Nearest Neighbor and Second Nearest Neighbor Distances in FCC Lattice?

[USI]

Homework Statement

An element crystalises in a face-centred cubic lattice with a basis group of two atoms at 000 and 1/4 1/4 1/4. The lattice constant is 3.55Angstroms.

i) what is the separation of nearest neighbor atoms
ii) how many nearest and second nearest neighbors does each atom have.

attempt

i) The nearest neighbor distance is just going to be from the 000 atom to the one 1/4 1/4 1/4 away from it? Which gives
$$\sqrt{3(\frac{1}{4}a)^3} = 1.54Angstroms$$

ii) If the guess for part i is right then each atom will have one nearest neighbor. I can't get my head around the second nearest neighbors.

Any help appreciated, thanks.

 

[bcjack]

see attachment...

1) distance = 1/2*(2)^(1/2) * lattice constant
2) 6 next-nearest neighbors at distance of lattice constant