The relationship between the radius (r) of th | vedclass

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(B) In a cubic close packing $(CCP)$ or face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.The length of the fac

Vedclass · Accepted Answer

$r = \frac{a}{2 \sqrt{2}}$

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(B) In a cubic close packing $(CCP)$ or face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.The length of the face diagonal is given by $\sqrt{2} a$,where $a$ is the edge length of the unit cell.Since the face diagonal consists of the radius of the corner atom,the diameter of the face-centered atom,and the radius of the other corner atom,we have: $4r = \sqrt{2} a$.Therefore,the radius $r$ is given by $r = \frac{\sqrt{2} a}{4} = \frac{a}{2 \sqrt{2}}$.

The relationship between the radius $(r)$ of the atom and the edge length $(a)$ of the cubic close packing $(CCP)$ structure is:

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