The mummified remains of
Ötzi the Iceman,
found frozen in the Ötztal Alps in 1991. Radiocarbon dating of his tissue and tools placed his death at about 3,300 BCE — more than 5,300 years ago. The technique relied on the slow, predictable decay of carbon-14: a single half-life of 5,730 years. Image: 120 / Wikimedia Commons —
File:Otzi-Quinson.jpg, CC BY-SA 3.0.
How can scientists determine the age of a 10,000-year-old bone
— without a time machine?
What is hidden inside an ancient artifact that counts how old it is?
Standards: HSA-SSE.A.1, HSA-SSE.B.3, HSF-LE.B.5, HSN-RN.A.1. This optional lesson is an application capstone for Section A. No new mathematical content — students apply fractional exponents and equal-factors-over-equal-intervals to radiocarbon dating problems. MP1 (sense-making) and MP2 (quantitative reasoning) are foreground.
Prep Checklist
☐ No printed materials needed — everything is on these slides.
☐ Calculators or spreadsheet access recommended for Activity 1 (table completion).
☐ If pacing is tight, skip Activity 2 (Old Manuscripts). The lesson synthesis covers the strategy menu without it.
☐ If skipping this lesson entirely (it is optional), proceed directly to Section B, Lesson 8.
Warm-Up5 min
Do Now
Complete on your own. Then confirm your answers with a partner.
A colony of microbes is growing exponentially and doubles in population every 6 hours. Explain why we could say the population grows by a factor of \(2^{1/6}\) every hour.
A bacteria population decreases exponentially by a factor of \(\tfrac{1}{3}\) every 4 hours. Explain why we could also say the population decays by a factor of \(\!\left(\tfrac{1}{3}\right)^{1/4}\) every hour.
5 min total for warm-up (slides 2–3). Students work for ~3 min, then pair-share. The aim is to re-activate the idea from Lesson 5: a fractional-interval factor is a root of the whole-interval factor — not a division.
Listen for students who say things like “the hourly factor must be such that, raised to the 6th power, it equals 2” (Problem 1) and “the hourly factor raised to the 4th power equals \(\tfrac{1}{3}\)” (Problem 2). That language is what to surface in the synthesis.
Common error to watch for: dividing 2 by 6, or dividing \(\tfrac{1}{3}\) by 4. The factor is multiplicative — equal factors over equal intervals — so we take a 6th root (or 4th root), not a quotient.
Warm-Up · Synthesis5 min
Halving and Doubling — Discussion
When a quantity multiplies by the same factor over equal intervals, the factor for a smaller interval is always a root — never a quotient.
“If the population doubles every 6 hours, what must the hourly factor satisfy?”If \(b\) is the hourly factor, then multiplying by \(b\) six times must give a doubling: \(b^{6} = 2\). The only positive number that works is \(b = 2^{1/6} \approx 1.122\). The population grows by about 12.2% every hour.
“Why isn't the hourly factor \(2 \div 6 = \tfrac{1}{3}\)?”Because growth here is multiplicative, not additive. Dividing would describe a constant amount added each hour — that is linear thinking. Equal factors over equal intervals means we repeat the same multiplication, six of which compound to 2.
“For the decay: what equation does the hourly factor \(b\) satisfy, and what is it?”\(b^{4} = \tfrac{1}{3}\), so \(b = \!\left(\tfrac{1}{3}\right)^{1/4} \approx 0.760\). Four hours of multiplying by 0.760 takes the population to one-third of its starting size. Each hour it keeps about 76% of the previous hour's population.
“What is the rule of thumb — in words?”If a quantity changes by a factor of \(k\) every \(n\) units, then in one unit it changes by a factor of \(k^{1/n}\). The \(n\)th root, never the \(n\)th quotient.
click to advance discussion ▶
Try Saying
The hourly factor must be \(b\) such that \(b^{n}\) equals ___, because ___.
Try Saying
Dividing the factor by \(n\) does/doesn't give the hourly factor because ___.
The deep idea here is the “equal factors over equal intervals” property of exponentials — restated for any interval. Today's activities apply this property to half-life: the factor over one half-life is \(\tfrac{1}{2}\), the factor over a fraction of a half-life is \((\tfrac{1}{2})^{\text{fraction}}\), and the factor over multiple half-lives is \((\tfrac{1}{2})^{\text{count}}\). All of today's calculations are the same move rehearsed here.
