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

Simran Bhatia , 12 Years ago
Grade 11
anser 2 Answers
Sakshi
For an fcc, if one says that the \sqrt{2 } a = 4r

thus, r = a/2 \sqrt{2 }
Thus, the maximum diameter of the atom to be occupied in the interstitial site is a/ \sqrt{2 }
Last Activity: 11 Years ago
Yeshwanth Talwar
Either octahedral voids –r 0.414r ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ or tetrahedral voids –r 0.225r ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ are occupied by interstitial sites inFCC where r+ is the radius of atom in interstitial sites and r– is the radius of atom forming the cubic lattice.Diameter of interstitial site = 117.1 = 2 0.414 24 a a = 400 pm
Last Activity: 9 Years ago
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