Question

Potassium has a bcc structure with nearest neighbour distance 4.52A. Its atomix weight is 39. Its densitt will be

Answer 1

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Answer 2

Answer : The density of the unit cell is, $0.1138g/Cm^{3}$

Solution : Given,

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

Nearest neighbor distance, r = $4.52A^o=4.52\times 10^{-8}cm$ $(1A^o=10^{-8}cm)$

Atomic mass (M) = 39 g/mole

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

First we have to calculate the edge length of unit cell.

Formula used :

$a=\frac{4r}{\sqrt{3}}$

where,

a = edge length of unit cell

r = nearest neighbor distance

Now put all the given values in this formula, we get :

$a=\frac{4\times 4.52\times 10^{-8}cm}{\sqrt{3}}=10.44\times 10^{-8}cm$

Now we have to calculate the density of unit cell.

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

$\rho=\frac{2\times (39g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (10.44\times 10^{-8}Cm)^3}=0.1138g/Cm^{3}$

Therefore, the density of the unit cell is, $0.1138g/Cm^{3}$