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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2021/paper-2/a9/b/i/solution

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Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 2 / A9 / b / i / Solution by

Codex  0 2026-10-20

Using

M_{Au} ≃ 196.97 g mo l^{- 1}

and the Avogadro constant, the areal atom count is

n_{Au} = \frac{1.56 \times 1 0 ^{- 6} g}{196.97 g mo l ^{- 1}} \frac{6.02214 \times 1 0 ^{23} mo l ^{- 1}}{1 0 ^{- 4} m ^{2}} ≃ 4.77 \times 1 0^{19} m^{- 2} .

(1)

One atom occupies each square cell of area

a^{2}

, so the inferred square lattice spacing is

a = n_{Au}^{- 1/2} ≃ 1.45 \times 1 0^{- 10} m = 0.145 nm .

(2)

This uses the given deposited mass and the assumed single square layer; it does not substitute a bulk gold lattice parameter for the measured areal count.

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