First slide

Crystal lattices and unit cells

Question

A metal has density of 4500 kg/m³ with an fcc crystal structure and edge length is 400 pm. The number of unit cells present in 1 kg of the metal is n × 10²⁴. Value of n is ________. (upto two decimal place)

NA

Solution

$Volume available = \frac{1 kg}{4500 kg / m^{3}} = \frac{1}{4500} m^{3}$
$Vol. of a unit cell = {(400 \times 10^{- 12})}^{3} m^{3}$
$No. of unit cell = \frac{1}{4500 \times {(400 \times 10^{- 12})}^{3}} = 3.47 \times 10^{24}$

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