Question

If the length of the unit cell is $5\mathring{A}$, the smallest distance in $\mathring{A}$ between the two neighbouring metal atoms in a face centred cubic lattice is:

• A. 250
• B. 5
• C. 7.07
• D. 3.535

Answer 1

Answer: (D)

3.535

If the length of the unit cell is 5A˚, the smallest distance in A˚ between the two neighbouring metal atoms in a face centred cubic lattice is
$\frac{a}{\sqrt{2}}=\frac{5}{\sqrt{2}}=3.535A^o$