Question

Taking the reference to be the corner particles for a fcc lattice, the number of second nearest atoms and the distance between them is, respectively :
Here,
r is the radius of the atom

• A. $8\text{and}\frac{4r}{\sqrt{2}}$
• B. $8\text{and}\frac{4r}{\sqrt{3}}$
• C. $6\text{and}\frac{4r}{\sqrt{2}}$
• D. $12\text{and}\frac{4r}{\sqrt{3}}$

Answer 1

The correct option is C $6\text{and}\frac{4r}{\sqrt{2}}$

In the fcc crytsal lattice, the atoms are present at corners of the cube and at the face-centres of the cube. Here, the corner atoms and the face-centre atoms are in contact along the face diagonal diagonal.
So, the distance between these atom is $\frac{\sqrt{2}a}{2}$. This is the nearest distance in fcc. Hence, the nearest atoms are the one which present at the face centres when the reference atom is at corner. There are 12 nearest atom in this unit cell.
The second nearest distance in the unit cell is 'a'
where, a is the edge length of the cube.
We know,
$2r=\frac{\sqrt{2}a}{2}$
$a=\frac{4r}{\sqrt{2}}$
Since, the second nearest atom is at distance 'a'
The number of second nearest atom is 6.