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Q. A face centred cubic lattice is made up of two types of atoms A and B, in which A occupies the comer positions and B occupies the face centres. If atom along an axis joining the diagonally opposite comers on a face are removed, the empirical formula of the remaining solid would be

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Solution:

A face of this solid would appear as
Missing $ A=\frac{1}{8}\times 2=\frac{1}{4} $ A present $ =1-\frac{1}{4}=\frac{3}{4} $ Missing $ B=\frac{1}{2} $ B present $ =3-\frac{1}{2}=\frac{5}{2} $ formula $ {{A}_{\frac{3}{4}}}{{B}_{\frac{5}{2}}}={{A}_{3}}{{B}_{10}} $

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