`Li` forms a body-centred cubic lattice. If the edge of the cube is `3.5 xx 10^(-10)m` and the density is `5.3 xx 10^(2) kg m^(-3)`, calculate the percentage occupancy of `Li` metal.

Options

A. 0.4
B. 0.6
C. 0.8
D. 0.98

Explanation

Since the Li forms a body centered cubic lattice, the effective number of atoms per unit cell`=8xx(1)/(8)+1=2`.
Volume of unit cell `=(3.50xx3.50xx3.50)(10^(-30))m^(2)=42.875xx10^(-30)m` Ideal density
`(2.324xx10^(-26))/(42.875xx10^(-30))=0.054xx10^(4)kgm^(3)=5.4xx10^(2)kgm^(3)`
Experimental density=% occupancy `=(5.3xx10^(3)kgm^(3))/(5.4xx10^(2)kgm^(3))xx100=98%`