Question

Consider a metal with atomic radius 1 angstrom, which crystallizes in a conventional fcc unit cell. To one decimal place and in angstroms, what is the unit cell edge length of the fcc lattice?

Answer 1

`a=2.8 angstroms`

Step-by-step explanation:

In the figure can be seen the Face centered cubic (FCC) structure. It contains 4 halves of an atom and 8 "1/8" of atom.

The side lengh:

`b=4*r`

Using pithagoras:

`b^2=a^2+a^2`

`(4*r)^2=2*a^2`

`a=\sqrt{((4*r)^2)/(2)}`

`a=√(8)*r`

`a=√(8)*1 ang.=2.8 angstroms`