Question

An element has a bcc structure with

unit cell edge length of 288 pm. How

many unit cells and number of atoms

are present in 200 g of the element?​

Answer 1

ANSWER

Volume of unit cell =(288 pm)

3

=(288×10

−10

cm)

3

=2.389×10

−23

cm

3

Volume of 208 g of the element =

Density

Mass

=

7.2

208

=28.89 cm

3

Number of unit cells =

Volume of a unit cell

Total Volume

=

2.389×10

−23

28.89

=12.09×10

23

For a BCC structure, number of atoms per unit cell =2

∴ Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells

=2×12.09×10

=24.18×10

23

=2.418×10

24