Question

An element having bcc geometry has atomic mass 50.Calculate the density of the unit cell, if the edge length is 290pm.

Answer 1

the answer to this is 29/5

Answer 2

Answer : The density of the unit cell is, $6.808g/Cm^{3}$

Solution : Given,

Number of atom in unit cell of BCC (Z) = 2

Edge length = $290pm=290\times 10^{-10}cm$     $(1pm=10^{-10}cm)$

Atomic mass of chromium (M) = 50 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

$\rho=\frac{2\times (50g/mol)}{(6.022\times 10^{23}mol^{-1}) \times(290\times 10^{-10}Cm)^3}=6.808g/Cm^{3}$

Therefore, the density of the unit cell is, $6.808g/Cm^{3}$