Calculate the packing efficiency of metal crystal that has simple cubic structure.

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Q. Calculate the packing efficiency of metal crystal that has simple cubic structure.

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Concept understanding — Packing Efficiency

Packing Efficiency: How Tightly Can We Pack Spheres?

Imagine you're filling a suitcase with tennis balls. No matter how carefully you arrange them, there will always be some empty space between the balls — you can't fill every corner. The packing efficiency is simply a measure of how much of the total space is actually occupied by the balls (the atoms), and how much is wasted as empty space.

In a crystal, atoms are modelled as hard spheres of equal size. They touch each other along specific directions, but they can't overlap. The unit cell is the box that contains the repeating pattern. Packing efficiency tells us: out of the total volume of that box, what fraction is actually filled by the atoms?

\text{Packing efficiency} = \frac{\text{Volume occupied by atoms in one unit cell}}{\text{Total volume of the unit cell}} \times 100\%

The key is to count how many whole atoms belong to a single unit cell, and then use the geometry of the cell to relate the atom's radius to the cell's edge length.

Simple Cubic (SCC) — 52.4%

In a simple cubic cell, atoms sit only at the eight corners. Each corner atom is shared by 8 neighbouring cells, so each contributes $\frac{1}{8}$ of an atom. Total atoms per cell: $8 \times \frac{1}{8} = 1$ .

The atoms touch along the edge of the cube. If the edge length is $a$ and the atomic radius is $r$ , then $a = 2r$ .

Volume of one atom: $\frac{4}{3}\pi r^3$ . Volume of the cube: $a^3 = (2r)^3 = 8r^3$ .

\text{Efficiency} = \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100\% = \frac{\pi}{6} \times 100\% \approx 52.4\%

That's barely more than half the space. The rest is empty — a very loose packing.

Body-Centred Cubic (BCC) — 68%

Now we add one atom right at the centre of the cube. The corner atoms still contribute 1 atom total, and the body-centred atom is entirely inside the cell — it's not shared. So total atoms per cell: $1 + 1 = 2$ .

The atoms do not touch along the edge. They touch along the body diagonal — from one corner, through the centre atom, to the opposite corner. The body diagonal length is $a\sqrt{3}$ , and it contains two full radii from the corner atoms plus two radii from the centre atom: $4r = a\sqrt{3}$ . So $a = \frac{4r}{\sqrt{3}}$ .

Volume of two atoms: $2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3$ . Volume of cube: $a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}}$ .

\text{Efficiency} = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100\% = \frac{\pi\sqrt{3}}{8} \times 100\% \approx 68\%

A significant improvement — the central atom fills a lot of the void.

Face-Centred Cubic (FCC) and Hexagonal Close-Packed (HCP) — 74%

In FCC, atoms sit at all eight corners and at the centre of each of the six faces. A face-centred atom is shared by 2 cells, so each contributes $\frac{1}{2}$ . Total atoms: $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4$ .

Atoms touch along the face diagonal. The face diagonal length is $a\sqrt{2}$ , and it contains 4 atomic radii: $4r = a\sqrt{2}$ , so $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}\,r$ .

Volume of four atoms: $4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3$ . Volume of cube: $a^3 = (2\sqrt{2}\,r)^3 = 16\sqrt{2}\,r^3$ . …

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Step 1. In a simple cubic cell the spheres touch along the edge, so $a=2r$ , i.e. $r=a/2$ .

Step 2. Volume of one sphere $=\tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi(a/2)^3 = \dfrac{\pi a^3}{6}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$Q.The packing fraction for body-centred cubic cell is$
(a) 0.42
(b) 0.53
(c) 0.68
(d) 0.82
›Reveal solutionSolution

A bcc cell packs 2 atoms with a packing fraction of about 0.68 (68%).

For a bcc cell the body diagonal = 4r, and body diagonal = a√3, so a = 4r/√3.

Atoms per cell = 2 (1 body centre + 8 corners x 1/8).

Packing fraction = (volume of 2 atoms)/(volume of cell) = [2 x (4/3)πr³] / a³. …

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CBSE 2025Set D1 markMCQ
$Q. The percentage of free space in a body-centred cubic unit cell is$
(a) 32
(b) 34
(c) 28
(d) 20
›Reveal solutionSolution

bcc packing efficiency = 68%, hence free space = 100 − 68 = 32%.

