Applications — radiocarbon dating, medical (Tc-99m, I-131), RTG (Pu-238 in spacecraft): Worked examples

The scenario matrix

Every exam question about applied radioactivity is one of these cells. We solve at least one example per cell.

Cell	What is unknown?	Which rearrangement of $N = N_{0} e^{- λ t}$ ?	Example
A. Find remaining amount	$N$ (given $t$ )	forward: $N = N_{0} e^{- λ t}$	Ex 1 (Pu-238), Ex 6
B. Find elapsed time	$t$ (given $N / N_{0}$ )	$t = \frac{1}{λ} ln \frac{N _{0}}{N}$	Ex 2 (C-14 dating)
C. Whole-number half-lives	$n$ (nice fractions)	$N / N_{0} = (1/2)^{n}$	Ex 3 (Tc-99m)
D. Zero / degenerate input	$t = 0$ , or $t \to \infty$	limiting behaviour	Ex 4 (limits)
E. Activity vs number	convert $A \leftrightarrow N$	$A = λ N$	Ex 5 (I-131 dose)
F. Real-world word problem	mixed, hidden data	pick the right form	Ex 7 (spacecraft)
G. Exam twist	two isotopes / ratio	two decay curves	Ex 8 (generator milking)
H. Find the rate itself	$λ$ or $t_{1/2}$ (given $N$ , $N_{0}$ , $t$ )	$λ = \frac{1}{t} ln \frac{N _{0}}{N}$	Ex 9 (unknown isotope)
I. Find the start	$N_{0}$ (given $N$ and $t$ )	$N_{0} = N e^{+ λ t}$	Ex 10 (back-calculate)

Definition Every symbol on this page, in plain words

$N_{0}$ = how many radioactive atoms you START with at $t = 0$ (a count, or anything ∝ count: mass, power, activity).
$N$ = ==how many are left after waiting time $t$ == (same kind of quantity as $N_{0}$ ).
$t$ = the elapsed time — how long you have waited since the start ( $t = 0$ ). It is the input in Cell A and the unknown in Cell B. Always measured in the SAME unit as $t_{1/2}$ .
$λ$ = the decay constant — chance per atom per unit time that it pops. Big $λ$ = fast decay. See Half-life and Decay Constant.
$t_{1/2}$ = time for HALF to go. Linked by $λ t_{1/2} = ln 2$ .
$A$ = activity — the number of decays per second (or per minute); $A = λ N$ , so it is always proportional to $N$ . $A_{0}$ = the activity at the start ( $t = 0$ ). See Activity and Units (Becquerel, Curie).
$P$ = thermal power (watts) an RTG puts out. Since heat comes from decays, $P \propto$ decay rate $\propto N$ , so $P$ obeys the exact same curve: $P = P_{0} e^{- λ t}$ , with $P_{0}$ the starting power.

Golden rule for units: whatever unit $t_{1/2}$ is in (years, hours, days), $λ$ comes out in "per that unit", and your times $t$ will be in that same unit. Keep them consistent and you cannot go wrong.

Figure — Applications — radiocarbon dating, medical (Tc-99m, I-131), RTG (Pu-238 in spacecraft)

The picture above is the master curve, with both axes labelled. Cell A reads up from a chosen time on the horizontal axis to the curve height (yellow arrow). Cell B reads across from a known fraction on the vertical axis to the time (white arrow). Cell C just counts the blue dashed halvings. Every example below points back to one of these three motions.

Cell A — find the remaining amount

Worked example Ex 1 · Pu-238 power after 40 years (Cell A)

An RTG starts at $P_{0} = 470 W$ of thermal power. Pu-238 has $t_{1/2} = 87.7$ yr. What thermal power $P$ remains after $t = 40$ years?

Forecast: 40 yr is less than half a half-life, so guess: still most of it — maybe ~ $70 - 80%$ ?

This is the yellow "read UP" arrow on the master curve: we pick $t = 40$ on the horizontal axis and want the curve's height.

Find $λ$ . Why this step? We need the "per year" decay rate before any exponential. $λ = \frac{l n 2}{87.7} = 7.90 \times 1 0^{- 3} yr^{- 1}$
Compute the exponent. Why? $e^{- λ t}$ is the surviving fraction; power heat ∝ decay rate ∝ $N$ , so $P$ follows the SAME curve. $λ t = 7.90 \times 1 0^{- 3} \times 40 = 0.3162$
Surviving fraction. Why? This is the height on the master curve at $t = 40$ . $e^{- 0.3162} = 0.729$
Multiply by starting power. Why? Fraction × start = actual watts. $P = P_{0} e^{- λ t} = 470 \times 0.729 = 342 W$

Verify: $40$ yr $\approx 0.456$ half-lives. After exactly 0.456 half-lives you keep $2^{- 0.456} = 0.729$ — matches. And $342 < 470$ but well above half ( $235$ W). Units: watts × dimensionless = watts. ✓

Cell B — find the elapsed time (dating)

Worked example Ex 2 · A charcoal sample (Cell B, real dating)

Fresh wood gives $A_{0} = 15.3$ counts/min/g of $^{14}$ C. A charcoal fragment gives $A = 9.2$ counts/min/g. $t_{1/2} = 5730$ yr. How old is the charcoal (elapsed time $t$ )? (See Activity and Units (Becquerel, Curie) — counts ∝ activity ∝ $N$ .)

