Question

If the radius of a metal atom is $2.00\stackrel{\circ }{A}$ and its crystal structure is fcc lattice, what is the volume (in $c{m}^3$) of one unit cell?

• A. $8.00\times {10}^{-24}$
• B. $0.5\times {10}^{-23}$
• C. $1.81\times {10}^{-22}$
• D. $4.45\times {10}^{-23}$

Answer 1

The correct option is C $1.81\times {10}^{-22}$

For a fcc lattice:


Along the face diagonal
$\sqrt{2}\times a=4\times r$
where
$a\text{is ledge length of unit cell}$
$r\text{is the radius of the atom}=2.00\stackrel{\circ }{A}=2\times {10}^{-8}cm$


$\therefore a=2\sqrt{2}r$

Volume (V) of a unit cell is given by:

$V=a^3=(2\sqrt{2}r{)}^{3}V=16\sqrt{2}{r}^3=16\times \sqrt{2}\times (2\times {10}^{-8}{)}^{3}V=1.81\times {10}^{-22}c{m}^3$