Chemistry · Ch 1 — Solid State

Packing efficiency of metal crystal in body-centred cubic lattice

1.7.2

Packing efficiency of metal crystal in body-centred cubic lattice

The packing efficiency of a metal crystal in the body-centred cubic lattice is obtained by the same four steps.

Step 1 : Radius of sphere (particle) : In a bcc unit cell the particle at the body centre touches the two corner particles that lie on the cube's body diagonal (Fig. 1.10). Consider the right-angled triangle $FED$ , with $\angle FED = 90^\circ$ . By the Pythagoras theorem, the face diagonal $FD$ satisfies

FD^2 = FE^2 + ED^2 = a^2 + a^2 = 2a^2 \qquad \text{...(1.8)}

Next consider the right-angled triangle $ADF$ , with $\angle ADF = 90^\circ$ . The body diagonal $AF$ satisfies

AF^2 = AD^2 + FD^2 = a^2 + 2a^2 = 3a^2 \qquad \text{...(1.9)}

AF = \sqrt{3}\,a \qquad \text{...(1.10)}

The body diagonal spans the radius of the corner sphere at $A$ , the full diameter of the body-centre sphere, and the radius of the corner sphere at $F$ , so $AF = 4r$ . Hence $\sqrt{3}\,a = 4r$ , that is

r = \frac{\sqrt{3}}{4}a \qquad \text{...(1.11)}

Figure 1.10bcc unit cell geometry: a perspective cube with vertices labelled A, D, E, F and the body-centre sphere C, showing the three spheres touching along the body diagonal AF and the dashed construction lines for the face diagonal FD used in the Pythagoras derivation of r = (root 3/4)a.

Fig. 1.10 — bcc unit cell geometry: a perspective cube with vertices labelled A, D, E, F and the body-centre sphere C, showing the three spheres touching along the body diagonal AF and the dashed construction lines for the face diagonal FD used in the Pythagoras derivation of r = (root 3/4)a.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A 3-D perspective cube with three shaded spheres drawn along its body diagonal: a corner sphere at $F$ , the body-centre sphere $C$ , and the opposite corner sphere at $A$ , touching one another in a line. Further vertices $D$ and $E$ are labelled, and dashed construction lines mark the face diagonal $FD$ and body diagonal $AF$ -- the right triangles $FED$ and $ADF$ used with the Pythagoras theorem to show $AF = \sqrt{3}\,a = 4r$ . Only the three diagonal s …

Step 2 : Volume of sphere : $\text{volume of one particle} = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi\left(\dfrac{\sqrt{3}}{4}a\right)^{3} = \dfrac{\sqrt{3}\,\pi a^3}{16}$

Step 3 : Total volume of particles : A bcc unit cell contains 2 particles, so the total occupied volume is

$2\times\frac{\sqrt{3}\,\pi a^3}{16} = \frac{\sqrt{3}\,\pi a^3}{8} \qquad \text{...(1.12)}$ …

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