Question

Copper had face-centered cubic (fcc) lattice with interatomic spacing equal to $2.54A^0$. The value of lattice constant for this lattice is:

• A. $1.27A^0$
• B. $5.08A^0$
• C. $2.54A^0$
• D. $3.59A^0$

Answer 1

The correct option is D $3.59A^0$

For FCC lattice, the inter atomic spacing$=\frac{a}{\sqrt{2}}$

where, $a=$ lattice constant

$\therefore a=$inter atomic spacing $\times \sqrt{2}$

$\therefore a=2.54\times 1.414$

$\therefore a=3.59A^0$