The unit cell of an element of atomic mass 108 and density '10.5 . g cm^-3' is a cube with edge 'length '409 pm'. Find the structure of the.crystal lattice 'N_A=6.022 xx 10^33'

Explanation

density. .=2 xx M/i/a^3 xx N_A.
.=Z xx 108 xx 10^-3 */(409 xx 10^-12)^3 xx 6.022 xx 10^23.
.Z=10.5 xx 10^-3 xx(409 xx 10^-12)^1 xx 6: 022 xx 10^23/10^-6 xx 108 xx 10^-3.
.=4.
Since there are 4 atoms per unit cell, the crysta lattice is faće centred cubic.