For a body-centred cubic lattice, atoms touch along the body diagonal, giving the relation 4r = √3 a (r = radius, a = edge). The unit cell has 2 atoms.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. What percentage of a body centred cubic structure is vacant?$
(a) 68%
(b) 52.4%
(c) 74%
(d) 32%
›Reveal solutionSolution

A bcc unit cell has 2 atoms per cell and packing efficiency 68% (=√3π/8), so the remaining 32% of the cell's volume is empty space.

In a body-centred cubic (bcc) arrangement, atoms touch along the body diagonal, giving edge length a related to atomic radius r by a = 4r/√3. There are 2 atoms per unit cell (1 corner-contribution + 1 body-centre atom). Packing efficiency = (volume occupi …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Efficiency of packing in Simple Cubic Lattice is:$
(a) 52.4%
(b) 62.4%
(c) 68%
(d) 74%
›Reveal solutionSolution

In simple cubic (SC) packing the atoms touch along the cell edge and occupy only about half the available volume, giving 52.4% packing efficiency - the least efficient of the common cubic packings.

In a simple cubic lattice, atoms are present only at the corners of the cube, and each corner atom touches the atoms in the adjacent cells along the edge. So the edge length a = 2r, where r is the radius of the atom.

Each unit cell contains 8 corner atoms x 1/8 = 1 atom.

Volume of the atom = (4/3)pi r^3

…

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CBSE 2020Set ANNUAL1 mark
$Q. Write the value of packing efficiency in body-centred cubic structure.$
›Reveal solutionSolution

In a bcc lattice atoms touch along the body diagonal, and working through the geometry gives a packing efficiency of (√3π)/8 ≈ 68%.

In a body-centred cubic unit cell, atoms touch each other along the body diagonal (not along the edge).

Step 1 — relate radius r to edge length a:

Body diagonal = √3·a = 4r ⟹ r = (√3/4)·a

Step 2 — atoms per unit cell (Z):

Corner atoms: 8 × 1/8 = 1

Body-centre atom: 1 × 1 = 1

Z = 2

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. Which of the following lattices has the highest packing efficiency (assuming that atoms are touching each other)?$
(a) Simple cubic
(b) Body centred cubic
(c) Face centred cubic
›Reveal solutionSolution
Face centred cubic (fcc) has the highest packing efficiency (74%) of the three cubic lattices.

Packing efficiency is the fraction of total volume of a unit cell occupied by the constituent spheres (atoms), assuming they touch each other along the relevant direction.

Simple cubic (sc): atoms touch along the cell edge; packing efficiency = 52.4%

Body centred cubic (bcc): atoms touch along the body diagonal; packing efficiency = 68% …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. Volume occupied by atoms in bcc is$
(a) 52.4%
(b) 74%
(c) 68%
(d) None
›Reveal solutionSolution

In a bcc unit cell, atoms occupy 68% of the total volume.

In a body-centred cubic (bcc) arrangement, atoms touch each other along the body diagonal of the cube (not along the edge).

Step 1 — Relation between radius and edge length.

For a cube of edge $a$ , the body diagonal length is $\sqrt{3}\,a$ . Along this diagonal, one corner atom, the body-centre atom, and the opposite corner atom touch, so the diagonal spans $4r$ (radius $r$ of two half-atoms at the ends plus the full diameter of the centre atom):

$\sqrt{3}\,a = 4r \implies r = \frac{\sqrt{3}}{4}a$

Step 2 — Number of atoms per unit cell.

Z = 8 corners × 1/8 + 1 body-centre × 1 = 1 + 1 = 2 atoms.

Step 3 — Packing fraction. …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Percentage of free space in bcc. unit cell is -$
(a) 32%
(b) 34%
(c) 28%
(d) 30%
›Reveal solutionSolution

bcc packing efficiency = 68% → free space = 32%.

In a bcc unit cell there are 2 atoms and the body diagonal = 4r = √3·a. Packing efficiency = (2 × ⁴⁄₃πr³)/a³ …

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Q.Calculate the packing efficiency of metal crystal that has simple cubic structure.

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