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(C) In a body-centered cubic $(BCC)$ unit cell,the atoms touch along the body diagonal.The length of the body diagonal is given by $\sqrt{3} \, a$,whe

Vedclass · Accepted Answer

$\frac{\sqrt{3} \, a}{2}$

Vedclass · Answer

(C) In a body-centered cubic $(BCC)$ unit cell,the atoms touch along the body diagonal.The length of the body diagonal is given by $\sqrt{3} \, a$,where $a$ is the edge length of the unit cell.Since the body diagonal consists of two radii of the corner atoms and one full diameter of the body-centered atom,the total length is $4r = \sqrt{3} \, a$.The distance between the nearest neighbors (the center atom and a corner atom) is the distance from the center of the unit cell to a corner,which is half of the body diagonal.Therefore,the distance $d = \frac{\sqrt{3} \, a}{2}$.

In a body-centered cubic $(BCC)$ structure like $CsCl$,the distance between the two nearest neighboring spheres is equal to:

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