Unit cell of the iron crystal has edge length of 288 pm and density of 7.86 g `cm^(-3)`. Determinte the type of crystal lattice. Atomic mass of Fe= 56

Correct Answer - Type of crystal lattice is bcc.
Given : Edge length =a= 288 pm `=2.88 xx 10^(-8)` cm
Density of crystal = d= 7.86 g `cm^(-1)`
A vogadro number = 6.022 `xx10^(-23) mol^(-1)`
Atomic mass of Fe= 56 g `mol^(-1)`
Type of crystal lattice = ?
Mass of one Fe atom = `(56)/(6.022 xx 10^(23)) = 9 .3 xx 10^(-23) g`
If there are z atoms in the unit cell , then
Mass of unit cell= mass of z atoms = z `xx 9.3 xx 10^(-23) g`
Volume of unit cell `= a^(3) = (2.88 xx 10^(-8) )^(3)`
`=23 .88 xx 10^(-24) cm^(3)`
Density of unit cell `d= ("mass of unit cell")/("Volume of unit cell")`
`7.86 = (z xx 9.3 xx 10^(-23))/(23. 88 xx 10^(-24))`
`:. z =(7.86 xx 23.88 xx 10^(-24))/(9.3xx 10^(-23)) = 2.01 =2`
Since the number of atoms in the unit cell is 2, the crystal lattice must be of bcc type.