The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in

Exercises · 2.37

Q. The diameter of zinc atom is 2.6 Å. Calculate

(a) radius of zinc atom in pm and

(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

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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 radius of a zinc atom is $130\ \text{pm}$ , and about $6.15 \times 10^{7}$ atoms fit side by side in a $1.6\ \text{cm}$ length.

Why this works — the Atomic Packing Scale

When we talk about atoms arranged "side by side lengthwise", we are essentially building a one-dimensional chain. The total length of the chain equals the number of atoms multiplied by the diameter of one atom. This is the simplest packing model — no gaps, no fancy crystal structures — just spheres touching each other in a straight line.

The only trick is keeping the units consistent. Atomic diameters are given in Ångströms ( $1\ \text{Å} = 10^{-10}\ \text{m}$ ), but exam questions often expect answers in picometres ( $1\ \text{pm} = 10^{-12}\ \text{m}$ ) or centimetres. Converting carefully is half the battle.

Step-by-step solution

1. Convert diameter to radius in picometres

The diameter of a zinc atom is $2.6\ \text{Å}$ .

First, recall the conversion:

1\ \text{Å} = 10^{-10}\ \text{m} = 100\ \text{pm}

So:

2.6\ \text{Å} = 2.6 \times 100\ \text{pm} = 260\ \text{pm}

The radius is half the diameter:

\text{radius} = \frac{260\ \text{pm}}{2} = 130\ \text{pm}

Tip

A quick mental shortcut: $1\ \text{Å} = 100\ \text{pm}$ , so $2.6\ \text{Å}$ is $260\ \text{pm}$ diameter, giving $130\ \text{pm}$ radius. No need to go through metres.

2. Find the number of atoms in a 1.6 cm length

We need the diameter in the same unit as the given length. Convert the diameter from Ångströms to centimetres.

1\ \text{Å} = 10^{-8}\ \text{cm}

So:

$\text{diameter} = 2.6\ \text{Å} = 2.6 \times 10^{-8}\ \text{cm}$ …

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Q.The diameter of zinc atom is 2.6 Å. Calculate

Why this works — the Atomic Packing Scale

Step-by-step solution

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