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In face centred cubic (fcc) crystal lattice edge length of the unit cell is 400 pm. Find the diameter fo the greatest sphere which can be fitted into the interstitial void without the distortion of lattice. |
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Answer» Solution :In fcc unit CELL : `4 R = SQRT(2)a` `therefore r=(a sqrt(2))/(4)=(400xxsqrt(2))/(4)=141.4p m` For octahedral void,`2(r + r^(1))=a` `therefore 2R^(1)=a - 2r` `= 400-2xx145` = 117.16 PM Hence diameter of greatest share = 117.6 pm. |
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