 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2019/2/d/solution

The leading coefficient fixes the atom count:

N = \frac{25}{3 k _{B}} = 6.04 \times 1 0^{23}, n_{mol} = \frac{25}{3 R} = 1.002 mol .

40 g mo l^{- 1}

, the sample mass is

m_{sample} ≃ 40.1 g .

One atom belongs to each primitive unit cell of monatomic argon on a face-centered cubic lattice, but the conventional cube contains four atoms. If its side is

a

N = 4 V / a^{3}

, and nearest neighbors lie across half a face diagonal. Therefore

d = \frac{a}{2} = \frac{1}{2} (\frac{4 V}{N})^{1/3} ≃ 3.66 \times 1 0^{- 10} m = 3.66 \overset{˚}{A},

using

V = 21 \times 1 0^{- 6} m^{3}

. The corresponding conventional lattice constant is

a ≃ 5.18 \overset{˚}{A}

. Treating the source's one-atom cell as the conventional cube would omit the factor four and give a wrong nearest-neighbor distance.

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