Activity 1 · Launch3 min
Radiocarbon Dating — the Big Idea
A radioactive element emits particles of energy as its atoms break down. Carbon-14 is one such element. Its decay rate is measured by its half-life — the time it takes for half of any sample to break down. The half-life of carbon-14 is 5,730 years.
Scientists measure how much carbon-14 remains in an ancient artifact, then use exponential decay to estimate when the carbon-14 was “reset” — the moment the plant or animal died.
How much C-14?
For every 1 part of carbon-14 in a living thing, there are about 1,000,000,000,000 parts of carbon-12 — one trillion to one. C-14 is rare, but the ratio is stable while alive.
After death
No new C-14 enters the body. What remains decays exponentially — half gone every 5,730 years, regardless of how much was there to start.
What are some ways we might get started on this problem? What quantities are we counting?
Launch (~3 min). Use the Three Reads protocol on the next slide. Do not reveal the table on slide 5 immediately — cover it or scroll the projector down so students see only the problem stem.
Read 1: What is this situation about? (Using an element to find the age of a fossil.) Read 2: What quantities can be counted or measured? (The half-life; the mass of C-14 in the fossil.) Read 3: Reveal the table. Ask: “What are some ways we might get started?”
Anchor: students should land on the idea that the table rows are spaced at half-life multiples (0, 5730, 11460, …) — and that 1,910 is one-third of 5,730.
Activity 1 · Work Time17 min
Radiocarbon Dating
A fossil had 3 picograms (a trillionth of a gram) of carbon-14 at a certain moment. Carbon-14 has a half-life of 5,730 years.
A scientist models the mass of C-14 in a different fossil with \(f(t) = 2.5 \cdot \!\left(\tfrac{1}{2}\right)^{t/5730}\), where \(t\) is the number of years after 20,000 BCE.
Years after fossil had 3 picograms of C-14
Mass of C-14 (picograms)
0
3
1,910
?
5,730
?
?
0.75
On your own, then compare with a partner
Complete the table with the missing mass and years.
What do the 2.5, the \(\tfrac{1}{2}\), and the 5,730 in \(f(t)\) mean in this situation?
Would more or less than 0.1 picogram of carbon-14 remain in this fossil today? Explain how you know.
17 min · Activity 1 Work Time. Leave this slide on screen during work time.
Table answers: Row 2 (1,910 years): \(3 \cdot (\tfrac{1}{2})^{1/3} \approx 2.38\) picograms — because 1,910 is one-third of 5,730. Row 3 (5,730 years): \(3 \cdot \tfrac{1}{2} = 1.5\) picograms (one half-life). Row 4 (mass 0.75): two half-lives, so \(t = 11{,}460\) years.
Meaning of constants: 2.5 is the picograms of C-14 in the sample at 20,000 BCE (initial value). The \(\tfrac{1}{2}\) and 5,730 together say the amount halves every 5,730 years.
Today question: About 22,000 years have passed since 20,000 BCE. That is less than 4 half-lives (4 × 5,730 = 22,920). After 4 half-lives, \(2.5 \cdot (\tfrac{1}{2})^{4} = 2.5/16 \approx 0.156\) picograms remain — more than 0.1. So today the fossil has more than 0.1 picogram.
Watch for students who mistake 1,910 as “one-third of the decay” and compute \(3 \cdot \tfrac{1}{3} = 1\). The correct move is the cube root of \(\tfrac{1}{2}\), not the third of \(\tfrac{1}{2}\). This is exactly the warm-up rehearsed in reverse.
Amplify · 4 inputs
Order: 4 separate Text→Math/Text Response→Note sequences, named answer1a, answer1b, answer1c, answer3. Problem 2 (interpretation) is a Text Response with no auto-grade.