Forecast: it dropped to about $60%$ — that's less than one halving, so guess ~ $4000$ – $5000$ yr.

This is the white "read ACROSS" arrow: we know the fraction $A / A_{0}$ on the vertical axis and want the time. Figure s02 draws this exact read-off.

$λ = ln 2/5730 = 1.210 \times 1 0^{- 4} yr^{- 1}$ . Why? Convert half-life to the per-year rate so we can invert the exponential.
Ratio $A_{0} / A = 15.3/9.2 = 1.663$ . Why? This "depletion factor" is what the exponential must equal; it's $e^{+ λ t}$ .
Take the log. Why? $ln$ is the tool that undoes the exponential — it turns " $e$ to the power of something" back into "something". That something is $λ t$ . $ln (1.663) = 0.5086$
Divide by $λ$ . Why? We have $λ t = 0.5086$ ; dividing peels off $λ$ to leave time. $t = \frac{1}{λ} ln \frac{A _{0}}{A} = \frac{0.5086}{1.210 \times 1 0 ^{- 4}} = 4204 yr$

Verify: plug back: $A = 15.3 e^{- λ t} = 15.3 e^{- 0.5086} = 15.3 \times 0.601 = 9.2$ ✓. Age is under one half-life (5730 yr), consistent with "more than half remains". ✓

Cell C — whole-number half-lives

Worked example Ex 3 · Tc-99m clearing overnight (Cell C, clean fractions)

Tc-99m has $t_{1/2} = 6$ h. A patient is injected at 9 a.m. What fraction of the activity $A$ is left at 9 a.m. the next day (elapsed time $t = 24$ h)?

Forecast: 24 h is a bunch of 6-hour halvings — maybe a few percent left?

This is Cell C's motion: count the blue dashed halvings on the master curve.

Count half-lives. Why? When time is a whole multiple of $t_{1/2}$ , we can skip $λ$ entirely and just halve repeatedly. $n = \frac{t}{t _{1/2}} = \frac{24}{6} = 4$
Halve four times. Why? Each half-life multiplies by $\frac{1}{2}$ ; four of them multiply by $(\frac{1}{2})^{4}$ . $(\frac{1}{2})^{4} = \frac{1}{16} = 0.0625$

Result: $6.25%$ remains — the isotope "self-limits" the dose.

Verify (via the general formula): $λ = ln 2/6 = 0.1155 h^{- 1}$ , then $e^{- 0.1155 \times 24} = e^{- 2.773} = 0.0625$ ✓. Whole-half-life shortcut and full formula agree exactly.

Cell D — zero and limiting inputs (the degenerate cases)

Worked example Ex 4 · What the formula says at the extremes (Cell D)

Check three boundary values of the elapsed time $t$ so you never panic when $t = 0$ or $t$ is huge.

Forecast: at $t = 0$ nothing has decayed (100%); at $t \to \infty$ everything is gone (0%); at exactly one half-life, 50%.

$t = 0$ : Why check? The clock just started — the formula must give back the start. $N = N_{0} e^{- λ \cdot 0} = N_{0} e^{0} = N_{0} \times 1 = N_{0} (100%)$
$t \to \infty$ : Why? Wait forever — must approach zero, never negative (you cannot have "negative atoms"). $e^{- λ t} \to 0 as t \to \infty \Rightarrow N \to 0^{+}$ The curve approaches the axis but never touches it — decay is asymptotic, which is exactly why " $^{14}$ C after 10 half-lives is unmeasurable, not literally zero" (see the dinosaur mistake in the parent note).
$t = t_{1/2}$ : Why? Sanity anchor. $e^{- λ t_{1/2}} = e^{- l n 2} = \frac{1}{2} (50%)$

Verify: all three read directly off the master curve — start at 1, halve at $t_{1/2}$ , hug zero at the right edge. ✓

Cell E — activity ↔ number of atoms

Worked example Ex 5 · How many I-131 atoms in a therapy dose? (Cell E)

A thyroid therapy dose has activity $A = 3.7 \times 1 0^{9}$ Bq (i.e. $3.7 \times 1 0^{9}$ decays/second). I-131 has $t_{1/2} = 8.0$ days. How many I-131 atoms $N$ is that?

