Question

Nickel has a face-centered cubic unit cell. The density of nickel is 6.84 g/cm3. Calculate a value for the atomic radius of nickel.
Please help! I do not understand the concept

Answer 1

i didn't know how to do this either, so i googled it, this is what came up:

"Assume that the ditsance from the cube centre to face is s - the atomic radius.

The volume of the cube is (2s)^3 or 8s^3.

AW of No is 58.7. Therefore 1 cm^3 is 6.84 / 58.7 = 0.1165 mol

Therefore 0.1165 x 6.023 x 10^23 x 8s^3 = 1 where 6.023 x10^23 is Avogadro's number

Soving for s leads to s = 1.21 x10^-8 cm or 0.121 nm"