Question

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

Substance Atomic Mass (u) Density (103 Kg m-3)

Carbon (diamond) 12.01 2.22

Gold 197.00 19.32

Nitrogen (liquid) 14.01 1.00

Lithium 6.94 0.53

Fluorine (liquid) 19.00 1.14

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

Answer 1

volume of spherical atom = 4/3 πr³
So, volume of one mole spherical atom = Avogadro's constant × 4/3 πr³
= 4/3 πr³ Na
=
Atomic mass of one mole = M
Density of 1 mole atom = D
Now,
Density = mass/volume
Volume = mass/density
4/3 πr³ Na = M/D
r = [ 3M/4πDNa]⅓

Now, use this formula for finding size of atom .

For carbon,
r =[ ( 3 × 12 × 10^-3/4 × 3.14 × 2.22 × 6.023 × 10²³ )]⅓
= 1.29 A°

For gold ,
M = 197 × 10^-3 Kg
D = 19.32 Kg/m³

Radius of gold = 1.59 A°
Similarly you can find out next all elements size by using formula .

If you do yourself , then developed hour skill.
For confidence I am giving answer
Radius of Nitrogen = 1.77A°
Radius of Lithium = 1.73 A°
Radius of fluorine = 1.88A°