Question

How many lithium atoms are present in a unit cell with edge length $3.5$ $\mathring{A}$ and density $0.53\,g\,cm^{-3}$ ?
(Atomic mass of $Li = 6.94$)

• A. $2$
• B. $1$
• C. $4$
• D. $6$

Answer 1

Answer: (A)

$d=\frac{Z\times M}{N_{A}\times a^{3}}$
$Z=\frac{a^{3}\times d\times N_{A}}{M}$
$Z=\frac{\left(3.5\times10^{-8}\right)^{3}\times0.53\times6.023\times10^{23}}{6.94}$
$=1.97\approx2$