Niobium crystallizes in body-centred cubic structure. If the density is `8.55 g cm^(-3)`, calculate the atomic radius of niobium using its atomic mass `93 u`.

Explanation

Calculation of edge length of unit cell.
No. of particles in b.c.c. type unit cell (Z) = 2
Atomic mass of the element (M) = `"93 g mol"^(-1)`
Mass of the unit cell = `(ZxxM)/(N_(0))=(2xx("93 g mol"^(-1)))/((6.022xx10^(23)"mol"^(-1)))=30.89xx10^(-23)g`
Density of unit cell `(rho)=8.55"g cm"^(-3)`
`"Volume of unit cell "(a^(3))=("Mass of unit cell")/("Density of unit cell")=((30.89xx10^(-23)g))/((8.55"g cm"^(-3)))`
`=3.613xx10^(-23)cm^(3)=36.13xx10^(-24)cm^(3)`
Edge length of unit cell (a) = `(36.13xx10^(-24)cm^(3))^(1//3)=3.31xx10^(-8)cm="331 pm"`
Step II. Calculation of radius of unit cell
For b.c.c. structure, `r=(sqrt(3))/(4)=(sqrt3xx(3.31xx10^(-8)cm))/(4)=1.43xx10^(-8)cm="143 pm".`