In a face-centred cubic (FCC) unit cell,what | vedclass

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(A) In an $FCC$ unit cell,the atoms touch along the face diagonal.The relationship between edge length $a$ and atomic radius $r$ is $a = 2\sqrt{2}r$.T

Vedclass · Accepted Answer

$0.293$

Vedclass · Answer

(A) In an $FCC$ unit cell,the atoms touch along the face diagonal.The relationship between edge length $a$ and atomic radius $r$ is $a = 2\sqrt{2}r$.The length covered by two atoms along the edge is $2r$.The fraction of the edge length not covered by atoms is given by $\frac{a - 2r}{a}$.Substituting $a = 2\sqrt{2}r$:$	ext{Fraction} = \frac{2\sqrt{2}r - 2r}{2\sqrt{2}r} = \frac{\sqrt{2} - 1}{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$.Using $\sqrt{2} \approx 1.414$,we get $1 - \frac{1}{1.414} = 1 - 0.707 = 0.293$.

In a face-centred cubic $(FCC)$ unit cell,what fraction of the edge length is not covered by the constituent atoms?

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