Decay law — N = N₀ e^(−λt), half-life, activity: Worked examples

Before anything, one reminder of the three tools we lean on, and why each:

Recall The three moves and when to use each

Multiplicative-halving ::: Use $N = N_{0} (1/2)^{t / T_{1/2}}$ when $t$ is a clean multiple of the half-life — no calculator needed. Exponential form ::: Use $N = N_{0} e^{- λ t}$ (or $A = A_{0} e^{- λ t}$ ) when $t$ is not a whole number of half-lives — the smooth curve handles any $t$ . Logarithm ::: Use $ln$ to undo the exponential when the unknown is trapped in the exponent (finding $t$ or $λ$ ). $ln$ is the question " $e$ to what power gives this?"

The scenario matrix

Every exam question about decay is one (or a mix) of these cells:

#	Case class	Given → Find	Right tool
A	Integer half-lives	$t = n T_{1/2}$ → fraction left	halving $(1/2)^{n}$
B	Non-integer half-lives	arbitrary $t$ → fraction left	exponential
C	Backwards for time	fraction left → $t$	logarithm
D	Backwards for $λ$ / $T_{1/2}$	two readings → $λ$	logarithm
E	Activity numbers	$N_{0}, T_{1/2}$ → $A_{0}$ in Bq	$A = λ N$ + unit care
F	Degenerate / limiting	$t = 0$ and $t \to \infty$	read the curve's ends
G	Real-world word problem	carbon dating style	logarithm
H	Exam twist — two isotopes	mixed sample → later ratio	two exponentials

The eight worked examples below cover one cell each. Together they leave no cell empty.

Figure — Decay law — N = N₀ e^(−λt), half-life, activity

Look at the curve: it starts at $N_{0}$ (left edge, $t = 0$ ), never touches the axis (that is cell F), and every case A–H is just "which point on this curve, or which slope, do we ask about?"

Case A — clean integer half-lives

Case B — non-integer half-lives

Worked example Same source,

$T_{1/2} = 8$ days. What fraction remains after 20 days? Forecast: 20 days is between 2 and 3 half-lives, so the fraction sits between $1/4$ and $1/8$ . Guess a number.

Step 1 — We cannot count whole halvings. $20/8 = 2.5$ — not an integer. So the halving shortcut starts the estimate but the exponential finishes it. Why this step? $(1/2)^{2.5}$ needs a calculator anyway; we go to the smooth curve.

Step 2 — Use the fractional exponent directly. $fraction = (\frac{1}{2})^{2.5} = 2^{- 2.5} .$ Why this step? $N = N_{0} (1/2)^{t / T_{1/2}}$ is valid for any real $t / T_{1/2}$ , integer or not.

Step 3 — Evaluate. $2^{- 2.5} = 2^{- 2} \cdot 2^{- 0.5} = \frac{1}{4} \cdot \frac{1}{2} \approx 0.25 \times 0.7071 = 0.1768$ .

Verify: Bounds check — $0.125 < 0.1768 < 0.25$ , sitting between the 3rd and 2nd half-life values exactly as forecast. ✓ Answer: $\approx 0.177 = 17.7%$ .

Case C — solve backwards for the time

Worked example A sample's activity falls to

30% of its start value. Its decay constant is $λ = 0.05 day^{- 1}$ . How long did that take? Forecast: Roughly 1.5 half-lives gives 35%, so guess: a bit more than 1.5 half-lives. $T_{1/2} = ln 2/0.05 \approx 13.9$ days, so guess ~24 days.

Step 1 — Write the activity ratio. $\frac{A}{A _{0}} = 0.30 = e^{- λ t}$ . Why this step? Activity obeys the same law $A = A_{0} e^{- λ t}$ , so its fractional drop equals the nuclei's fractional drop.

Step 2 — The unknown $t$ is stuck in the exponent — take $ln$ . $ln (0.30) = - λ t .$ Why $ln$ ? It is the inverse of $e^{(\cdot)}$ ; it pulls the exponent down so we can isolate $t$ .

Step 3 — Solve. $t = \frac{- ln ( 0.30 )}{λ} = \frac{1.2040}{0.05} = 24.1$ days. Why the minus sign survives? $ln (0.30)$ is negative (number below 1); the two minuses make $t$ positive, as time must be.

Verify: Substitute back: $e^{- 0.05 \times 24.1} = e^{- 1.204} = 0.300$ . ✓ Units: $day / day^{- 1}$ ... actually $[ln (\cdot)] / [day^{- 1}] = day$ . ✓ Answer: $\approx 24.1 days$ .

