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| Substance | Atomic Mass (u) | Density (103 Kg m-3) |
| Carbon (diamond) Gold Nitrogen (liquid) Lithium Fluorine (liquid) | 12.01 197.00 14.01 6.94 19.00 | 2.22 19.32 1.00 0.53 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].
Asked by Abhigna Last Modified
Answer | Substance | Radius (Å) |
| Carbon (diamond) Gold Nitrogen (liquid) Lithium Fluorine (liquid) | 1.29 1.59 1.77 1.73 1.88 |
Atomic mass of a substance = M
Density of the substance = ρ
Avogadro’s number = N = 6.023 × 1023
Volume of each atom 
Volume of N number of molecules N … (i)
Volume of one mole of a substance = 
… (ii)
N =
For carbon:
M = 12.01 × 10–3 kg,
ρ = 2.22 × 103 kg m–3
= 1.29 Å
Hence, the radius of a carbon atom is 1.29 Å.
For gold:
M = 197.00 × 10–3 kg
ρ = 19.32 × 103 kg m–3
= 1.59 Å
Hence, the radius of a gold atom is 1.59 Å.
For liquid nitrogen:
M = 14.01 × 10–3 kg
ρ = 1.00 × 103 kg m–3
= 1.77 Å
Hence, the radius of a liquid nitrogen atom is 1.77 Å.
For lithium:
M = 6.94 × 10–3 kg
ρ = 0.53 × 103 kg m–3
= 1.73 Å
Hence, the radius of a lithium atom is 1.73 Å.
For liquid fluorine:
M = 19.00 × 10–3 kg
ρ = 1.14 × 103 kg m–3
= 1.88 Å
Hence, the radius of a liquid fluorine atom is 1.88 Å.
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