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

Quick symbol reminder (all from the parent):

$N$ = number of undecayed nuclei still present; $N_{0}$ = that number at the start ( $t = 0$ ).
$λ$ = decay constant, the constant probability per second that any one nucleus decays. Units $s^{- 1}$ .
$T_{1/2} = \frac{ln 2}{λ}$ = half-life, the time to halve $N$ .
$τ = \frac{1}{λ}$ = mean life.
$A = λ N$ = activity, decays per second (Bq).

Level 1 — Recognition

L1.1 A radioactive source has decay constant $λ = 2.0 \times 1 0^{- 3} s^{- 1}$ . Write down its mean life $τ$ and half-life $T_{1/2}$ .

Recall Solution L1.1

What we do: pick the right one-liners. $τ = 1/ λ = 1/ (2.0 \times 1 0^{- 3}) = 500 s$ . $T_{1/2} = ln 2/ λ = 0.693/ (2.0 \times 1 0^{- 3}) = 346.6 s$ . Sanity: $τ > T_{1/2}$ always — the long-tail survivors pull the average up. ✓

L1.2 After exactly 4 half-lives, what fraction of the original nuclei remains?

Recall Solution L1.2

Each half-life multiplies by $\frac{1}{2}$ . After $n = 4$ : fraction $= (\frac{1}{2})^{4} = \frac{1}{16} = 0.0625$ , i.e. $6.25%$ . Why multiply, not add? Halving repeatedly is repeated multiplication — see the L1 mistake below.

L1.3 A sample undergoes $8.0 \times 1 0^{9}$ decays per second. Express this activity in becquerel and in curie.

Recall Solution L1.3

$1 Bq = 1$ decay/s, so activity $= 8.0 \times 1 0^{9} Bq$ . $1 Ci = 3.7 \times 1 0^{10} Bq$ , so activity $= \frac{8.0 \times 1 0 ^{9}}{3.7 \times 1 0 ^{10}} = 0.216 Ci$ .

Level 2 — Application

L2.1 A sample starts with $N_{0} = 6.0 \times 1 0^{18}$ nuclei and has $T_{1/2} = 8.0$ hours. How many remain after 24 hours?

Recall Solution L2.1

Step 1: $n = t / T_{1/2} = 24/8 = 3$ half-lives (whole number → use the halving form). Step 2: $N = N_{0} (1/2)^{3} = 6.0 \times 1 0^{18} \times \frac{1}{8} = 7.5 \times 1 0^{17}$ nuclei.

L2.2 For a source with $T_{1/2} = 10$ hours and $N_{0} = 1.0 \times 1 0^{20}$ nuclei, find the initial activity $A_{0}$ in Bq.

Recall Solution L2.2

Step 1 (why seconds): activity is per second, so convert $T_{1/2}$ : $10 \times 3600 = 36000 s$ . $λ = 0.693/36000 = 1.925 \times 1 0^{- 5} s^{- 1}$ . Step 2: $A_{0} = λ N_{0} = (1.925 \times 1 0^{- 5}) (1.0 \times 1 0^{20}) = 1.925 \times 1 0^{15} Bq \approx 1.9 \times 1 0^{15} Bq$ .

L2.3 How long does it take activity to fall to $30%$ of its initial value if $λ = 0.05 day^{- 1}$ ?

Recall Solution L2.3

Why the exponential, not halving: $30%$ is not a nice power of $\frac{1}{2}$ , so use $A = A_{0} e^{- λ t}$ . $0.30 = e^{- λ t} \Rightarrow ln 0.30 = - λ t \Rightarrow t = \frac{- ln 0.30}{0.05} = \frac{1.204}{0.05} = 24.1 days$ .

Level 3 — Analysis

L3.1 The activity of a source drops from $8000 Bq$ to $1000 Bq$ in $9.0$ minutes. Find the half-life.

Recall Solution L3.1

Step 1 (recognise the ratio): $A / A_{0} = 1000/8000 = 1/8 = (1/2)^{3}$ . So 3 half-lives passed. Step 2: $3 T_{1/2} = 9.0 min \Rightarrow T_{1/2} = 3.0 min$ . Cross-check via logs: $λ = ln (8) /9 = 2.079/9 = 0.231 min^{- 1}$ , and $T_{1/2} = 0.693/0.231 = 3.0 min$ . ✓

L3.2 A Geiger counter reads $A_{0} = 2400$ counts/s at $t = 0$ and $A = 300$ counts/s at $t = 30$ minutes. Find $λ$ (in $min^{- 1}$ ) and the mean life $τ$ .

Recall Solution L3.2

Step 1: $A / A_{0} = 300/2400 = 1/8 = (1/2)^{3}$ → 3 half-lives in 30 min → $T_{1/2} = 10$ min. Step 2: $λ = 0.693/10 = 0.0693 min^{- 1}$ . Step 3: $τ = 1/ λ = 1/0.0693 = 14.43 min$ (indeed $τ > T_{1/2}$ ). ✓

L3.3 Look at the decay curve in the figure below. Reading off the graph, one full half-life corresponds to which horizontal gap, and how does the slope (activity) change from the first half-life interval to the second?

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

Recall Solution L3.3

Reading it: the count halves ( $100 \to 50 \to 25 \to 12.5$ ) over equal horizontal steps — each step is one $T_{1/2}$ (here 1 unit). Slope: the curve is steepest at the start and flattens. Since $A = λ N$ and the slope magnitude is the activity, and $N$ halves each interval, the average steepness also halves each interval. The curve never touches zero (asymptotic).