Format examples: Example: 2.38 · Example: 1.5 · Example: 11460 · Example: more
Problem 1a: Mass of C-14 after 1,910 years (table row 2)
content:
when answer1a.submitted and answer1a.numericValue > 2.378 and answer1a.numericValue < 2.385
"Correct! 1,910 is one-third of 5,730, so the factor is (1/2)^(1/3). Then 3 · (1/2)^(1/3) ≈ 2.38 picograms."
when answer1a.submitted and answer1a.numericValue > 0.999 and answer1a.numericValue < 1.001
"Hmm... 1 picogram looks like 3 · (1/3) — dividing because 1,910 is one-third of 5,730. The factor for a fractional interval is a *root*, not a division. Try 3 · (1/2)^(1/3)."
when answer1a.submitted and answer1a.numericValue > 1.499 and answer1a.numericValue < 1.501
"Almost — 1.5 is the mass after a *full* half-life (5,730 years), not after 1,910 years. The factor for 1,910 years is (1/2)^(1910/5730) = (1/2)^(1/3)."
when answer1a.submitted and answer1a.numericValue > 0.499 and answer1a.numericValue < 0.501
"I see what happened — 0.5 looks like the factor (1/2) treated as the mass. Multiply 3 by (1/2)^(1/3), not by 1/2."
when answer1a.submitted and not isBlank(answer1a.latex)
"You're close — 1,910 years is one-third of one half-life, so the factor is (1/2)^(1/3). Multiply 3 picograms by that factor. What do you get?"
otherwise ""
Problem 1b: Mass of C-14 after 5,730 years (table row 3)
content:
when answer1b.submitted and answer1b.numericValue > 1.499 and answer1b.numericValue < 1.501
"Correct! 5,730 years is exactly one half-life, so the mass halves: 3 · (1/2) = 1.5 picograms."
when answer1b.submitted and answer1b.numericValue > 0.749 and answer1b.numericValue < 0.751
"Not quite — 0.75 is the mass after *two* half-lives (11,460 years). After one half-life, the mass is 3 · (1/2) = 1.5 picograms."
when answer1b.submitted and answer1b.numericValue > 2.999 and answer1b.numericValue < 3.001
"Hmm — 3 is the starting mass. One half-life later, half of it is gone. What is half of 3?"
when answer1b.submitted and not isBlank(answer1b.latex)
"Take another look — 5,730 years is one half-life. The mass halves over one half-life. Start with 3 picograms."
otherwise ""
Problem 1c: Number of years when mass is 0.75 picogram (table row 4)
content:
when answer1c.submitted and answer1c.numericValue > 11459 and answer1c.numericValue < 11461
"Correct! 0.75 is one-quarter of 3, which is *two* halvings. Two half-lives is 2 · 5,730 = 11,460 years."
when answer1c.submitted and answer1c.numericValue > 5729 and answer1c.numericValue < 5731
"Almost — after 5,730 years (one half-life), the mass is 1.5, not 0.75. You need *two* halvings to go from 3 to 0.75. How many years is that?"
when answer1c.submitted and answer1c.numericValue > 17189 and answer1c.numericValue < 17191
"Hmm — 17,190 is three half-lives. After three half-lives the mass is 3 · (1/2)^3 = 0.375, not 0.75. You only need two halvings."
when answer1c.submitted and answer1c.numericValue > 22919 and answer1c.numericValue < 22921
"Not quite — 22,920 is four half-lives, taking the mass to 3/16 ≈ 0.1875. You only need to get to 0.75. How many halvings is 3 → 1.5 → 0.75?"
when answer1c.submitted and not isBlank(answer1c.latex)
"Good start — go step by step: 3 picograms halves to 1.5, which halves to 0.75. That is *two* halvings. Each halving takes 5,730 years."
otherwise ""
Problem 3: More or less than 0.1 picogram of C-14 today? (Type 'more' or 'less')
content:
when answer3.submitted and answer3.content matches "more|More|MORE"
"Correct! About 22,000 years have passed since 20,000 BCE — less than 4 half-lives (4 × 5,730 = 22,920). After 4 half-lives, 2.5 · (1/2)^4 = 2.5/16 ≈ 0.156 picograms remain. So more than 0.1."
when answer3.submitted and answer3.content matches "less|Less|LESS"
"Take another look — count half-lives. From 20,000 BCE to today is about 22,000 years. One half-life is 5,730 years. How many half-lives fit in 22,000? Then compute 2.5 · (1/2)^(that many)."
when answer3.submitted and isBlank(answer3.content)
"Type 'more' or 'less'. Then explain your reasoning out loud with your partner."
when answer3.submitted
"Type 'more' or 'less'. Hint: how many half-lives fit between 20,000 BCE and today?"
otherwise ""
Activity 1 · Synthesis5 min
Radiocarbon Dating — Strategies Share
Three rows of the table required three different moves. Surface each one explicitly — and connect the “today” question to the structure of half-lives.
