Ar solid –189℃ fcc density 1.66 g/cm3 radius

Ar is a solid below –189℃

with a face-centered cubic array of atoms

and a supposed density of 1.66 g/cm3.

Assuming that the atoms are spheres

in contact along the face diagonal,

what is the radius of a Ar atom (in Angstroms)?

An angstrom = 10^(-10) m or 10^(-8) cm.

Hint, first find the length of an edge of the unit cell.

(Argon’s actual density is 1.65 g/cm3.)

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face-centered cubic = fcc

d = [M × Z] / [N_A × a^3]

= [M × Z] / [N_A × V]

( 참고 https://ywpop.tistory.com/12439 )

> M = 39.95 g/mol

> Z = 4 atoms (fcc 이므로,)

> N_A = 6.022×10^23 atoms/mol

V = a^3 = [M × Z] / [N_A × d]

= [39.95 × 4] / [(6.022×10^23) × 1.66]

= 1.5986×10^(-22) cm3

a = (1.5986×10^(-22) cm3)^(1/3)

= 5.4273×10^(-8) cm

(5.4273×10^(-8) cm) (1 Å / (10^(-8) cm))

= 5.4273 Å

fcc의 edge length(a)

a = 4r / √2

r = a × √2 / 4

= 5.4273 Å × (2)^(1/2) / 4

= 1.9188 Å

---> radius of a Ar atom

답: 1.92 Å

[키워드] Ar solid –189℃ fcc density 2.32 g/cm3 radius 기준, Ar 원자의 반지름 기준, 아르곤 원자의 반지름 기준, 고체화학 기준