[kimisizer]

I vaguely remember this type of question, but do not remember how to do it. The question is:
The density of nickel is 8.92g/cm^3. Nickel has a face-centered cubic unit cell. What is the radius of a nickel atom?

I remember that face-centered has 4 atoms and Avogadro's number was needed but other than that I forget. Can anyone help me? The answer is supposed to be 1.25x10^-8 cm

[opti384]

You should first determine the volume of the unit cell by using the density.

[vmelkon]

I can't give the answer because you have to show that you have at least attempted the problem but I can give a start

Besides knowing the density, you also need to know the molar mass
Density = 8.92g/cm^3
Molar mass = 58.71 g/mol

From that, you can calculate how many mol a single cm^3 has.
From that, you can calculate how many atoms a single cm^3 has.

As for the face-center, we need a picture
http://en.wikipedia.org/wiki/Face-centered_cubic

so a single cell has 14 atoms.
Some of those atoms are shared by neighboring cells.
An corner atom is shared by 4 cells, therefore it's "whatever you call it" is 1/4.
A face atom is shared by 2 cells, therefore it is 1/2.
When we count all the atoms in a cell :
1/4+1/4+1/4+1/4+1/4+1/4+1/4+1/4+1/2+1/2+1/2+1/2+1/2+1/2= 5 atoms

and I'll leave the rest to you.
I hope I didn't make a mistake with the atoms/cell calculation.