“How did you fill in the 5,730-year row?”5,730 years is one half-life. The mass is cut in half: \(3 \cdot \tfrac{1}{2} = 1.5\) picograms. No exponents needed — just “half of the previous amount.”
“How did you fill in the 1,910-year row?”1,910 is one-third of 5,730. So the factor is \((\tfrac{1}{2})^{1/3}\) — the same idea as the warm-up. Then \(3 \cdot (\tfrac{1}{2})^{1/3} \approx 2.38\) picograms. The factor over a fractional interval is a root, never a quotient.
“How did you find the years column when mass = 0.75?”Going backward: \(3 \to 1.5 \to 0.75\) is two halvings. Two half-lives is \(2 \times 5{,}730 = 11{,}460\) years. Or by equation: \((\tfrac{1}{2})^{n} = 0.25\) gives \(n = 2\), so \(t = 2 \cdot 5{,}730 = 11{,}460\).
“What does the function \(f(t) = 2.5 \cdot (\tfrac{1}{2})^{t/5730}\) tell us?”The 2.5 is picograms in the sample at 20,000 BCE. The \(\tfrac{1}{2}\) is the half-life factor. The 5,730 is the years per halving — it sets the time unit for the exponent. Together, \((\tfrac{1}{2})^{t/5730}\) means “the fraction remaining after \(t/5{,}730\) half-lives.”
“Today — more or less than 0.1 picogram?”More. About 22,000 years separate today from 20,000 BCE. That is fewer than 4 half-lives (4 × 5,730 = 22,920). After 4 half-lives the mass is \(2.5 \cdot (\tfrac{1}{2})^{4} = \tfrac{2.5}{16} \approx 0.156\) picograms — already more than 0.1. Today's mass is higher still.
click to advance discussion ▶
Try Saying
The factor over 1,910 years is ___ because 1,910 years is ___ of a half-life.
Try Saying
The constant ___ in \(f(t)\) tells me ___ about the sample.
Run as cold-call. Select students you watched solving each row differently — ideally one who used the cube-root reasoning, one who counted half-lives, and one who extended a pattern in a table or spreadsheet. The mathematical core: equal factors over equal intervals applied with fractional and whole-number exponents. The “today” question pays off MP1 — making sense of an awkward number (22,000 years) by relating it to a known structure (4 half-lives).
Activity 2 · Launch2 min
Old Manuscripts
Rhind Mathematical Papyrus, Egypt c. 1550 BCE — Wikipedia
Pythagoras (c. 570–495 BCE) — Wikipedia
Papyrus is a writing surface made from the papyrus plant — the ancient equivalent of paper. Once papyrus is harvested and turned into a page, its carbon-14 starts decaying. Pythagoras lived in ancient Greece between 600 BCE and 500 BCE. The half-life of carbon-14 is about 5,730 years.
Roughly how long ago did Pythagoras live? About what fraction of a carbon-14 half-life is that?
Quick launch (~2 min). The point of this slide is to settle the facts students will need: papyrus carbon-14 starts decaying when the plant is harvested; Pythagoras lived ~2,550–2,650 years ago; the half-life of C-14 is 5,730 years. Don't solve the activity here — just make sure no one is hunting for these facts mid-problem.
The discussion prompt seeds Activity 2 Problem 1: 2,600 years is a little less than half of one half-life. That is what students will explain.
Activity 2 · Work Time8 min
Old Manuscripts
The half-life of carbon-14 is about 5,730 years.
On your own — explain your reasoning in writing
Pythagoras lived between 600 BCE and 500 BCE. Explain why the age of a papyrus from the time of Pythagoras is about half of a carbon-14 half-life.
Someone claims they have a papyrus scroll written by Pythagoras. Testing shows the scroll has 85% of its original amount of carbon-14 remaining. Explain why the scroll is likely a fake.
8 min · Activity 2 Work Time.
