In an FCC unit cell,what fraction of the edge | vedclass

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(B) In an $FCC$ unit cell,the atoms touch along the face diagonal.The relationship between edge length $a$ and atomic radius $r$ is $4r = a\sqrt{2}$,w

Vedclass · Accepted Answer

$0.293$

Vedclass · Answer

(B) In an $FCC$ unit cell,the atoms touch along the face diagonal.The relationship between edge length $a$ and atomic radius $r$ is $4r = a\sqrt{2}$,which implies $r = \frac{a}{2\sqrt{2}}$.Along the edge of the unit cell,two atoms are present at the corners,each contributing a radius $r$ to the edge length.The total length covered by atoms along the edge is $2r$.The uncovered length along the edge is $a - 2r$.The fraction of the edge not covered is $\frac{a - 2r}{a} = 1 - \frac{2r}{a}$.Substituting $r = \frac{a}{2\sqrt{2}}$ into the expression:Fraction $= 1 - \frac{2(a / 2\sqrt{2})}{a} = 1 - \frac{1}{\sqrt{2}}$.Since $\frac{1}{\sqrt{2}} \approx 0.707$,the fraction $= 1 - 0.707 = 0.293$.

In an $FCC$ unit cell,what fraction of the edge length is not covered by atoms?

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