Question

An element with density 10g/cc forms a cubic cell with edge length of 3 x 10-8cm. what is the nature of cubic unit cell if the atomic mass of the element is 81g/mol.

Answer 1

body centred cubic lattice

Answer 2

Answer : The nature of the cubic unit cell is, BCC (Body centered cubic unit cell) and the number of atoms present in unit cell is, 2

Solution : Given,

Density = 10 g/cc

Edge length = $3\times 10^{-8}cm$

Atomic mass of an element(M) = 81 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

$10g/cm^3=\frac{Z\times (81g/mol)}{(6.022\times 10^{23}mol^{-1}) \times(3\times 10^{-8}Cm)^3}$

$Z=2.0073=2$

The number of unit cell present in the unit cell is 2 that means the nature of cubic unit cell is BCC (Body centered cubic unit cell)

Therefore, the nature of the cubic unit cell is, BCC (Body centered cubic unit cell) and the number of atoms present in unit cell is, 2