Problem 1 target: Pythagoras lived about 2,500–2,600 years ago. One half-life is 5,730 years. \(2{,}600 / 5{,}730 \approx 0.45\) — a little less than half of a half-life. (Any reasoning that estimates ~2,500–2,650 years and divides by 5,730 is acceptable.)
Problem 2 target: After half of one half-life, the remaining fraction is \((\tfrac{1}{2})^{1/2} = \tfrac{1}{\sqrt{2}} \approx 0.707\), or about 70.7%. So a real Pythagoras-era papyrus should have about 70% of its C-14 left, not 85%. If 85% remains, the papyrus is much younger — the scroll is likely a fake.
Watch for students who compute \(\tfrac{1}{2} \cdot 0.5 = 0.25\) (multiplying as if half a half-life means a quarter) — that's the same warm-up confusion: roots, not multiplications.
Amplify · 2 inputs
Order per problem: Text component (problem text) → Text Response named answer1 / answer2 → Note with CL.
Both problems require written explanations, so use answer.content with keyword matching, not numericValue.
Problem 1: Why is a Pythagoras-era papyrus about half a half-life old?
content:
when answer1.submitted and answer1.content matches "2[,.]?[56]\d\d|half"
"On the right track — Pythagoras lived around 2,550–2,650 years ago. One half-life is 5,730 years. 2,600 ÷ 5,730 ≈ 0.45 — a little less than half of a half-life."
when answer1.submitted and answer1.content matches "5[,.]?730|half-life|halflife"
"Good — you're connecting to the half-life. Make sure your answer states *how many years* Pythagoras lived ago and *what fraction* that is of 5,730."
when answer1.submitted and isBlank(answer1.content)
"Start with how long ago Pythagoras lived (around 2,500–2,600 years ago). Then compare to the C-14 half-life of 5,730 years."
when answer1.submitted
"Take another look — Pythagoras lived around 2,500–2,600 years ago. Divide that by 5,730 (the half-life). What fraction of a half-life is it?"
otherwise ""
Problem 2: Why is the 85% scroll likely a fake?
content:
when answer2.submitted and answer2.content matches "70|0\.70|0\.71|70\.7|too young|not old enough|younger|fake"
"Correct reasoning — after half a half-life, the fraction remaining is (1/2)^(1/2) ≈ 0.707, or about 70%. A genuine Pythagoras-era scroll should have about 70% of its C-14, not 85%. 85% means the scroll is much younger — likely a fake."
when answer2.submitted and answer2.content matches "50|half"
"Almost — half of a half-life does NOT mean half of the carbon-14. Half remaining happens after one *full* half-life. After half of a half-life, the factor is (1/2)^(1/2) ≈ 0.707."
when answer2.submitted and answer2.content matches "25|quarter|0\.25"
"Hmm — 25% is what you get after *two* half-lives. After half of a half-life, the factor is (1/2)^(1/2) ≈ 0.707 (about 70%)."
when answer2.submitted and isBlank(answer2.content)
"Find the C-14 fraction remaining after half of a half-life: (1/2)^(1/2). Then compare to 85%."
when answer2.submitted
"Take another look — after half of a half-life, the remaining fraction is (1/2)^(1/2) ≈ 0.707. Is that closer to 85%, or is 85% way too high for a real Pythagoras-era scroll?"
otherwise ""
Activity 2 · Synthesis3 min
Old Manuscripts — Discussion
Carbon dating is not a single calculation but a chain of estimates — each one a fractional-exponent factor away.
“About how old is a Pythagoras-era papyrus? How many half-lives?”About 2,600 years, or about \(\tfrac{1}{2}\) of one half-life (since 5,730 / 2 = 2,865 is close to 2,600).
“By what factor does C-14 decrease in one half-life? In half a half-life?”In one half-life, the factor is \(\tfrac{1}{2}\). In half a half-life, the factor is \((\tfrac{1}{2})^{1/2} = \tfrac{1}{\sqrt{2}} \approx 0.707\). So about 70% of the carbon-14 remains after one half of one half-life — not 85%, and certainly not 50%.
“Carbon-14 dating only works for objects younger than about 50,000 years. About how many half-lives is that?”50,000 / 5,730 \(\approx\) 8.7, or about 9 half-lives. After 9 half-lives, the fraction left is \((\tfrac{1}{2})^{9} = \tfrac{1}{512} \approx 0.002\) — about 0.2%. Below that, the signal is so small that other dating methods are needed.
click to advance discussion ▶
Try Saying
The scroll is/isn't authentic because the carbon-14 reading is ___ what we expect for ___ years.
