Question

A metallic element 'X' exists as a cubic lattice Each edge of the unit cell is 2.90Ȧ and density of
metal is 7.20 g cm-3
.How many unit cell will be present in 100g of the 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 = 103.57d=

N∗a

3

Z∗A

i.e.7.2=

(6.022∗10

23

)∗(2.88∗10

−8

)

3

1∗A

i.eA=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} } =

103.57∗1.667∗10

−24

100

= 5.792*10^{23}10

23

number of unit cells