An element has a bcc structure with a unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (Density of an element = 14.44 g/cm3 )
Solution
Given: Type of unit cell is bcc.
Edge length (a) = 288 pm = 2.88 × 10–8 cm,
Mass of element (x) = 200 g
To find:
i. Number of unit cells in 200 g of element
ii. Number of atoms in 200 g of element
Formulae:
i. Number of atoms in x g of element = `"x n"/(rho "a"^3)`
ii. Number of unit cells in x g of element = `"x"/(rho "a"^3)`
Calculation:
i. For bcc unit cell, n = 2.
Using formula (i),
Number of atoms in 200 g of element
`= (200 "g" xx 2)/(14.44 "g cm"^-3 xx (2.88 xx 10^-8 "cm")^3) = 1.16 xx 10^24`
ii. Using formula (ii),
Number of unit cells in 200 g element
`= (200 "g")/(14.44 "g cm"^-3 xx (2.88 xx 10^-8 "cm")^3) = 5.80 xx 10^23`
∴ Number of atoms in 200 g element is 1.16 × 1024.
∴ Number of unit cells in 200 g element is 5.80 ×1023.
