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 (10^{3
}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

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 × 10^{23}
Volume of each atom = (4/3)πr^{3}
Volume of N number of molecules = (4/3)πr^{3}N ....(i)
Volume of one mole of a substance = M / ρ ....(ii)
(4/3)πr^{3}N = M / ρ
∴ r = (3M / 4πρN)^{1/3}
For carbon:
M = 12.01 × 10^{–3} kg,
ρ = 2.22 × 10^{3} kg m^{–3}
Substituting the values and solving, we get r = 1.59 A^{0}
Hence, the radius of a gold atom is 1.59 Å.
For liquid nitrogen:
M = 14.01 × 10^{–3} kg
ρ = 1.00 × 10^{3} 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 × 10^{3} 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 × 10^{3} 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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