Question -
| 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 ofAvogadro’s number. You should, however, not take the actual numbers you obtainfor various atomic sizes too literally. Because of the crudeness of the tightpacking approximation, the results only indicate that atomic sizes are in therange of a few Å].
Answer -
If r is the radius of the atom then the volume of each atom =(4/3)πr3
Volume of all the substance = (4/3)πr3 xN = M/ρ
M is the atomic mass of the substance
ρ is the density of the substance
One mole of the substance has 6.023 x 1023 atoms
r = (3M/4πρ x 6.023 x 1023)1/3
For carbon, M = 12. 01 x 10-3 kgand ρ = 2.22 x 103 kg m-3
R = (3 x 12. 01 x 10-3/4x 3.14 x 2.22 x 103 x 6.023 x 1023)1/3
= (36.03 x 10-3 /167.94x 1026)1/3
1.29 x 10 -10 m= 1.29 Å
For gold, M = 197 x 10-3 kgand ρ = 19. 32 x 103 kg m-3
R = (3 x 197 x 10-3/4x 3.14 x 19.32 x 103 x 6.023 x 1023)1/3
= 1.59 x 10 -10 m= 1.59 Å
For lithium, M = 6.94 x 10-3 kg and ρ = 0.53 x 103 kg/m3
R = (3 x 6.94 x 10-3/4x 3.14 x 0.53 x 103 x 6.023 x 1023)1/3
= 1.73 x 10 -10 m= 1.73 Å
For nitrogen (liquid), M = 14.01 x 10-3 kg and ρ = 1.00 x 103 kg/m3
R = (3 x 14.01 x 10-3/4x 3.14 x 1.00 x 103 x 6.023 x 1023)1/3
= 1.77 x 10 -10 m= 1.77 Å
For fluorine (liquid), M = 19.00 x 10-3 kg and ρ = 1.14 x 103 kg/m3
R = (3 x 19 x 10-3/4x 3.14 x 1.14 x 103 x 6.023 x 1023)1/3
= 1.88 x 10 -10 m= 1.88 Å