An element crystallizes in fcc lattice having edge length `400 pm` Calculate the maximum diameter of an atom which can be place in interstitial site without distorting the structure.

Explanation

In a cubic crystal system, there are two types of voids known as octahedral and tetrahedral voids. If `r_(1)` is the radius of void and r2 is radius of atom creating these voids then
`((r_(1))/(r_(2)))_("octa") = 0.414 and ((r_(1))/(r_(2)))_("tetra") = 0.255`
The above radius ratio values indicate that octahedral void has larger radius, hence for maximum diameter of atom to be present in interstitial space :
`r_(1) = 0.414 r_(2)`
also in fcc, `4r_(2) = sqrt(2)` a
`rArr` Diameter required `(2r_(1)) = (2r_(2)) xx 0.414`
` = (a)/(sqrt(2)) xx 0.414 = (400 xx 0.414)/(sqrt(2))` = 117 pm