In fcc, no of atoms=4
volume occupied atoms `=4[(4)/(3)pir^(3)]=(16)/(3)pir^(3)`

Correct Answer – D
Volume of atom in a cell `=(4)/(3)pi r^(3)xxn`
`(n=4 fo r f.c.c.)`
`=(4)/(3)pi r^(3)xx4=(16)/(3)pi r^(3)`

Correct Answer – D

Correct Answer – B
`a=2sqrt2r`
Volume of the cell =`a^3=(2sqrt2r)^3=16sqrt2r^3`
No. of spheres in fcc = `8xx1/8+6xx1/2=4`
Volume of 4 spheres = `4xx4/3pir^3=16/3pir^3`

Correct Answer – D
Total no. of atoms in fcc unit cell `=4`
Total volume `=4xx(4pir^(3))/3=16/3 pir^(3)`

Correct Answer – 4
`v=(4)/(3)pir^(3)` In fcc `v=4xx(4)/(3)pir^(3)=(16)/(3)pi r^(3)`

Correct Answer – D
For fcc, `Z_(eff) = 4//”unit cell”`
Volume of atoms in the unit cell `= 4/3 pir^(3) xx Z_(eff)`
`= 4/3 pi r^(3) xx 4`
`= (16)/(3) pir^(3)`