Question

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.

Answer 1

In fcc, there are octahedral voids and tetrahedral voids.
For tetrahedral voids, `r//R = 0.225`
For octahedral voids, `r//R = 0.414`
For maximum diameter of atom in interstitial site, octahedral voids must be taken because they can accomodate a larger atom.
`(r)/(R) = 0.414`
For fcc,
`R = (a)/(2sqrt2) = (400)/(2sqrt2)`
(r)/(R) = 0.414`
`r = 0.414 xx (400)/(2sqrt2)`
Diameter `= 2r = 2 xx 0.414 xx (400)/(2sqrt2) = 117.1` pm