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प्रश्न
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].
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उत्तर १
| Substance | Radius (Å) |
| Carbon (diamond) | 1.29 |
| Gold | 1.59 |
| Nitrogen (liquid) | 1.77 |
| Lithium | 1.73 |
| Fluorine (liquid) | 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 pi r^3`
Volume of N number of molecules = `4/3 pir^3N` ... (i)
Volume of one mole of a substance = `M/rho` ... (ii)
`4/3 pir^3 N = M/rho`
`:. r = root(3)((3m)/(4pirhoN))`
For carbon
M = 12.01 × 10–3 kg,
ρ = 2.22 × 103 kg m–3
`:. r = ((3xx12.01xx10^(-3))/(4pixx2.22xx10^3xx6.023xx 10^23))^(1/3)` = 1.29 Å
For gold:
M = 197.00 × 10–3 kg
ρ = 19.32 × 103 kg m–3
:. r = `((3xx197xx10^3)/(4pixx19.32xx10^3xx6.023xx10^23))^(1/3)` = 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
`:. r = ((3xx14.01xx10^(-3))/(4pixx1.00xx10^3xx 6.23 xx10^23))^(1/3)` = 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
`:. r= ((3xx6.94xx10^(-3))/(4pixx0.53xx10^3xx6.23xx10^23))^(1/3)` = 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
`:. r= ((3xx19xx10^(-3))/(4pixx1.14xx10^3xx6.023xx10^23))^(1/3)` = = 1.88 Å
Hence, the radius of a liquid fluorine atom is 1.88 Å.
उत्तर २
In one mole of a substance, there are 6.023 x 1023 atoms
`:. (4/3 pi R^3) xx 6.023 xx 10^23 = M/rho`
or ` R = [(3M)/(4pirho xx 6.023xx10^23)]^(1/3)`
For carbon = `12.01 xx 10^(-3)` kg and `rho = 2.22 xx 10^3 "kg m"^(-3)`
`:. R = [(3xx12.01xx10^(-3))/(4xx3.14xx2.22xx10^3xx6.023xx10^23)]^(1/3)`
= `1.29 xx 10^(10) m` = 1.29 Å
For gold, M = `197 xx10^(-3)` kg and `rho = 19.32 xx 10^3 "kg m"^(-3)`
`:. R = [(3xx197xx10^(-3))/(4xx3.14xx19.32xx10^3 xx 6.023 xx 10^(23))]^(1/3)`
= 1.59 xx 10^(-10) m = 1.59 Å
For nitrogen (liquid), `M= 14.01 xx 10^(-3)` kg and `rho = 1.00 xx 10^3 "kg m"^(-3)`
`R = [(3xx14.01xx10^(-3))/(4xx3.14xx100xx10^3xx6.023xx10^23)]^(1/3)`
= `1.77 xx 10^(-10)m` = 1.77 Å
For lithium, M= `6.94 xx 10^(-3)` kg, `rho = 0.53 xx 10^3 "kg m"^(-3)`
`:. R = [(3xx6.94xx10^(-3))/(4xx3.14xx0.53xx10^3xx6.023xx10^(23))]^(1/3)`
`= 1.73 xx 10^(-10)` = 1.73 Å
For fluorine (liquid), M = `19.00 xx 10^(-3)` kg, `rho = 1.14 xx 10^3 "kg m"^(-3)`
`:. R = [(3xx19.00xx10^(-3))/(4xx3.14xx1.14xx10^3xx6.023xx10^(23))]^(`1/3)`
`= 1.88 xx 10^(-10)m` = 1.88 Å
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