Answer & explanation
The official answer, with Fermi’s reasoning step by step.
Answer & explanation
Official answer
56,000
atoms
How to estimate it
Fermi’s official “Napkin math”
A hair is about a tenth of a millimeter wide, and foil is maybe a fifth of that. How wide is one atom? Way under a nanometer, call it a third of one.
- Hair width
- ~0.1 mm
- Foil, a fifth of that
- ≈ 0.02 mm = 20,000 nm
- Atoms per nanometer
- × 3
- 20,000 × 3
- ≈ 60,000
A hair is about 3 thousandths of an inch, and foil is maybe a fifth of that. How wide is one atom? Call it a third of a nanometer, and an inch is about 25 million of those.
- Hair width
- ~0.003 in
- Foil, a fifth of that
- ≈ 0.0006 in ≈ 15,000 nm
- Atoms per nanometer
- × 3
- 15,000 × 3
- ≈ 45,000
Rounded assumptions may give a slightly different estimate. The official answer above is the game’s reference value.
Source calculation & reference
- Standard household foil thickness (Reynolds)
- 0.016 mm = 16,000 nm
- Center-to-center spacing of aluminum atoms (lattice 0.405 nm ÷ √2)
- ≈ 0.286 nm
- 16,000 ÷ 0.286
- ≈ 55,900
- ≈
- 56,000
Reference cited by Fermi: Reynolds Kitchens foil FAQ (standard foil 0.016 mm thick) · aluminum crystal lattice measurements (Arblaster 2018, lattice constant 404.93 pm)
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