Question

The closest distance between two atoms (in terms of edge length) would be highest for which of the unit cell, assuming the edge length of each unit cell to be 'a'?

• A. FCC unit cell
• B. BCC unit cell
• C. Diamond unit cell
• D. Primitive cubic unit cell (SCC)

Answer 1

The correct option is D Primitive cubic unit cell (SCC)

The closest distance between two atoms is given by $2r$.
In FCC, the nearest atoms are the one which present at the face centres when the reference atom is at corner. Here, the corner atoms and the face-centre atoms are in contact along the face diagonal diagonal.
So, the distance between these atom is $\frac{\sqrt{2}a}{2}$.
$\therefore$
$2r=\frac{\sqrt{2}a}{2}$

In BCC, the nearest atom is the one which present at the body centre when the reference atom is at the corner. Here, the corner atoms and the centre atoms are in contact along the body diagonal.we know, distance along body diagonal is $\sqrt{3}a$.
Thus, half the distance of body diagonal is $\frac{\sqrt{3}a}{2}$
$\therefore$
$2r=\frac{\sqrt{3}a}{2}$

In diamond, carbon particles occupy half of tetrahedral voids and occupy fcc sites. The structure of diamond is identical to zinc blende $\left(ZnS\right)$ structure.
Here,
$\frac{\sqrt{3}a}{4}=2r$

In primitive cubic, the first nearest atom for any atom in a unit cell is the atom located at adjacent corner of it.
$\therefore$
$2r=a\left(\text{Highest value}\right)$
So, the highest value is for primitive cubic cell.