Total volume of atoms present in a face-centred cubic unit cell of a metal is (r is atomic radius)
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Q: Why do we take the contribution of a corner atom as 1/8?

Imagine a single corner of a cube in a 3D lattice. That specific corner point is the meeting junction for 8 identical cubes (4 on the bottom layer, 4 on the top layer). Therefore, the atom sitting exactly at that corner is divided equally among those 8 cubes.

Q: Why is the face-centered atom's contribution 1/2?

A face of a cube is shared exactly between two adjacent cubes in a lattice. So, an atom placed right at the center of that face will have half of its volume in one cube and the other half in the neighboring cube.

Q: What if the question asked for a Body-Centered Cubic (BCC) unit cell?

For a BCC unit cell, you have 8 corner atoms (contribution = 1) and 1 atom completely inside the body center (contribution = 1). So, $Z = 2$. The total volume would be $2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3$.

Q: Do we need to know the edge length 'a' for this question?

No! The question specifically asks for the volume in terms of the atomic radius '$r$'. If it asked for the packing efficiency or the volume of the unit cell itself, then we would need the relationship between '$a$' and '$r$' ($a = 2\sqrt{2}r$ for FCC).

Q: Is the volume of an atom always $\frac{4}{3}\pi r^3$?

Yes, in the hard-sphere model of crystallography, we assume atoms are perfect, incompressible spheres. The volume of a sphere is universally $\frac{4}{3}\pi r^3$.

Question

Total volume of atoms present in a face-centred cubic unit cell of a metal is ($r$ is atomic radius)

Options:

(A) $\frac{20}{3} \pi r^3$

(B) $\frac{24}{3} \pi r^3$

(C) $\frac{12}{3} \pi r^3$

(D) $\frac{16}{3} \pi r^3$

Sigma Percentile · Accepted Answer

### The Anatomy of an FCC Unit Cell

Imagine you are holding a tiny, microscopic cube. This cube represents the fundamental building block of a crystal lattice, known as a unit cell. In a **Face-Centred Cubic (FCC)** lattice, the atoms are arranged in a very specific and highly packed manner.

If you look closely at this cube, you will find atoms sitting exactly at all the $8$ corners. But that's not all! As the name suggests, there is also an atom sitting right in the center of every single face of the cube. Since a cube has $6$ faces, there are $6$ face-centered atoms.

### Counting the Atoms: The Corners

Now, here is where the magic of crystallography comes in. An atom at the corner of a unit cell does not belong entirely to that single cell. In a massive 3D crystal lattice, that specific corner is the meeting point for $8$ adjacent cubes.

Because the atom is shared equally among these $8$ cubes, only a fraction of it actually resides inside our specific unit cell.

$$ 	ext{Contribution per corner} = \frac{1}{8} $$

Since there are $8$ corners in total, the effective number of atoms contributed by all the corners combined is:

$$ 	ext{Total corner atoms} = 8 	imes \frac{1}{8} = 1 $$

### Counting the Atoms: The Faces

Next, let's shift our focus to the atoms located at the centers of the faces. A face of a cube acts as a wall separating exactly two adjacent unit cells. Therefore, an atom sitting on this wall is shared equally between these two cells.

$$ 	ext{Contribution per face} = \frac{1}{2} $$

A cube has $6$ faces, so the effective number of atoms contributed by all the face centers is:

$$ 	ext{Total face atoms} = 6 	imes \frac{1}{2} = 3 $$

### The Grand Total and Volume

To find the total effective number of atoms (often denoted by $Z$) in an FCC unit cell, we simply add the contributions from the corners and the faces.

$$ Z = 1 + 3 = 4 $$

So, an FCC unit cell effectively contains $4$ complete atoms.

Now, we need to find the total volume occupied by these atoms. In the hard-sphere model, we assume each atom is a perfect sphere with a radius $r$. The volume of a single sphere is a standard geometric formula:

$$ 	ext{Volume of one atom} = \frac{4}{3} \pi r^3 $$

Since we have $4$ effective atoms in our unit cell, the total volume occupied by them is simply $4$ times the volume of a single atom:

$$ 	ext{Total Volume} = 4 	imes \left( \frac{4}{3} \pi r^3 ight) $$

$$ 	ext{Total Volume} = \frac{16}{3} \pi r^3 $$

And there we have it! The total volume of atoms present in a face-centred cubic unit cell is $\frac{16}{3} \pi r^3$, which corresponds perfectly to option (d).

Animated Solution for Chemistry - States of Matter: Total volume of atoms present in a face-centred cubic unit cell of a metal is ( $r$ is atomic radius)

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Visualized Solution

FCC Unit Cell Structure

Contribution of Corner Atoms

Contribution of Face-Centered Atoms

Total Effective Atoms ( $Z$ )

Volume of One Atom