Question

Copper has face centered cubic (fcc) lattice with interatomic spacing equal to $2.54\mathring{A}$, the value of lattice constant for this lattice is :

• A. $1.27\mathring{A}$
  
• B. $5.08\mathring{A}$
• C. $2.54\mathring{A}$
• D. $3.59\mathring{A}$

Answer 1

Answer: (D)

$3.59\mathring{A}$

For FCC lattice,

$4r=\sqrt{2}a$

Given- interatomic distance, $2r=2.54\mathring{A}$

$a=\frac{2\times 2r}{\sqrt{2}}=\sqrt{2}\times 2.54=3.59\mathring{A}$