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प्रश्न
An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm-3. calculate the number of atoms in 108 g of the element.
उत्तर
The volume of a unit cell
=(300 pm)3
=(3.00 × 10-8 cm)3
= 2.7 × 10-23 cm3
Volume of 108 g of element
= `"mass"/"density"= (108 "g")/(10.8 "cm"^-3) = 10 "cm"^3`
Number of unit cells in this volume =`(10"cm"^3)/(2.7xx10^-23 "cm"^3//"unit cell")= ( 10^24)/(2.7) "Unit cells"`
Since each FCC unit cell contains 4 atoms, therefore the total number of atoms in 108 g 4 atoms/unit cell × `10^24/2.7`unit cell
= 1.48 × 1024 atoms
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