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# How many atoms are there in a face-centered cubic unit cell of an atomic crystal in which all atoms are at lattice points?

##4##

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** ##8## lattice points, one for each corner of the cube.

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 ##1/8"th"## of an atom.

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 ##1/2## of an atom.

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")##