Question #ee4f4

The idea here is that you need to use the configuration of a face-centered cubic unit cell to find a relationship between the length of the cell, which is usually labeled `a`, and the radius of an atom of gold, which is usually labeled `r`.

So, a face-centered cubic unit cell looks like this

![https://classconnection.s3.amazonaws.com/38/flashcards/512038/png/fcc1316360426857png]()

Now, you should know that the diagonal of a square is equal to

`"diagonal" = "side" xx sqrt(2)`

In this case, the diagonal of a face has a length of

`"diagonal" = r + 2r + r = 4r`

Here `r` represents the radius of the two quarter of an atom that occupy the two corners of the face and `2r` represents the diameter of the atom that occupies the middle of the face.

![)

This means that the length of the side of the unit cell

`"side" = "diagonal"/sqrt(2)`

will be equal to

`a = (4r)/sqrt(2) = (4r sqrt(2))/2 = 2sqrt(2) * r`

Plug in your value to find

`a = 2 sqrt(2) * "0.144 nm" = color(darkgreen)(ul(color(black)("0.407 nm")))`

The answer is rounded to three sig figs.