In a compound formed by atoms A and B, atoms | vedclass

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(B) In a $CCP$ structure, atoms $A$ are at the corners and face centers. The center of the cube is the body center.Distance of a corner atom from the

Vedclass · Accepted Answer

$81.3$

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(B) In a $CCP$ structure, atoms $A$ are at the corners and face centers. The center of the cube is the body center.Distance of a corner atom from the body center is $\frac{\sqrt{3}a}{2}$.Distance of a face-centered atom from the body center is $\frac{a}{\sqrt{2}}$.Since the question asks for the distance of $A$ from the center, and $A$ occupies both corners and face centers, the distance of the corner atom is $\frac{\sqrt{3} 	imes 115}{2} \approx 99.5 \ pm$ and the distance of the face-centered atom is $\frac{115}{\sqrt{2}} \approx 81.3 \ pm$.Given the options, $81.3 \ pm$ is the correct distance for the face-centered atom $A$ from the center of the cube.

In a compound formed by atoms $A$ and $B$, atoms $A$ form a cubic close-packed $(CCP)$ structure, while atoms $B$ are located at the center of each edge. If the edge length of the cubic unit cell is $115 \ pm$, the distance of atom $A$ from the center of the cube is .......... $pm$.

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