Answer:
The radius the Argon atom in [tex]2.072\ \AA [/tex]
Solution:
As per the question:
Density of the solid, [tex]\rho = 1.32\ g/cm^{3}[/tex]
No. of atoms present in FCC lattice, Z = 4
Now,
Mass of Argon, m = 40 g/mol
Density of the unit cell is given by the formula:
[tex]\rho = \frac{m\times Z}{N_{o}\times a^{3}}[/tex] (1)
where
a = edge length
[tex]N_{o}[/tex] = Avogadro's number
Now,
[tex]a^{3} = \frac{40\times 4}{6.022\times 10^{23}\times 1.32} = 2.012\times 10^{- 22}[/tex]
[tex]a = (2.012\times 10^{- 22})^{\frac{1}{3}} = 5.86\times 10^{- 8}\ cm[/tex]
Now,
In FCC unit cell, the relation between the radius, R and the edge length, a is given by:
[tex]R = \frac{a}{2\sqrt{2}}[/tex]
[tex]R = \frac{5.86\times 10^{- 8}}{2\sqrt{2}} = 2.07\times 10^{- 8}\ cm = 2.072\ \AA [/tex]