Quick discussion (~3 min). The big takeaway: C-14 dating is an approximation driven by where the artifact's age falls between half-lives. The 50,000-year limit is a useful concrete fact — tell students this if they ask why we never see C-14 dating used for dinosaurs (dinosaurs are ~65 million years old, far beyond C-14's range; potassium-argon dating handles that).
Lesson Synthesis~3 min
A menu of strategies for exponential problems
Today's problems could be solved several different ways. The point is not which way is “right” but to choose the way that fits the question.
Write an expression
Substitute into \(f(t) = a \cdot b^{t/h}\) and evaluate. Best when the input is specific.
Use a table
Multiply by the factor row by row. Best when half-life multiples align with the question.
Count half-lives
Divide elapsed time by the half-life. Best for “more or less than” estimates.
Reason backward
Start from the target mass and divide. Best when you have an output and want a time.
“Which strategy fits: how much C-14 is in a 4,000-year-old fossil that started with 6 picograms? (C-14 half-life is 5,730 years.)”“Write an expression.” The input is specific (4,000 years). Compute \(6 \cdot (\tfrac{1}{2})^{4000/5730} \approx 6 \cdot 0.616 \approx 3.7\) picograms. Calculator: 6*(1/2)^(4000/5730)
“Which strategy fits: how many years to drop from 8 picograms to 1?”“Count half-lives” or “reason backward”. \(8 \to 4 \to 2 \to 1\) is three half-lives, so 3 × 5,730 = 17,190 years.
“Which strategy fits: a 32-mg sample has a half-life of 10 years — how much remains after 30 years?”
Years
Amount (mg)
0
32
10
?
20
?
30
?
“Use a table.” Multiply by \(\tfrac{1}{2}\) each row: \(32 \to 16 \to 8 \to 4\). After 30 years (three half-lives), 4 mg remain.
click to advance discussion ▶
Try Saying
For this problem, I would use ___ because the question gives me ___ and asks me to find ___.
I Can… (SWBAT)
Use the half-life of an element to calculate how much of the element remains over time.
Describe in writing what the parameters \(a\), \(b\), and the input scaling \(t/h\) mean in an exponential decay expression.
Lesson Synthesis (~3 min). Cold-call through the two strategy-choice questions. Emphasize: there is rarely one right method. Equations work, tables work, half-life counting works. Real scientists use all three. The judgment is knowing which one is fastest for the question on the page.
SWBAT belongs here, not on the title slide. Read aloud after the discussion concludes.
Cool-Down5 min
Half Gone, Again and Again
A scientist models the number of grams of a radioactive substance with the expression
\[16 \cdot \!\left(\tfrac{1}{2}\right)^{d/5},\]
where \(d\) is the number of days since the sample was measured.
On your own — show your reasoning
What is the half-life of this radioactive substance? Explain your reasoning.
After 20 days, will there be more or less than 1 gram of the substance left? Explain your reasoning.
Cool-Down · 5 min. Exit ticket. Students must work independently.
Problem 1: Half-life is 5 days. The exponent \(d/5\) increases by 1 every time \(d\) increases by 5, so the factor multiplies by \(\tfrac{1}{2}\) every 5 days — that is the definition of half-life.
Problem 2: The IM expected answer is less than 1 gram. At exactly 20 days, \(16 \cdot (\tfrac{1}{2})^{4} = 1\) gram. The IM cool-down treats “after 20 days” as strictly after, so the substance has already decayed past 4 half-lives and the mass is less than 1 gram. Accept either “less” with that reasoning, or “exactly 1 gram at 20 days, less thereafter” with computation.
Watch for students who write “more” because they don't see that the exponent \(d/5\) counts half-lives. The point is reading the exponent structure, not the base.
Amplify · 2 inputs
Order per problem: Text component (problem text + Example: 5) → Math Response (Problem 1, named answer1) or Text Response (Problem 2, named answer2) → Note with CL.
Problem 1: numericValue check. Problem 2: answer2.content matching ‘less’ or ‘more’.
