Question

Metallic gold crystallizes in a face-centered cubic lattice, with one au atom per lattice point. if the edge length of the unit cell is found to be 408 pm, what is the metallic radius of au in pm?

Answer 1

The metallic radius (r) of au is 144.2pm

A face-centered cubic unit cell structure is made up of atoms arranged in a cube with six additional whole atoms placed in the centre of each cube face and a fraction of an atom at each of the cube's four corners. Eight additional unit cells share the atoms at the cube's corner.

Atoms in an FCC crystal are in contact along the unit cell's diagonal.          4r = √2 a, where r is the radius of an atom and a is the edge length of a unit cell.

Putting the values in the above expression,we get

r = √2×408 / 4

r = $\frac{1.414*408}{4}$

r = $\frac{576.9}{4}$

r = 144.2 pm

Therefore, The metallic radius (r) of au is 144.2pm

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