Chromium metal crystallizes with a body-centred cubic lattice. The edge length of the unit cell is found to be 287 pm. Calculate the atomic radius. What would be the density of chromium in g `cm^(-3)` ? (atomic mass of Cr = 52.99)

Explanation

For body centred cubic cell atomic radius,
`r=(sqrt(3)xx"edge length")/(4)`
`=(sqrt(3)xx287)/(4)=124.27" pm"`
Thus, atomic radius of chromium metal is 124.27 pm.
Density of the unit cell of the solid
`=("Mass of unit cell")/("Volume of unit cell")`
Mass of unit cell
= No. of atoms in unit cell (Z) `xx` Mass of each atom (m)
For body centered cubic lattice, the number of atoms in unit cell (Z) is
`=1+((1)/(8)xx8)=1+1=2`
Mass of each atom
`(m)=("Atomic mass")/("Avogadro.s no.")`
`=(52.99)/(6.023xx10^(23))`
The volume of the cubic lattice
`=("Edge length")^(3)`
`=(287xx10^(-10))^(3)`
`therefore` Density of the unit cell is
`=(2xx52.99)/(6.023xx10^(23)xx(287)^(3)xx10^(-30))gm//cm^(3)`
`=7.44g//cm^(3)`
Thus, the density of chromium is `7.44g//cm^(3)`.