If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by

Exercises · 2.35

Q. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

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Concept understanding — Atomic Packing Scale

Atomic Packing Factor (APF) — First Principles

Imagine you're packing oranges into a crate. No matter how carefully you arrange them, there will always be some empty space between the spheres. The Atomic Packing Factor (APF) is simply the fraction of that crate's volume that is actually filled by the oranges — except here, the "oranges" are atoms, and the "crate" is the unit cell of a crystal.

The Intuition

Atoms in a solid are not tiny cubes that stack perfectly. They're spheres. And spheres, when packed together, always leave gaps. The APF tells you how efficiently a given crystal structure uses space. A higher APF means atoms are packed more tightly together — less wasted volume.

For example, if you stack cannonballs in a pyramid, they pack more densely than if you just pile them randomly. Different crystal structures (like simple cubic, body-centered cubic, face-centered cubic) have different packing efficiencies. The APF is the number that quantifies this.

The Precise Definition

\text{APF} = \frac{\text{Volume of atoms in a unit cell}}{\text{Volume of the unit cell}}

That's it. A ratio between 0 and 1 (or 0% to 100%). For most metals, APF values range from about 0.52 to 0.74.

How to Calculate It — Step by Step

You need three things:

Number of atoms per unit cell — Count carefully. Atoms at corners are shared by 8 cells, atoms on faces by 2 cells, atoms at edges by 4 cells, and atoms fully inside belong entirely to that cell.
Radius of the atom — Usually given as $R$ .
Volume of the unit cell — Depends on the crystal structure. For a cube of side $a$ , it's $a^3$ . But $a$ itself depends on $R$ through the geometry of how atoms touch.

Watch out

The most common mistake: forgetting that atoms are spheres, not cubes. The volume of one atom is $\frac{4}{3}\pi R^3$ , not $R^3$ or $(2R)^3$ .

Worked Example: Simple Cubic (SC)

In a simple cubic cell:

8 corner atoms, each shared by 8 cells → $8 \times \frac{1}{8} = 1$ atom per cell
Atoms touch along the cube edge: $a = 2R$
Volume of cell: $a^3 = (2R)^3 = 8R^3$
Volume of one atom: $\frac{4}{3}\pi R^3$

\text{APF} = \frac{\frac{4}{3}\pi R^3}{8R^3} = \frac{\pi}{6} \approx 0.524

So only about 52.4% of the space is filled. Nearly half is empty — that's why simple cubic is rare in real metals.

Common APF Values

Crystal Structure	APF	Atoms per cell
Simple Cubic (SC)	0.524	1
Body-Centered Cubic (BCC)	0.680	2
Face-Centered Cubic (FCC)	0.740	4
Hexagonal Close-Packed (HCP)	0.740	6

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The key idea is to convert all lengths to the same unit (metres) and then divide the total length by the diameter of one atom. The number of carbon atoms that can be placed side by side in a straight line across 20 cm is $\boxed{1.33 \times 10^{9}}$ .

Why This Works: The Atomic Packing Scale

When you place atoms side by side in a straight line, you are essentially stacking them like beads on a string. The number of atoms that fit is simply the total length divided by the diameter of one atom — provided both lengths are in the same unit. This is a pure scaling problem: you are comparing a macroscopic length (20 cm) to a microscopic length (0.15 nm). The trick is to handle the enormous difference in scale without losing track of powers of ten.

Watch out

The most common mistake here is forgetting to convert units. You cannot divide 20 cm by 0.15 nm directly — the units must match. Always convert everything to metres (or any single consistent unit) before dividing.

Step-by-Step Solution

1. Write down what is given.

Diameter of a carbon atom: $d = 0.15 \text{ nm}$
Total length: $L = 20 \text{ cm}$

2. Convert both lengths to metres (the SI base unit).

$1 \text{ nm} = 10^{-9} \text{ m}$ , so $d = 0.15 \times 10^{-9} \text{ m} = 1.5 \times 10^{-10} \text{ m}$ .
$1 \text{ cm} = 10^{-2} \text{ m}$ , so $L = 20 \times 10^{-2} \text{ m} = 0.2 \text{ m}$ .

Tip

You could also convert both to nanometres: $20 \text{ cm} = 20 \times 10^{7} \text{ nm} = 2 \times 10^{8} \text{ nm}$ . Then divide by $0.15 \text{ nm}$ . Either way works — just be consistent.

3. The number of atoms is the total length divided by the diameter of one atom.

Since the atoms are placed side by side with no gaps, the number $N$ is:

N = \frac{L}{d}

4. Substitute the values in metres. …

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Q.If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Why This Works: The Atomic Packing Scale

Step-by-Step Solution

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