Question

Argon crystallizes in a face-centred cubic lattice with one atom at each lattice point. If the unit cell length is $5.311\mathring{A}$ at $0K$, the nearest neighbour distance in $Ar$ at $0K$ is:

• A. $3.755\mathring{A}$
• B. $4.75\mathring{A}$
• C. $6.25\mathring{A}$
• D. $7.33\mathring{A}$

Answer 1

The correct option is A $3.755\mathring{A}$

Given that the unit cell length $\left(a\right)$ is $5.311\mathring{A}$.

For FCC unit cell, $r=\frac{a}{2\sqrt{2}}=0.3535a=0.3535\times 5.311=1.8774\mathring{A}$

Hence, the nearest neighbor distance is $2\times 1.8774\mathring{A}=3.755\mathring{A}$