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)].

An element having atomic mass 107.9 u has FCC lattice. The edge length

1.	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)].
Answer» Solution :(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)`