An element has a bcc structure with unit cell edge length of 288 pm.

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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?

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Given: Types of unit cell = bcc, for bcc unit cell n = 2

Edge length (a) = 288 pm = 2.88 x 10-8 cm3

Mass of element (x) =200g

Density of element (r) = 7.2 gm/cm3

Number of unit cell in x g of element = $$\frac{x}{ρa^3}$$

= $$\frac{200}{7.2×(2.88×10^{-8})^3}$$ = 1.16 x 1024

Number of atoms in x g of element = $$\frac{nx}{ρa^3}$$

= $$\frac{2x}{ρa^3}$$= 2 × 1.16 x 1024 = 2.32 x 1024

Ans. Number of unit cell in x g of element = 1.16 x 1024

Number of atoms in x g of element = 2.32 x 1024