Platinum crystallizes in a face-centered cubi | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In a face-centered cubic $(FCC)$ unit cell,atoms touch each other along the face diagonal.The length of the face diagonal is $\sqrt{2}a$,where $a$ is the edge length of the unit cell.The distance between nearest neighbours $(d)$ is equal to the distance between the centers of two touching atoms,which is $2r$.Since the atoms touch along the face diagonal,$4r = \sqrt{2}a$.Therefore,the distance between nearest neighbours $d = 2r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.

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

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