# Question #f3e7d

##### 1 Answer

#### Explanation:

The first thing to do here is to determine *how many* lattice points you get for a **face-centered cubic** unit cell.

For starters, all **three** types of cubic lattice unit cells, i.e. *simple cubic*, *body-centered cubic*, and *face-centered cubic*, have lattice points in the **corners** of the cubic unit cell.

So right from the start you know that the face-centered cubic cell has **at least**

Now, the characteristic of a face-centered cubic cell is that it also contains lattice points in the *centers* of the cube's faces.

Since a cube has a total of **six faces**, it follows that the total number of lattice points you get per unit cell is

#"no. of lattice points" = overbrace("8 lat. pts.")^(color(blue)("corners")) + overbrace("6 lat. pts.")^(color(red)("faces")) = "14 lat. pts."#

To determine the number of *atoms* that you can fit into a face-centered cubic unit cell, you need to look at how the lattice structure is arranged.

For a given unit cell, pick one of its corners. This corner is being shared by a total of **eight** unit cells. This means that *each corner* in a unit cell will contain

Simply put, eight unit cells share one atom *per corner*.

For a given cell, pick one of its faces. This face is being shared by a total of **two** unit cells, which means that *each face* in a unit cell will contain

In other words, two unit cells share one atom *per face*.

So, the total number of atoms that can fit into a face-centered cubic unit cell will be

#"no. of atoms" = overbrace( 1/8 xx 8)^(color(blue)("8 corners")) + overbrace(1/2 xx 6)^(color(red)("6 faces")) = color(green)("4 atoms")#