Question #ee4f4

1 Answer
Jul 8, 2017

#"0.407 nm"#

Explanation:

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.

http://www.chemteam.info/Liquids&Solids/WS-fcc-AP.html

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.