question_answer

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

A)  \[{{A}_{2}}{{B}_{5}}\]                   

B)  \[{{A}_{3}}{{B}_{7}}\]

C)  \[{{A}_{3}}{{B}_{10}}\]                 

D)  \[{{A}_{7}}{{B}_{24}}\]

Correct Answer: C

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\[{{A}_{\frac{3}{4}}}{{B}_{\frac{5}{2}}}={{A}_{3}}{{B}_{10}}\]