Question

A metallic element exists as a cubic lattice each edge of the cubic lattice is 2.88 amstrong the density of the metal is 7.2g/cc . How many unit cell are there in 100g of metal .

Answer 1

Assuming that it is a simple cubic lattice the effective number of atoms per unit cell (Z)=1

Atomic mass of the element is A

Avogadro's number N = 6.022*10^23

Edger length a = 2.88*10^(-8)cm

Density d = 7.20 g/cc

now

d=\frac{Z*A}{N* a^{3} } 
i.e. 7.2= \frac{1*A}{(6.022* 10^{23})*(2.88* 10^{-8}) ^{3}} 
i.e A = 103.57


so

mass of one atom ≡ 1 unit cell

i.e. 103.57u ≡ 1 unit cell

or 103.57 * 1.667 * 10^(-24) g ≡ 1 unit cell

or 100 g metal⇒ \frac{100}{103.57*1.667* 10^{-24} } =5.792* 10^{23} number of unit cells