Case D — solve backwards for $λ$ and $T_{1/2}$

Worked example At

$t = 0$ a counter reads $A_{0} = 8000$ Bq. Two hours later it reads $A = 2000$ Bq. Find $λ$ and $T_{1/2}$ . Forecast: Activity dropped by a factor of 4 in 2 h. Factor 4 = two halvings. So guess $T_{1/2}$ = 1 hour.

Step 1 — Ratio the two readings. $\frac{A}{A _{0}} = \frac{2000}{8000} = \frac{1}{4} = e^{- λ t}$ with $t = 2$ h. Why ratio? The unknown $A_{0}$ cancels — we only need how the reading changed, not absolute counts.

Step 2 — Take $ln$ to free $λ$ . $ln \frac{1}{4} = - λ (2) \Rightarrow λ = \frac{- ln ( 1/4 )}{2} = \frac{ln 4}{2} = \frac{1.3863}{2} = 0.6931 h^{- 1} .$ Why this works: $ln$ converts the multiplicative drop into an additive one that $t$ scales linearly.

Step 3 — Convert $λ$ to $T_{1/2}$ . $T_{1/2} = \frac{ln 2}{λ} = \frac{0.6931}{0.6931} = 1.00$ h.

Verify: One half-life = 1 h, and 2 h = 2 half-lives → factor $(1/2)^{2} = 1/4$ , matching $2000/8000$ . Forecast confirmed. ✓ Answer: $λ = 0.693 h^{- 1}, T_{1/2} = 1.0 h$ .

Case E — activity in becquerels (unit care)

Case F — degenerate and limiting inputs

Worked example For

$N = N_{0} e^{- λ t}$ : (i) What is $N$ at $t = 0$ ? (ii) What does $N$ approach as $t \to \infty$ ? (iii) Can $N$ ever equal exactly zero? Forecast: Predict all three before reading. Especially (iii) — yes or no?

Step 1 — Plug $t = 0$ . $N = N_{0} e^{0} = N_{0} \times 1 = N_{0}$ . Why? $e^{0} = 1$ ; at the start nothing has decayed. This is the left edge of the figure.

Step 2 — Let $t \to \infty$ . As $t$ grows, $- λ t \to - \infty$ and $e^{- λ t} \to 0$ , so $N \to 0$ . Why the curve hugs the axis: the exponential shrinks but is always positive.

Step 3 — Ask if $N = 0$ is ever reached. $e^{- λ t} > 0$ for every finite $t$ . So $N > 0$ always; zero is only the limit, never an actual value. Why this matters: it kills the "after 2 half-lives it's gone" myth — the curve is asymptotic.

Verify: From the figure, the red curve never crosses the horizontal axis; it only flattens toward it. Numerically, at $t = 20 T_{1/2}$ a fraction $2^{- 20} \approx 9.5 \times 1 0^{- 7}$ still survives — tiny but nonzero. ✓ Answer: $N (0) = N_{0}, N (\infty) \to 0, N = 0 never exactly reached$ .

Case G — real-world word problem (dating)

Case H — exam twist: a two-isotope mixture

Worked example A sample starts with

equal numbers $N_{0}$ of two isotopes: X with $T_{1/2} = 1$ h and Y with $T_{1/2} = 2$ h. After 2 hours, what is the ratio (nuclei of X) : (nuclei of Y)? Forecast: X halves faster, so after 2 h there should be fewer X than Y — guess a ratio below 1, like 1:2.

Step 1 — Track each isotope separately. They decay independently, each with its own $λ$ . $N_{X} = N_{0} (\frac{1}{2})^{2/1} = N_{0} (\frac{1}{2})^{2} = \frac{N _{0}}{4} .$ $N_{Y} = N_{0} (\frac{1}{2})^{2/2} = N_{0} (\frac{1}{2})^{1} = \frac{N _{0}}{2} .$ Why separately? Decay of one nucleus is unaffected by the presence of another type — no interaction.

Step 2 — Form the ratio. $\frac{N _{X}}{N _{Y}} = \frac{N _{0} /4}{N _{0} /2} = \frac{1}{2}$ . Why $N_{0}$ cancels? Both started equal, so the common factor drops out.

Verify: X ran 2 half-lives → 1/4 left; Y ran 1 half-life → 1/2 left. Ratio $(1/4) / (1/2) = 1/2$ , matching the forecast that X is scarcer. ✓ Answer: $N_{X} : N_{Y} = 1 : 2$ .

One picture that unifies all cases

The mint curve (slow, $T_{1/2} = 2$ h) always sits above the coral curve (fast, $T_{1/2} = 1$ h) — that is Case H in one glance. The dashed markers show where Cases A, C, and G read the height off the axis.

Connections

Parent: Decay law
Exponential functions and natural log — the $ln$ move in Cases C, D, G
First-order chemical kinetics — identical maths
Carbon dating — Case G in the wild
Radioactivity — alpha, beta, gamma decay