Question

An element having bcc geometry has atomic mass 50 you calculate the density of the unit cell if it says length is 290 picometre

Answer 1

Hope it helps..............

Answer 2

Answer:

The density of the unit cell is $6.809 g/cm^3$.

Explanation:

Number of atom in BCC unit cell = Z = 2

Density of platinum = $?$

Edge length of cubic unit cell= a  = 290 pm = $2.90\times 10^{-8} cm$

$1 pm = 10^{-10} cm$

Atomic mass of element= 50

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where:

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$\rho=\frac{2\times 50 g/mol}{6.022\times 10^{23} mol^{-1}\times (2.90\times 10^{-8} cm)^{3}}$

$\rho = 6.809 g/cm^3$

The density of the unit cell is $6.809 g/cm^3$.