Question

Calculate the percentage efficiency of packing in the case of a body-centred cubic crystal.

Calculate the percentage of the occupied space in a body-centred cubic unit cell.

Answer 1

In a body-centred cubic arrangement, the spheres are touching along the leading diagonal of the cube, as shown in the figure.

In ∆ABC

Ac² = AB² + BC²

AC = `sqrt((AB)^2 + (BC)^2)`

AC = `sqrt(a^2 + a^2)`

= `sqrt(2a^2)`

= `sqrt 2a`

In ∆ACG

AG² = AC² + CG²

AG = `sqrt((AC)^2 + (CG)^2)`

AG = `sqrt((sqrt 2a)^2 + a^2)`

AG = `sqrt(2a^2 + a^2)`

= `sqrt(3a^2)`

= `sqrt 3a`

i.e., `sqrt 3a` = 4r

r = `sqrt3/4 a`

∴ Volume of the sphere with radius ‘r’

= `4/3πr^3`

= `4/3π (sqrt3/4 a)^3`

= `sqrt 3/16πa^3`

Number of spheres belong to a unit cell in bcc arrangement is equal to two and hence the total volume of all spheres

= `2 xx ((sqrt 3πa^3)/16)`

= `(sqrt 3πa^3)/8`

Packing fraction = `"Total volume occupied by spheres in a unit cell"/"Volume of the unit cell" xx 100`

Packing fraction = `(((sqrt 3πa^3)/8))/((a^3)) xx 100`

= `(sqrt 3π)/8 xx 100`

= `sqrt 3π xx 12.5`

= 1.732 × 3.14 × 12.5

= 68%

i.e., 68% of the available volume is occupied. The available space is used more efficiently than in simple cubic packing.

Answer 2

A body-centered cubic unit cell contains two atoms. Let r represent the radius of an atom.

The volume occupied by two atoms

= `2 xx 4/3 pi r^3`

= `8/3 pi r^3`    ...(i)

As seen in Fig., the corners of a body-centred cubic unit cell have atoms that touch the body-centred atom but not each other. Let a be the unit cell’s edge length. Fig. shows the body diagonal.

AD = 4 r   ...(ii)

The face diagonal in the ABC right-angled triangle

AC = `sqrt (AB^2 + BC^2)`

= `sqrt (a^2 + a^2)`

AC = `sqrt2  a`    ...(iii)

The body diagonal in the right-angled triangle ACD

AD = `sqrt (CD^2 + AC^2)`

= `sqrt (a^2 + 2 a^2)`

AD = `sqrt 3  a`    ...(iv)

From eqs. (ii) and (iv), we have

`sqrt 3  a = 4 r`,

or, a = `4/sqrt 3 r`    ...(v)

The volume of the unit cell = a³

= `(4/sqrt 3 r)^3`

= `(64 r^3)/(3 sqrt 3)`    ...(vi)

∴ `"Packing fraction" = "Volume occupied by atoms"/"Volume of unit cell"`

= `(8/3 pi r^3)/((64 r^3)/(3 sqrt 3))`

= `(sqrt 3 pi)/8`

= 0.68

Hence, the percentage of the occupied space = 68%.