Why we need exponentials: You already know that a sample loses half its un-decayed nuclei every half-life t½. That only tells you N at 1, 2, 3… whole half-lives. To find N at any time — say 5 days into an 8-day half-life — you need the smooth exponential law. That is the HL upgrade for this topic.
Decay is random but constant-rate: each nucleus has the same fixed probability of decaying per second, so the number left falls by the same proportion in equal times. That proportional fall is exactly what an exponential describes.
Two curves, same shape: Both the number of un-decayed nuclei N and the activity A (decays per second) follow the same exponential curve, because activity is just λ times the number of nuclei. Whatever fraction of nuclei is left, the activity has fallen to that same fraction.
- number of un-decayed nuclei now / at t = 0
- activity now / at t = 0 (becquerel, Bq)
- decay constant — probability of decay per unit time (s⁻¹ or day⁻¹)
- elapsed time (same unit as 1/λ)
Activity is the measurable one: A detector counts decays per second — it measures A, not N directly. Because A = λN, the activity is the most convenient thing to track, and it obeys the very same exponential as N.
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λ and t½ are two ways of saying the same thing: The decay constant λ and the half-life t½ both describe how fast a sample decays — they are locked together. A big λ means each nucleus is very likely to decay each second, so the half-life is short.
- decay constant (s⁻¹ or day⁻¹)
- half-life — time for half the nuclei to decay
- ≈ 0.693 (constant)
Worked example — decay constant from a half-life
A radioactive source has a half-life of 8.0 days. Find its decay constant λ in day⁻¹.
Solution
- Write the given relation between λ and t½:
- Put in t½ = 8.0 days:
- Work it out — keep the unit:
Final answer
λ = 0.087 day⁻¹. (Match the time unit of λ to the time unit of t you will use later.)
Watch the time unit: If λ is in day⁻¹, then t in must be in days. Mixing days with seconds is the classic slip — keep the units of λ and t the same.
Once you have λ you can find N (or A) at any time, not just at whole half-lives. Substitute λ and t into the exponential — the result is a fraction of the original that you can also check against the (½)ⁿ picture.
Worked example — fraction left after 24 days
For the same source (t½ = 8.0 days, so λ = 0.087 day⁻¹), find the fraction of nuclei remaining after t = 24 days.
Solution
- Start from the given decay law:
- Substitute λ = 0.087 day⁻¹ and t = 24 days:
- Evaluate the exponential:
Final answer
N = 0.125 N₀ = N₀/8. Check: 24 days = 3 half-lives, and (½)³ = ⅛ — the exponential and the 'n half-lives' picture agree.
Whole half-lives? Use the fast route: If t is a whole number n of half-lives, skip the exponential: the fraction left is just . Save for awkward times that are not a whole number of half-lives.
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Where it shows up: Quantitative decay is HL only (E.3):
- Paper 1A — a one-step 'find λ from t½', or 'what fraction / activity remains?'. - Paper 2 — determine an activity or a time using , often after first finding λ from the half-life.
Three easy marks: (1) Find λ = ln2/t½ first and keep its time unit. (2) Match t to that unit before substituting. (3) At a whole number of half-lives, just use (½)ⁿ — it is faster and avoids a calculator slip.
IB-style question — activity after three half-lives
The source above (t½ = 8.0 days) has an initial activity A₀ = 4.0 × 10⁶ Bq. Determine its activity after 24 days.
Solution
- Recognise 24 days = 3 half-lives, so use the given activity law:
- With λ = 0.087 day⁻¹ and t = 24 days the bracket is e−2.08 = ⅛:
- Work it out — keep the unit:
Final answer
A = 5.0 × 10⁵ Bq. (Activity falls in exactly the same proportion as the number of nuclei: ⅛ of the original.)