An element crystallises in `f.c.c.` lattice having edge length `400 p m`. Calculate the maximum diameter, which can be placed in interstitial sites without disturbing the structure.

Explanation

Either octahedral voids `((r_(1))/(r_(2))=0.414)` or tetrahedral voids `((r_(1))/(r_(2))=0.225)`
are occupied by interstitial site in `f.c.c.,` where `r_(1)` is radius of atom in interstitial site and `r_(2)` is radius of atoms arranged in `f.c.c.`.
Since in `f.c.c.` atoms along face diagonal are touching, thus `4r_(2)=sqrt(2a)`
Required diameter of interstitial sites
`2r_(1)=2xx0.414r_(2)`
`=(2xx0.414xxsqrt(2a))/(4)`
`=(2xx0.414xxsqrt(2)xx400)/(4)=117.1p m`