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An element crystallizes in fcc lattice having edge length 400 pm. Calculate the maximum diameter which can be placed in interstitial sites without disturbing the structure .

Answer»

Solution :Interstitital sites can be either tetrahedral or octahedral voids. As octahedral voids are bigger than tetrahedral voids, MAXIMUM diameter can fit into octahedral voids. If R is the size of atoms in the fcc packing and r is the size of the octahedral void, r=0.414 R
In fcc lattice, face diagonal =`sqrt2a`
As in fcc, atoms along the face diagonal are touching each other .
`4R=sqrt2a "or" R=(sqrt2a)/4`
REQUIRED diameter of the interstitial site =2 r =2 x 0.414 R =2 x 0.414 x `sqrt2/4` x 400 PM
=117.1 pm (`because` a=400 pm )


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