Problem 1

The Avogadro constant

N_A

can be determined by three independent methods. (Method A) X-ray diffraction shows that gold crystallises in a face-centred cubic unit cell of edge

a = 0.408

nm; the density of Au is

1.93 \times 10^{4}\ \text{kg m}^{-3}

and

M(\mathrm{Au}) = 196.97\ \text{g mol}^{-1}

. (Method B, Rutherford 1911) A purified sample of 192 mg of

\ce{^{226}Ra}

stands 40 days, after which its total

\alpha

-decay rate is 27.7 GBq; sealing it for a further 163 days produces 10.4

\text{mm}^3

of He measured at 101325 Pa and 273 K (

\alpha

particles become He atoms). (Method C, Perrin 1909) Colloidal spheres of radius

2.12 \times 10^{-7}

m and density

1.206 \times 10^{3}\ \text{kg m}^{-3}

suspended in water (

\rho = 999\ \text{kg m}^{-3}

) at 15 \°C show a Boltzmann height distribution: a plot of

\ln(n_h/n_{h_0})

versus

(h - h_0)

has slope

-0.0235\ \mu\text{m}^{-1}

. Determine

N_A

from each method.

Step 2 of 3: Radioactive decay method

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Intuition

Rutherford counted helium atoms produced and decays measured — the ratio is atoms per mole.

N_{\alpha} = 27.7 \times 10^{9}\ \text{s}^{-1} \times 163 \times 86400\ \text{s} = 3.90 \times 10^{17};\quad n(\ce{He}) = \dfrac{pV}{RT} = 4.64 \times 10^{-7}\ \text{mol};\quad N_A = \dfrac{3.90 \times 10^{17}}{4.64 \times 10^{-7}} = 8.4 \times 10^{23}\ \text{mol}^{-1}

Analysis

Every $\alpha$ particle becomes one He atom. The decay rate integrated over 163 days gives $3.90 \times 10^{17}$ particles, and $pV/RT$ converts the measured He volume to moles. The ratio gives $N_A \approx 8.4 \times 10^{23}$ mol $^{-1}$ (higher than the true value because the decay series had not fully reached steady state).