In a cubic lattice each edge of the unit cell is 400 pm. Atomic weigth of the element is 60 and its density is` 625g//c.c.`Avogadro number =`6times10^(23)`.The crystal lattice is

Options

A. Face centred
B. Primitive
C. Body centred
D. End centred

Answer

A

Explanation

Let the number of atoms in a unit cell =x
Mass of x atoms i.e., one unit cell =`(60timesx)/(6times10^(23))`
Volume of the unit cell =`("edge length")^(3)`
=`(400times10^(-12)times100)^(3)`
=`(400times10^(-10)cm)^(3)`
=`(4times10^(-8)cm)^(3)`=`64times10^(-24)cm^(3)`
`"Density"=6.25=("Mass of unit cell")/("Volume of unit cell")`
`therefore6.25=(60timesx)/(6times10^(23)times64times10^(-24))`
`x=(6.25times6times64times10^(-1))/60=4`
Since the unit cell contains 4 atoms, so it is
face centred cubic unit cell.