In a compound made of atoms A and B,atoms A f | vedclass

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(C) $1$. In an $fcc$ unit cell,the number of atoms $A$ at corners is $8 	imes (1/8) = 1$ and at face centers is $6 	imes (1/2) = 3$. Total $A = 4$.$

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$A_3B_4$

Vedclass · Answer

(C) $1$. In an $fcc$ unit cell,the number of atoms $A$ at corners is $8 	imes (1/8) = 1$ and at face centers is $6 	imes (1/2) = 3$. Total $A = 4$.$2$. Number of tetrahedral voids = $2 	imes 4 = 8$. Number of octahedral voids = $4$.$3$. Atoms $B$ occupy $25\%$ of tetrahedral voids $(8 	imes 0.25 = 2)$ and $50\%$ of octahedral voids $(4 	imes 0.50 = 2)$. Total $B = 4$.$4$. If atoms along one axis are removed,two face-centered atoms are removed. New number of $A = 4 - 2 = 2$. The corner atoms remain $1$. Total $A = 3$.$5$. The number of voids remains unchanged as the lattice framework is intact. Total $B = 4$.$6$. The resulting formula is $A_3B_4$.

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