Level 4 — Synthesis

L4.1 A source contains two independent isotopes:

Isotope X: $N_{0 X} = 1.0 \times 1 0^{12}$ nuclei, $T_{1/2, X} = 2.0$ hours.
Isotope Y: $N_{0 Y} = 4.0 \times 1 0^{12}$ nuclei, $T_{1/2, Y} = 8.0$ hours.

Find the total activity at $t = 8.0$ hours (in decays/hour).

Recall Solution L4.1

Why add activities: decays are independent events, so total $A = A_{X} + A_{Y}$ . Decay constants (per hour): $λ_{X} = 0.693/2 = 0.3466 h^{- 1}$ ; $λ_{Y} = 0.693/8 = 0.0866 h^{- 1}$ . Nuclei left at $t = 8$ h:

X: 4 half-lives → $N_{X} = 1.0 \times 1 0^{12} (1/2)^{4} = 6.25 \times 1 0^{10}$ .
Y: 1 half-life → $N_{Y} = 4.0 \times 1 0^{12} (1/2)^{1} = 2.0 \times 1 0^{12}$ . Activities: $A_{X} = λ_{X} N_{X} = 0.3466 \times 6.25 \times 1 0^{10} = 2.166 \times 1 0^{10}$ /h. $A_{Y} = λ_{Y} N_{Y} = 0.0866 \times 2.0 \times 1 0^{12} = 1.733 \times 1 0^{11}$ /h. Total: $A = 2.166 \times 1 0^{10} + 1.733 \times 1 0^{11} = 1.949 \times 1 0^{11}$ decays/hour.

L4.2 A parent nuclide decays with $λ = 0.10 day^{- 1}$ . Its activity today is $5.0 \times 1 0^{6} Bq$ . What was its activity exactly 5 days ago?

Recall Solution L4.2

Key idea (run the clock backward): $A (t) = A_{0} e^{- λ t}$ . Going back 5 days means the earlier activity was larger by $e^{+ λ \cdot 5}$ . $A_{past} = A_{now} e^{λ (5)} = 5.0 \times 1 0^{6} \times e^{0.10 \times 5} = 5.0 \times 1 0^{6} \times e^{0.5}$ . $e^{0.5} = 1.6487$ , so $A_{past} = 8.24 \times 1 0^{6} Bq$ .

Level 5 — Mastery

L5.1 In a sample, the activity is $A_{0}$ at $t = 0$ . What fraction of all the decays that will ever happen occur during the first mean life $τ$ ?

Recall Solution L5.1

Total decays ever: every one of $N_{0}$ nuclei eventually decays → total $= N_{0}$ . Decays in $[0, τ]$ : that's $N_{0} - N (τ)$ . Since $N (τ) = N_{0} e^{- λ τ} = N_{0} e^{- 1}$ (because $λ τ = 1$ ), decays so far $= N_{0} (1 - e^{- 1})$ . Fraction $= 1 - e^{- 1} = 1 - 0.3679 = 0.6321$ , i.e. $\approx 63.2%$ . Meaning: almost two-thirds of all decays happen before one mean life — the process front-loads its activity.

L5.2 A sample's activity is measured at three times: $A_{0}$ at $t = 0$ , then at $t = T$ it is $A_{0} /2$ . Predict the activity at $t = 3 T$ , and show the general rule $A (k T) = A_{0} (1/2)^{k}$ follows purely from the constant-fraction property, without ever computing $λ$ .

Recall Solution L5.2

Given: in time $T$ the activity halves — so $T$ is the half-life here. Constant-fraction property: in every equal interval the same fraction survives. Over one $T$ , factor $\frac{1}{2}$ . Over $k$ intervals of length $T$ : factor $(\frac{1}{2})^{k}$ . Therefore $A (3 T) = A_{0} (1/2)^{3} = A_{0} /8$ . Why no $λ$ needed: the exponential $A_{0} e^{- λ t}$ can be rewritten $A_{0} (1/2)^{t / T}$ ; at $t = k T$ the exponent is exactly $k$ . The physics ("equal times → equal fractions") gives the answer directly.

L5.3 A radioactive nuclide is being produced at a constant rate $R = 1.0 \times 1 0^{9}$ nuclei/s while it also decays with $λ = 1.0 \times 1 0^{- 3} s^{- 1}$ . Find the steady-state number of nuclei $N_{eq}$ (the count at which production balances decay).

Recall Solution L5.3

What "steady state" means: the number stops changing, so $d N / d t = 0$ . Balance equation: $\frac{d N}{d t} = R - λ N$ . Set to zero: $R = λ N_{eq}$ . $N_{eq} = \frac{R}{λ} = \frac{1.0 \times 1 0 ^{9}}{1.0 \times 1 0 ^{- 3}} = 1.0 \times 1 0^{12}$ nuclei. At equilibrium the activity equals the production rate: $A = λ N_{eq} = R = 1.0 \times 1 0^{9} Bq$ — every nucleus made is (on average) matched by one decaying. This is the "secular equilibrium" idea used in reactor and medical-isotope production.

Connections

Parent: Decay Law — $N = N_0 e^{-\lambda t}$, Half-life, Activity
Exponential functions and natural log — the log step in L2.3, L4.2.
First-order chemical kinetics — identical math (rate ∝ amount).
Carbon dating — L2.3-style "solve for $t$ " problems.
Radioactivity — alpha, beta, gamma decay

Recall One-line self-test

Why does the two-isotope sample (L4.1) have no single half-life? ::: Its total count is a sum of two exponentials with different $λ$ ; only a single exponential has a single half-life.