Metallic magnesium has a hexagonal close-packed structure and a density of `1.74gcm^(-3)`. Assuming magnesium atoms to be spherical, calculate the radius of magnesium atom. (Atomic mass of Mg = 24.3 u)

Options

A. `1.6xx10^(-8)` cm
B. `2.5xx10^(-8)` cm
C. `3.5xx10^(-8)` cm
D. `4.5xx10^(-8)` cm

Answer

A

Explanation

Mass of `1cm^(3)` of Mg = volume `xx` density = `1xx1.74` = 1.74 g
Number of atoms in 1.74 g of Mg = `(6.02xx10^(23)xx1.74)/(24.3)=4.31xx10^(22)` atoms
Volume occupied by Mg = 74.1% (for hcp) = 0.741 `cm^(3)`.
So, `4.31xx10^(22)` atoms will have volume = 0.741 `cm^(3)`.
Volume occupied by 1 atom = `(0.741)/(4.31xx10^(22))=1.72xx10^(-23)cm^(3)`
Let r be the radius of Mg atom, then
`4/3pir^(3)=1.72xx10^(-23)rArrr^(3)=1.72xx10^(-23)xx3/4xx1/3.14rArrr=(4.108xx10^(-24))^(1//3)=1.6xx10^(-8)cm`