Forecast: activity is decays per second but the atoms live days, so there should be a HUGE number of atoms per unit of activity.

Activity links to number by $A = λ N$ . Why this tool? Activity is "how many pop per second"; that equals the per-atom rate $λ$ times how many atoms exist. Rearranged: $N = A / λ$ . See Radioactive Decay Law.
Get $λ$ in per-second (match the Bq!). Why seconds? Bq is per second, so $λ$ must also be per second or the units clash. $t_{1/2} = 8.0 d = 8.0 \times 86400 = 6.912 \times 1 0^{5} s$ $λ = \frac{l n 2}{6.912 \times 1 0 ^{5}} = 1.003 \times 1 0^{- 6} s^{- 1}$
Divide. Why? $N = A / λ$ gives the standing population. $N = \frac{3.7 \times 1 0 ^{9}}{1.003 \times 1 0 ^{- 6}} = 3.69 \times 1 0^{15} atoms$

Verify: dimension check $\frac{decays/s}{1/s} = decays$ = atoms ✓. Reasonableness: $\sim 1 0^{15}$ atoms is a sub-microgram of iodine — tiny mass, big activity, exactly as expected. ✓

Cell A again — a "how much decayed" twist

Cell F — full real-world word problem

Worked example Ex 7 · Will the spacecraft still power its heaters? (Cell F)

A deep-space probe needs at least $200$ W of thermal power to keep instruments warm. Its Pu-238 source starts at $P_{0} = 290$ W. Mission planners want it alive at Pluto, reached $t = 9.5$ years after launch, and ideally through an extended mission to $t = 30$ years. Does it clear $200$ W at both times? ( $t_{1/2} = 87.7$ yr.)

Forecast: Pu-238 is slow (88 yr), so after only 9.5 yr almost nothing lost; even at 30 yr it should still beat 200 W.

One $λ$ for everything. Why? Same isotope → same decay constant for both times. $λ = \frac{l n 2}{87.7} = 7.90 \times 1 0^{- 3} yr^{- 1}$
Power at 9.5 yr. Why? Heat ∝ $N$ , so power obeys $P = P_{0} e^{- λ t}$ . $P = 290 e^{- 7.90 \times 1 0^{- 3} \times 9.5} = 290 e^{- 0.0750} = 290 \times 0.9277 = 269 W$
Power at 30 yr. Why? Check the extended-mission requirement too. $P = 290 e^{- 7.90 \times 1 0^{- 3} \times 30} = 290 e^{- 0.2369} = 290 \times 0.7891 = 229 W$

Result: $269$ W at Pluto and $229$ W at 30 yr — both above 200 W, mission is powered throughout.

Verify: monotonic decrease $290 \to 269 \to 229$ ✓ (later time = less power, never more). Both exceed $200$ W ✓. Units W throughout. This is exactly why real RTGs are launched with power margin — they fade slowly but predictably.

Cell G — the exam twist (two isotopes)

Worked example Ex 8 · Milking the Mo-99 / Tc-99m generator (Cell G)

A generator holds $^{99}$ Mo ( $t_{1/2} = 66$ h) which decays into $^{99 m}$ Tc. After a technician "milks" all the Tc out at 9 a.m., how much of the parent Mo-99 is left to keep making Tc after elapsed time $t = 48$ h? And what fraction of Mo-99's activity remains after one week ( $t = 168$ h)?

Forecast: 48 h is under one Mo half-life (66 h), so most of the "cow" survives; a week is a bit over 2.5 half-lives, so maybe ~15% left.

Mo-99 decay constant. Why? The parent's own decay sets how long the generator stays useful; use Mo's $t_{1/2}$ , not Tc's. $λ_{Mo} = \frac{l n 2}{66} = 1.050 \times 1 0^{- 2} h^{- 1}$
Fraction of Mo after 48 h. Why? Same forward formula, parent's clock. $e^{- 1.050 \times 1 0^{- 2} \times 48} = e^{- 0.5041} = 0.604 (60.4%)$
Fraction of Mo after 168 h (one week). Why? Shows the generator's shelf life. $e^{- 1.050 \times 1 0^{- 2} \times 168} = e^{- 1.764} = 0.171 (17.1%)$

Result: $\approx 60%$ of the parent survives the first two days; only $\approx 17%$ after a week — which is why hospitals replace the generator roughly weekly.

