An element having atomic mass 107.9 u has FCC lattice. The edge length of its unit cell is 408.6 pm. Calculate density of the unit cell.
`["Given, "N_(A)=6.022xx10^(23)"mol"^(-1)]`.

Explanation

(a) In simple cubic unit cell, atoms are located only at the corners of the cube. The particles touch one another along the edge.
If the edge length of the cube = a, and radius of each particle is r, the a is related to r as,` a = 2r`
The volume of the cubic unit cell `=a^3 = (2r)^3 = 8r^3`
Since a simple cubic unit cell contains only 1 atom
The volume of the occupied space `= 4/3 pi r^3`
Packing Efficiency = `("Volume of space occupied by atom")/("Volume of cubic unit cell") xx 100`
`= (4/3 pi r^2)/(8 r^2) xx 100`
`= (4pi r^3)/(3 xx 8 r^3) xx 100`
`= pi/6 xx 100 = 52.36%`
(b) `d = (z.M)/(a^3.N_A)`
`= (4 xx 107.9 xx 10^3)/((408.6 xx 10^(-12))^(3) xx 6.022 xx 10^(23))= 10.5 xx 10^(3) kg m^(-3)`