From X-ray diffraction study it is observed that solid Ar possesses cubic close packed structure. If the density of solid Ar is `1.70g*cm^(-3)`, then find (1) edge length of the unit cell of Ar crystal (2) the radius of Ar atom.

Explanation

Since solid Ar has ccp structure, its crystal structure possesses fcc unit cell.
We know, `rho=(ZxxM)/(Nxxa^(3))` Given: `rho=1.7g*cm^(-3)`.
For Ar, M = `40g*mol^(-1)` and for fcc unit cell, Z = 4.
`therefore" "a^(3)=(ZxxM)/(rhoxxN)=(4xx40)/(1.7xx6.022xx10^(23))=1.562xx10^(-22)cm^(3)`
`therefore" "a=5.38xx10^(-8)cm=5.38Å`
(1) Hence, edge-length of the unit cell of Ar crystal is 5.38Å.
(2) In case of a fcc unit cell, r = `a/(2sqrt2)=(5.38)/(2sqrt2)=1.90Å`
`therefore" "`Radius of Ar atom = 1.90Å