In an alloy of metals M and N,N atoms form a | vedclass

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(D) $1$. In a $bcc$ structure,$N$ atoms are at the corners ($8$ corners $	imes 1/8 = 1$ atom) and the center ($1$ atom). Total $N = 2$. $2$. $M$ atom

Vedclass · Accepted Answer

$M_4N$

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(D) $1$. In a $bcc$ structure,$N$ atoms are at the corners ($8$ corners $	imes 1/8 = 1$ atom) and the center ($1$ atom). Total $N = 2$. $2$. $M$ atoms are at the centers of the $6$ faces ($6 	imes 1/2 = 3$ atoms). $3$. $A$ body diagonal passes through $2$ corners and the body center. $4$. Removing atoms along one body diagonal: - $N$ atoms: $2$ corners are removed ($2 	imes 1/8 = 1/4$ atom) and $1$ body center is removed ($1$ atom). Remaining $N = 2 - (1/4 + 1) = 3/4$. - $M$ atoms: Face centers are not on the body diagonal,so $M$ remains $3$. $5$. Ratio $M:N = 3 : 3/4 = 12 : 3 = 4 : 1$. $6$. The formula is $M_4N$.

In an alloy of metals $M$ and $N$,$N$ atoms form a $bcc$ structure,while $M$ atoms occupy the center of each face. If the cube is cut such that all atoms along one body diagonal are removed,the formula of the resulting solid will be ...........

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