Verify: $168/66 = 2.545$ half-lives $\Rightarrow 2^{- 2.545} = 0.171$ ✓. Trap avoided: do NOT use Tc's 6 h half-life for the cow's lifetime — that's the milk, not the cow. Parent lifetime uses parent $λ$ . ✓

Cell H — find the rate itself ( $λ$ and $t_{1/2}$ )

Worked example Ex 9 · Measuring an unknown isotope's half-life (Cell H)

In a lab you count a fresh sample and get $N_{0} = 8000$ decays in the first minute. Exactly $5.0$ hours later the same sample gives $N = 500$ decays per minute. You do NOT know the isotope. Find its decay constant $λ$ and half-life $t_{1/2}$ .

Forecast: the count fell from 8000 to 500 — that's $8000/500 = 16 = 2^{4}$ , so about 4 halvings in 5 h, hinting $t_{1/2} \approx 1.25$ h.

This is the inverse of Cell A: here $N$ , $N_{0}$ and $t$ are all known, and the RATE is unknown. We must peel $λ$ out of the exponent.

Start from the law and isolate the exponential. Why? We want $λ$ , which currently sits inside $e^{- λ t}$ . $\frac{N}{N _{0}} = e^{- λ t} \Rightarrow \frac{N _{0}}{N} = e^{+ λ t}$
Take the log to free $λ t$ . Why? $ln$ undoes $e$ , exactly as in Cell B, giving $λ t$ . $ln \frac{N _{0}}{N} = λ t \Rightarrow λ = \frac{1}{t} ln \frac{N _{0}}{N}$
Plug in the numbers. Why? $N_{0} / N = 8000/500 = 16$ and $t = 5$ h. $λ = \frac{1}{5} ln 16 = \frac{2.7726}{5} = 0.5545 h^{- 1}$
Convert $λ$ to half-life. Why? $t_{1/2} = ln 2/ λ$ turns the rate into the more intuitive "time to halve". $t_{1/2} = \frac{l n 2}{0.5545} = \frac{0.6931}{0.5545} = 1.25 h$

Verify: $5$ h $/1.25$ h $= 4$ half-lives, and $2^{- 4} = 1/16$ , so $8000/16 = 500$ ✓ — matches the measured count exactly. Forecast nailed. Units: $ln$ of a ratio is dimensionless, divided by hours gives per-hour. ✓

Cell I — find the starting amount $N_{0}$

Worked example Ex 10 · Back-calculating the original dose (Cell I)

A vial of Tc-99m ( $t_{1/2} = 6$ h) is measured $t = 9.0$ h after it was prepared and now reads an activity $A = 250 MBq$ . What was the activity $A_{0}$ when it was freshly prepared?

Forecast: 9 h is $1.5$ half-lives, so the sample dropped to about $2^{- 1.5} \approx 0.35$ of the start — meaning the original was roughly $3 \times$ bigger, near $700$ MBq.

This is Cell A run backwards in time: we know a LATER value and want the earlier start, so we multiply UP by $e^{+ λ t}$ .

Rearrange the law for $N_{0}$ (here $A_{0}$ ). Why? $A = A_{0} e^{- λ t}$ ; divide both sides by $e^{- λ t}$ , which flips the sign in the exponent. $A_{0} = A e^{+ λ t}$
Find $λ$ . Why? We need the per-hour rate to build the exponent. $λ = \frac{l n 2}{6} = 0.1155 h^{- 1}$
Build the growth factor. Why? Going backward multiplies UP, so the exponent is positive. $e^{+ λ t} = e^{0.1155 \times 9} = e^{1.0397} = 2.828$
Multiply. Why? Start = now × (how much it has shrunk by), which is the growth factor. $A_{0} = 250 \times 2.828 = 707 MBq$

Verify: run it forward — $A = 707 e^{- 0.1155 \times 9} = 707 \times 0.3536 = 250$ MBq ✓. Also $9$ h $= 1.5$ half-lives, $2^{1.5} = 2.828$ ✓, and $707 \approx 3 \times$ the reading, matching the forecast. ✓

Recall check

Recall Which rearrangement for which unknown?

Given $t$ , want remaining amount ::: forward: $N = N_{0} e^{- λ t}$ Given the ratio $N / N_{0}$ , want elapsed time ::: $t = \frac{1}{λ} ln \frac{N _{0}}{N}$ Given $N$ , $N_{0}$ and $t$ , want the rate ::: $λ = \frac{1}{t} ln \frac{N _{0}}{N}$ , then $t_{1/2} = ln 2/ λ$ Given a later $N$ and $t$ , want the start $N_{0}$ ::: $N_{0} = N e^{+ λ t}$ (multiply UP, positive exponent) Ratio is a clean power of $\frac{1}{2}$ , want half-lives ::: $n$ where $N / N_{0} = (1/2)^{n}$ Convert activity (Bq) to number of atoms ::: $N = A / λ$ (with $λ$ in per-second) Fraction that has DECAYED after time $t$ ::: $1 - e^{- λ t}$ (one minus survived)