Atomic radius of this metal is 0.85 x 10^23
Given:
density of metal = 7. 24 g/cm3
metal has the bcc crystal structure
atomic weight = 48. 9 g/mol
To Find:
atomic radius of this metal
Solution: Atomic radius is defined as half the distance between the centers of two neighboring atoms. the atomic radius of a simple cube and hcp is a/2 respectively, whereas it is a√2/4 and a√3/4 for fcc and bcc respectively.
density = Z x M/N• x a^3
a = edge length
Z = 2 atoms = numeber of atoms per unit cell
N• = 6.023 x 10^23 atoms/mol = Avogadro's number
a^3 = density x N•/Z x M
a^3 = 7.24 x 6.023 x 10^23/ 2 x 48. 9
a = 3√0.44 x 10^23
a = 1.98 x 10^23
4R = √3a
R = √3 x 1.98 x 10^23/4
R = 0.85 x 10^23
Atomic radius of this metal is 0.85 x 10^23
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