Kinetic Theory NCERT Solutions Class 11 Physics - Solved Exercise Question 13.14

Question 13.14:
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)
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].
Solution:

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 = (4/3)πr3
Volume of N number of molecules = (4/3)πr3N   ....(i)
Volume of one mole of a substance = M / ρ     ....(ii)  
(4/3)πr3N  =  M / ρ
∴ r = (3M / 4πρN)1/3

For carbon:
M = 12.01 × 10–3 kg,
ρ = 2.22 × 103 kg m–3
Substituting the values and solving, we get r = 1.59 A0
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
Substituting the values and solving, we get r = 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
Substituting the values and solving, we get r = 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
Substituting the values and solving, we get r = 1.88 Å
Hence, the radius of a liquid fluorine atom is 1.88 Å.

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