Problem 1: What is the half-life of the substance? (Enter the number of days.)
content:
when answer1.submitted and answer1.numericValue > 4.995 and answer1.numericValue < 5.005
"Correct! The exponent d/5 grows by 1 every 5 days, so the mass multiplies by 1/2 every 5 days. That is the half-life: 5 days."
when answer1.submitted and answer1.numericValue > 15.995 and answer1.numericValue < 16.005
"Hmm — 16 is the *starting* mass (the value when d = 0), not the half-life. The half-life is the number of days it takes for the mass to halve. Look at the exponent d/5: when does d/5 equal 1?"
when answer1.submitted and answer1.numericValue > 0.495 and answer1.numericValue < 0.505
"Almost — 1/2 is the *factor* per half-life, not the half-life itself. The half-life is measured in days. When d = ?, does d/5 = 1?"
when answer1.submitted and answer1.numericValue > 19.995 and answer1.numericValue < 20.005
"Not quite — 20 is the number of days in the second question, not the half-life. The exponent d/5 increases by 1 every time d increases by 5. So the mass halves every 5 days."
when answer1.submitted and answer1.numericValue > 9.995 and answer1.numericValue < 10.005
"Almost — 10 days is *two* half-lives. After 10 days the mass is 16 · (1/2)^2 = 4 grams. The half-life itself is the number of days for *one* halving."
when answer1.submitted and isBlank(answer1.latex)
"Enter the number of days in one half-life. Look at the exponent d/5: when does it equal 1? Example: 5"
when answer1.submitted
"Take another look — for what value of d does the exponent d/5 equal 1? That is one half-life."
otherwise ""
Problem 2: After 20 days, more or less than 1 gram? (Type 'more' or 'less'.)
content:
when answer2.submitted and answer2.content matches "less|Less|LESS"
"Correct reasoning — 20 days is exactly 4 half-lives: 16 · (1/2)^4 = 16/16 = 1 gram. After 20 days (strictly past 4 half-lives), the mass has dipped below 1 gram. The IM answer key says 'less'."
when answer2.submitted and answer2.content matches "exactly|equal|1 gram"
"Good observation — at d = 20, the mass is exactly 1 gram (4 half-lives). The IM expected answer is 'less' because 'after 20 days' is read as strictly after. Either explanation is mathematically sound."
when answer2.submitted and answer2.content matches "more|More|MORE"
"Take another look — count half-lives. 20 days ÷ 5 days per half-life = 4 half-lives. After 4 half-lives the mass is 16 · (1/2)^4 = 16/16 = 1 gram. So at 20 days the mass is at *most* 1 gram — never more."
when answer2.submitted and isBlank(answer2.content)
"Type 'more' or 'less'. Hint: how many half-lives fit in 20 days, given a half-life of 5 days?"
when answer2.submitted
"Take another look — 20 days is 4 half-lives. After 4 half-lives, 16 · (1/2)^4 = 1 gram. Is the answer 'more' or 'less' than 1?"
otherwise ""
Lesson Summary
Half-life — the clock inside an ancient thing
Some substances change over time through radioactive decay. Their rate of decay can be measured by their half-life — the time it takes for half of any sample to break down. The same decay equation models them all.
Half-life ~3 years. Start with 4 ng. After 1 year: \(4 \cdot (\tfrac{1}{2})^{1/3} \approx 3.2\) ng. The cube-root factor handles a single year.
Reaching 0.015 ng?
\(0.015 \approx 4 \cdot (\tfrac{1}{2})^{8}\). Eight half-lives is 24 years — the answer found by counting halvings.
Lesson Summary
Archaeologists, geologists, and scientists use exponential functions with half-lives to estimate the age of ancient things. The same equation \(a \cdot (\tfrac{1}{2})^{t/h}\) describes the decay of every radioactive isotope — only the constants \(a\) and \(h\) change.
Lesson Summary slide. Read aloud or have a student read. Stress the universality: carbon-14 (h = 5,730 years), sodium-22 (h \(\approx\) 3 years), uranium-238 (h \(\approx\) 4.5 billion years) — same equation, different \(h\). The choice of isotope depends on the time scale of the thing being dated. Tomorrow (Section B), students begin solving for the exponent — the move that unlocks logarithms in Section C.
