Platinum crystallizes in a face-centered cubi | vedclass

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Question

(C) In a face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal. The length of the face diagonal is $d = a\sqrt{2}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In a face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal. The length of the face diagonal is $d = a\sqrt{2}$. Since the face diagonal consists of two radii of the corner atoms and the diameter of the face-centered atom,we have $4r = a\sqrt{2}$,where $r$ is the radius of the atom. The distance between the nearest neighbour atoms is equal to $2r$. From $4r = a\sqrt{2}$,we get $2r = \frac{a\sqrt{2}}{2} = a \frac{\sqrt{2}}{2}$.

Platinum crystallizes in a face-centered cubic $(FCC)$ crystal with a unit cell edge length $a$. The distance between nearest neighbour atoms is

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