Evaporated Gold¶
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Section 1, session 6. The same session starts the group effort on Collective Reproduction.
The problem
Reconstructed from the syllabus. No statement for this problem survives: the syllabus gives the bare name, after the three skills listed for the day and before the group effort on Collective Reproduction. What follows is the editors' reading of that name — gold that has been evaporated onto something — not Winfree's wording.
A physicist puts a bead of pure gold, mass 1.00 g, into a tungsten boat at the centre of an evacuated bell jar and heats it until the gold has entirely evaporated. A clean glass microscope slide, 2.5 cm by 7.5 cm, is fixed inside the jar 15 cm from the boat, facing it. When the jar is opened, the slide carries a gold-coloured film.
- Before calculating, write down every assumption you will need, marking each as something you know, assume, or could look up.
- Estimate the film's thickness, in metres and in atoms.
- Check that answer by two independent routes that share no arithmetic. (One useful fact: a gram of gold beats out to about a square metre of leaf.)
- Would the film look gold? Would light pass through it? What thickness would let light through, and what colour would it be?
- Which assumptions, if wrong, would change the answer by more than a factor of ten? Which by less than two?
You are graded on the care of your assumption list and your checks, not on the number you get.
The reconstructed set-up. Drawn for this site (CC BY 4.0).
Why it is in the course¶
Session 6 closes Section 1, "Detecting Nonsense, Error Checking, False Assumptions, Cherishing Mistakes". The readings due are a packet on valuing mistakes and Ehrlich's introduction on judging a crazy idea. The syllabus then names three skills for the hour — "Some ways to check for errors", "hidden assumptions", "what is 'understand'?" — and lists Evaporated Gold straight after them. That placement suggests it was the worked example. It is a placement, not evidence.
The reconstruction above is built to exercise those three skills. List your assumptions before you compute, while you can still attack them. Cross-check by a route that shares no arithmetic with the first — harder than it sounds, as the resolution shows. Then ask whether you understand the number or have merely produced one. The same day the class begins Collective Reproduction, where the checking has to be done on someone else's reasoning.
Where it comes from¶
We do not know. No puzzle of this name has been found outside Winfree's own schedule.
The physics behind the likeliest reading is ordinary. Vacuum evaporation — heating a metal in a chamber pumped down to roughly 10^-4^ Pa, so atoms fly in straight lines and condense on whatever they hit — is how thin metal films are made, on electron-microscope specimens, mirrors and electrodes. Gold suits order-of-magnitude work because two of its numbers are famous: a density near 19.3 g/cm3, and a gram that beats out to about a square metre of leaf. Thin gold is also optically strange, which is where part 4 comes from: Michael Faraday's 1857 Bakerian Lecture examined how gold films thin enough to transmit light behave.
Hints
- Write the assumptions down first, apart from the calculation; the ones you did not write down are the likeliest to be wrong.
- Thickness is volume over area. The volume follows from the density; the area is the hard part.
- Ask what would look different if the film were ten times thicker, or ten times thinner. An answer you can recognise as absurd is an answer you can check.
- Atoms leaving a hot source in vacuum fly in straight lines in every direction. What fraction can reach a slide of that size, that far away?
Resolution
This resolves the reconstruction above, not a recovered original.
The hidden assumption. At 19.3 g/cm3, 1.00 g of gold occupies about 0.052 cm3. If all of it landed on the slide (18.75 cm2), the film would be 28 micrometres thick, some 100,000 atoms. But the vapour leaves in every direction, spreading over roughly a sphere of radius 15 cm, area about 2,800 cm2, of which the slide intercepts 0.7 %. Mean thickness there: about 180 nanometres, roughly 600 atoms at gold's 0.29 nm spacing. A boat standing on the baseplate radiates into something nearer a hemisphere, so call the geometry good to a factor of two. The gap between that quibble and the factor-of-150 blunder is the lesson.
Which checks are really checks. Gold leaf looks like an independent route and is not. A gram spread over a square metre is 52 nm thick; our sphere has 3.5 times less area, so 3.5 times the thickness, 180 nm. That is the same volume divided by an area — the answer re-derived, not tested. Counting atoms against Avogadro's number tells the same story: it checks the arithmetic, never the geometry, which is where the error was. Two checks do bite. Real specimen coatings run 5 to 20 nm and use far less than a gram, so a 28-micrometre film is absurd on its face. And ask where the rest went: if 99.3 % misses the slide, the jar must come out coated — a prediction the apparatus can falsify.
Appearance. At 180 nm the film is opaque and mirror-bright. Gold turns semi-transparent only below a few tens of nanometres, and then transmits greenish-blue, as Faraday found in 1857 — so the colour of the slide tells you almost nothing beyond "thicker than about 50 nm".
Sources¶
- Arthur T. Winfree, "The Art of Scientific Discovery": course handout and retrospective syllabus (479/479H/579, Ecology and Evolutionary Biology, University of Arizona), archived capture (2001) — Wayback Machine 🔓
- Arthur T. Winfree, faculty home page, archived 25 Dec 2002 — Wayback Machine 🔓
- Michael Faraday, "The Bakerian Lecture: Experimental Relations of Gold (and Other Metals) to Light", Philosophical Transactions of the Royal Society of London 147, 145–181 (1857) — public-domain scan at Wikimedia Commons 🔓
- Wikipedia contributors, "Evaporation (deposition)" — en.wikipedia.org 🔓
- Wikipedia contributors, "Gold" — en.wikipedia.org 🔓
- Wikipedia contributors, "Fermi problem" — en.wikipedia.org 🔓
- Wikipedia contributors, "Fritz Haber" — en.wikipedia.org 🔓
- Fritz Haber, "Das Gold im Meerwasser", Zeitschrift für angewandte Chemie (now Angewandte Chemie) 40(11), 303–314 (1927) — doi:10.1002/ange.19270401103 🔒
- 911 Metallurgist (reprinting an older mining-engineering article), "How much Gold in Sea Water" — 911metallurgist.com 🔓
- Robert Ehrlich, Nine Crazy Ideas in Science: A Few Might Even Be True (2001) — Princeton University Press 🔒
- Arthur T. Winfree, The Art of Scientific Discovery: original course syllabus — PDF 🔓
How sure are we that this is Winfree's problem?
Not at all. The syllabus gives the name and nothing else, the archived handout carries no statement for it, Winfree's other archived pages say nothing about gold, and no published puzzle of this name turned up elsewhere. Three readings were weighed, and all three are low confidence.
- A thin-film thickness estimate, used above. "Evaporated gold" is the standard laboratory phrase for vapour-deposited gold, and the syllabus promises exercises "made from elementary mathematics so as to require no lab setup". Two things in the syllabus cut against it. The exercises are meant to "depend as little as possible on knowledge of any particular subject area", and this one leans on physics. And the puzzles are "many of them silly", meant "to slow you down for a few minutes"; every Section 1 neighbour is short and whimsical, while this reconstruction is a multi-part calculation.
- Gold from seawater. Old assays put tens of milligrams of gold in a tonne of seawater. Fritz Haber's 1920s scheme to extract it for German reparations found the true figure a thousand times smaller: the old numbers came from contaminated reagents. A fine "valuing mistakes" story — but the syllabus says nothing of the sea.
- A Fermi estimate: spread all the gold ever mined over the Earth and say how thick the layer is. Nothing ties it to Winfree.
If you know the original, the editors would like to hear from you.
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