Metallic magnesium has a hexagonal close packed structure and a density of `1.74 g//cm^3`. Assuming magnesium atoms to be spherical, calculate the volume of each atom and atomic radius of Mg atom (Atomic mass of Mg =24)

Explanation

As Mg has hexagonal close packed structure , no. of atoms per unit cell = 6 ,i.e, Z = 6
`p= (Z xx M)/( a^(3)xxN_(0)) = (Z xxM)/(V xx N_(0)) " " (V =a^(3)`= volume of the unit cell)
` 1.74 = ( 6xx24)/( V xx (6.023 xx10^(23))) or V = 1.374 xx 10^(-22) cm^(3)`
Now ,as in hexagonal close packing, 74% of the space is occupied by atoms, therefore, volume occiped by atoms ` = 74/100 xx V`
` = 74/100 xx 1.734 xx 10^(-22) = 1.018 xx 10^(-22) cm^(3)`
As a unit cell contains 6 Mg atom, therefore, volume of each Mg atom.
`( 1.018 xx 10^(-22))/6 = 1.697 xx 10^(-23) cm^(3)`
As Mg is assumed to be spherical.
` 4/3 pir^(3) = 1.697 xx 10^(-23)`
on solving we get ` r = 1.594 xx 10^(-8) cm= 1.594Å`