Compute the percentage void space per unit volume of unit cell in zinc-sulphide structure. If the  answer is `X^(2)`?, then the value of X is 

Answer

5

Explanation

Anions are in FCC position and half of the T.V are occupied by cations three are 4 anions 2 & 8 T.V per unit cell. Here Face diagonal `=sqrt2a therefore 4r=sqrt2a, a=2sqrt2r`
Volume of the unit cell `=a^(3)rArr (2sqrt2r)^(3)=16sqrt2r^(3)`
`therefore " P.F"=(4xx((4)/(3)pi_(-)^(3))+(1)/(2)xx8xx((4)/(3)pir_(+)^(3)))/(16sqrt2r^(3))rArr (pi)/(3sqrt2){1+((r_(+))/(r_(-)))^(3)}" Since for T.V "(r_(+))/(r_(-))=0.225`
`therefore" P.F "=(pi)/(3sqrt2)[1+(0.255)^(3)]=0.7485" per unit cell"`
`therefore" Void space "=1-07485=0.2515~~25%=5^(2),x=5`