A face-centred cubic (FCC) unit cell is made | vedclass

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(D) In an $FCC$ unit cell,there are $8$ corners and $6$ face centres.Initially,the number of atoms of $A$ at corners $= 8 	imes \frac{1}{8} = 1$.The

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$AB_4$

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(D) In an $FCC$ unit cell,there are $8$ corners and $6$ face centres.Initially,the number of atoms of $A$ at corners $= 8 	imes \frac{1}{8} = 1$.The number of atoms of $B$ at face centres $= 6 	imes \frac{1}{2} = 3$.One body diagonal passes through $2$ corners and the body centre. However,in an $FCC$ lattice,there are no atoms at the body centre.When atoms along one body diagonal are removed,$2$ corner atoms are removed.Remaining atoms of $A = 1 - (2 	imes \frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4}$.Since no face-centred atoms lie on the body diagonal,the number of $B$ atoms remains $3$.The ratio of $A:B = \frac{3}{4} : 3 = 3 : 12 = 1 : 4$.Therefore,the empirical formula is $AB_4$.

$A$ face-centred cubic $(FCC)$ unit cell is made up of two types of atoms $A$ and $B$,in which $A$ occupies the corner positions and $B$ occupies the face centres. If atoms along one of the body diagonals are removed,what is the empirical formula of the remaining solid?

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