What is the atomic packing factor? Refer to the diagram below. Given that the lattice constant for a face-centred cubic unit cell a = 2√2r, show that the atomic packing efficiency for face-centred cubic is 74%.

An FCC unit cell has 8 atoms at the corners (each contributing $\frac{1}{8}$) and 6 atoms at the faces (each contributing $\frac{1}{2}$).

$$N = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \text{ atoms}$$

Assuming atoms are spheres of radius $r$, the volume of one atom is $V_{atom} = \frac{4}{3}\pi r^3$.
The total volume of atoms in the unit cell is:

$$V_{atoms} = N \times V_{atom} = 4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3$$

For a cubic unit cell, the volume is $V_{cell} = a^3$, where $a$ is the lattice constant.
The problem states that for an FCC unit cell, $a = 2\sqrt{2}r$. Substitute this into the volume formula:

$$V_{cell} = (2\sqrt{2}r)^3 = (2^3)(\sqrt{2}^3)r^3 = 8 \times (2\sqrt{2})r^3 = 16\sqrt{2}r^3$$

The APF is the ratio of the volume of atoms to the volume of the unit cell:

$$APF = \frac{V_{atoms}}{V_{cell}} = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2}r^3}$$ $$APF = \frac{16\pi r^3}{3 \times 16\sqrt{2}r^3} = \frac{\pi}{3\sqrt{2}}$$

$$APF = \frac{\pi}{3\sqrt{2}} \approx \frac{3.14159}{3 \times 1.41421} \approx \frac{3.14159}{4.24263} \approx 0.74048$$ $$APF \approx 0.74048 \times 100\% \approx 74.048\%$$

Rounding to the nearest whole number, the atomic packing efficiency for a face-centered cubic unit cell is approximately 74%.

The atomic packing efficiency for a face-centered cubic unit cell is indeed $\boxed{74\%}$.