What fraction of calcium atoms lies on the surface of a cubie crystal that is 1.00 cm in length ? Calcium has fcc lattice with edge length 0.556 nm.

Explanation

The number of unit cell present in the crystal =`(V_(("crystal")))/(V_(("unit cell")))=((1.0cm)^(3))/((0.556xx10^(-7)cm)^(3))=5.82xx10^(21)`
Number of atoms present (Z = 4) = `4xx5.82xx10^(21)=23.28xx10^(21)`
Number of unit cells on the surface `=("Area of six faces of crystal")/("Area of unit cell")`
`(6xx(1cm)^(2))/((0.556xx10^(-7)cm)^(2))=1.94xx10^(15)`
The number of atoms present on the surface of the unit cell = face centred atom + 1/4th of each of the four corner atoms = 2
Thus, the number of atoms present on the surface `=1.94xx10^(15)xx2=3.88xx10^(15)`
Fraction of atoms on the surface `=(N_(("surface")))/(N_(("crystal")))=(3.88xx10^(15))/(23.28xx10^(21))=1.67xx10^(-7).`