign shows the number of nuclei is decreasing.
Think of λ as each nucleus’s personal “chance of popping” per second. If you have N nuclei each with that chance, the total number popping per second is λN — that’s the activity. As nuclei decay, N shrinks, so the activity shrinks too. That’s exactly why decay slows down over time.
The exponential decay equation
Because the activity depends on how many nuclei are left, the number of nuclei doesn’t fall in a straight line — it falls exponentially:
Exponential decay of nucleiN = N0e−λtN0 = initial number of nuclei • N = nuclei remaining at time t
The same exponential shape applies to activity (A = A0e−λt) and count rate, because all three are proportional to the number of nuclei.
Linking half-life and decay constant
Here’s the key relationship. Half-life is the time for N to fall to half of N0. Substituting N = ½N0 into the exponential equation and rearranging gives a clean link:
Deriving the link½N0 = N0e−λt½ → ½ = e−λt½take ln of both sides: ln(½) = −λt½ → λt½ = ln 2
Half-life and decay constantt½ = ln 2 / λ
This shows half-life and decay constant are inversely proportional: the shorter the half-life, the larger the decay constant, and the faster the decay.
Large λ
means
Short half-life
so
Fast decay
The two quantities are just two ways of saying the same thing. A high decay constant means each nucleus is very likely to decay soon, so the sample halves quickly — a short half-life. ln 2 (about 0.693) is simply the conversion factor between them. Keep λ and t½ in matching time units (both per second, or both per year) and you’ll never go wrong.
The log graph
Curves are hard to read accurately, so we straighten the exponential out with logs. Taking the natural log of N = N0e−λt gives a straight line:
Log form — a straight lineln N = −λt + ln N0 ↔ y = mx + cgradient = −λ • y-intercept = ln N0
Plotting ln N against time straightens the exponential. The gradient is −λ and the y-intercept is ln N0.
WE 1
In a 1 g sample of radium-226, 2.22 × 1012 atoms decay in 1 minute. A second sample of 3.2 × 1022 radium-226 atoms has an activity of 12 Ci. (a) Find the value of 1 curie (Ci) in Bq. (b) Find the decay constant of radium-226. (1 Ci = 3.7 × 1010 Bq)
(a) Step 1 — activity in decays per secondA = ΔN/Δt = 2.22 × 10¹² ÷ 601 Ci = 3.7 × 10¹⁰ Bq(b) Step 2 — activity of the second sample in BqA = 12 × (3.7 × 10¹⁰) = 4.44 × 10¹¹ BqStep 3 — decay constant from A = λNλ = A/N = (4.44 × 10¹¹) ÷ (3.2 × 10²²)λ = 1.4 × 10⁻¹¹ s⁻¹Convert the minute to 60 s first. Then use A = λN → λ = A/N. Keep everything in seconds so the decay constant comes out per second.
WE 2
Strontium-90 has a half-life of 28.0 years. (a) Calculate its decay constant in year−1. (b) Determine the fraction of a sample remaining after 50 years.
(a) Step 1 — rearrange t½ = ln2/λλ = ln 2 / t½ = ln 2 / 28λ = 0.025 year⁻¹(b) Step 2 — exponential decay for the fractionN/N₀ = e⁻λᵗ = e⁻⁽⁰․₀₂₅₋⁵₀⁾N/N₀ = e⁻¹․²₅ = 0.28728.7% remains after 50 yearsBecause 50 years isn’t a whole number of half-lives, the (½)ⁿ trick won’t give a clean answer — the exponential equation handles it exactly. Keep λ and t in the same unit (years here).
⚛ Working a decay constant question
Given half-life, need λ? Use λ = ln 2 / t½ (or the reverse).
Given activity and N? Use A = λN.
Amount after any time? Use N = N0e−λt (works for any t).
Keep units matched: λ per second with t in seconds, or per year with years.
From a log graph? gradient = −λ.
💡 Top tips
t½ = ln 2 / λ — the one equation to memorise here.
Half-life and decay constant are inversely proportional.
Keep λ and t in the same time unit.
Use the exponential equation for non-whole numbers of half-lives.
On a log graph, the gradient is −λ, intercept is ln N0.
⚠ Common mistakes
Mixing time units — λ in s−1 with t in years
Forgetting the minus sign in the exponent of e−λt
Reading the log-graph gradient as +λ instead of −λ
Using the halving rule for a non-whole number of half-lives
Confusing N (nuclei left) with ΔN (nuclei decayed)
Quick recap: The decay constantλ is the per-second probability a nucleus decays. Activity is A = λN, and the number of nuclei decays exponentially as N = N0e−λt. Half-life links to λ by t½ = ln 2 / λ — they’re inversely proportional. A log plot of ln N against t is a straight line with gradient −λ.
You’ve now got both tools: the quick halving rule and the exact exponential equation. The final page pulls it all together into the formal law of radioactive decay — the single statement that the rate of decay is proportional to how many nuclei remain, and everything else that flows from it. Next page: The Radioactive Decay Law.
Decay constant equations feeling fiddly?
Book a free meeting and we’ll drill t½ = ln2/λ, the exponential equation, and log graphs until they click.
