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Question
An atom crystallizes in fcc crystal lattice and has a density of 10 g cm−3 with unit cell edge length of 100 pm. calculate the number of atoms present in 1 g of crystal.
Solution
Given, Density = 10 g cm−3
mass = 1 g
Edge length of unit cell = 100 pm
Volume = `"mass"/"density" = (1 "g")/(10 "g cm"^-3)` = 0.1 cm3
Volume of unit cell = a3
= 100 × 10−10cm3
= 1 × 10−24 cm3
Number of unit cell in 1 g of crystal,
= `"Total volume"/"Volume of unit cell"`
= `(0.1 "cm"^3)/(1 xx 10^-24 "cm"^3)`
The given unit cell is of fcc type. Therefore. it contains 4 atoms.
0.1 × 1024 unit cells will contain 4 × 0.1 × 1024 = 4 × 1023atoms
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