Potassium metal crystallizes in a face-center | vedclass

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(A) For a face-centered cubic $(FCC)$ lattice,the atoms touch along the face diagonal.The relationship between the edge length $(a)$ and the atomic ra

Vedclass · Accepted Answer

$0.4059$

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(A) For a face-centered cubic $(FCC)$ lattice,the atoms touch along the face diagonal.The relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $4r = \sqrt{2}a$.The shortest distance between two nuclei in an $FCC$ lattice is the distance between the center of an atom at the corner and the center of an atom at the face center,which is equal to half of the face diagonal.Shortest distance $(d)$ = $\frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.Given $a = 0.574 \ nm$.$d = \frac{0.574}{1.414} \ nm = 0.4059 \ nm$.

Potassium metal crystallizes in a face-centered cubic $(FCC)$ lattice. If the edge length of the unit cell is $0.574 \ nm$,what is the shortest distance between any two potassium nuclei (in $nm$)?

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