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Given below are densities of some solids and liquids .Give rough estimates of the sizes of the atoms.
Substance Atomic Mass(u) Density (${10^3}\;kg\;{m^{ - 3}}$)
Carbon(diamond)12.012.22
Gold197.0019.32
Nitrogen(liquid) 14.011.00
Lithium6.940.53
Fluorine(liquid) 19.001.14


[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 the atomic sizes are in the range of few ${A^ \circ }$]

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Last updated date: 17th Jun 2024
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Answer
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Hint: For a one mole of a substance of atomic radius r and density $\rho $ having mass M is the mass is equal to volume of the atom times density. All the other parameters are given and the radius of the atom means the size would be found out.

Complete step by step answer:
We know that for a one mole of a substance with atomic radius r and density $\rho $ having mass M . Let us suppose the atoms to be spherical in nature.
Avogadro’s number $ = 6.022 \times {10^{23}}$
We know that density is equal to mass divided by volume. Then $\rho = \dfrac{m}{V}$
$ \Rightarrow M = \rho \times V$
Where M is the mass, $\rho $ is the density and $V$ is the volume of the spherical atom
Then, $M = {N_A}\dfrac{4}{3}\pi {r^3}\rho $
$ \Rightarrow r = {(\dfrac{{3M}}{{4{N_A}\pi \rho }})^{\dfrac{1}{3}}}$
Now for carbon atom mass M = $12.01 \times {10^{ - 3}}\;kg$
Density, $\rho = 2.22 \times {10^3}\;kg\;{m^{ - 3}}$
$\therefore r = {(\dfrac{{3 \times 12.01 \times {{10}^{ - 3}}}}{{4\pi \times 2.22 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$
On doing the above calculations , we get
$r = 1.29\;{A^ \circ }$
So the radius of the carbon atom is $1.29\;{A^ \circ }$
Now for gold M $ = 197.00 \times {10^{ - 3}}\;kg$
$\rho = 19.32 \times {10^3}\;kg\;{m^{ - 3}}$
$\therefore r = {(\dfrac{{3 \times 197 \times {{10}^{ - 3}}}}{{4\pi \times 19.32 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$
$ \Rightarrow r = 1.59\;{A^ \circ }$
Hence the radius of the gold atom is $1.59\;{A^ \circ }$
Now for liquid nitrogen M$ = 14.01 \times {10^{ - 3}}\;kg$
density, $\rho = 1.00 \times {10^3}\;kg\;{m^{ - 3}}$
$ \Rightarrow r = 1.77\;{A^ \circ }$
Hence the radius of the liquid nitrogen atom is $1.77\;{A^ \circ }$
$\therefore r = {(\dfrac{{3 \times 14.01 \times {{10}^{ - 3}}}}{{4\pi \times 1.00 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$
Now for lithium atom M$ = 6.94 \times {10^{ - 3}}\;kg$
Density, $\rho = 0.53 \times {10^3}\;kg\;{m^{ - 3}}$
$\therefore r = {(\dfrac{{3 \times 6.94 \times {{10}^{ - 3}}}}{{4\pi \times 0.53 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$
$ \Rightarrow r = 1.73\;{A^ \circ }$
Hence the radius of lithium atom is $1.73\;{A^ \circ }$
Now for liquid fluorine atom M $ = 19.00 \times {10^{ - 3}}\;kg$
Density, $\rho = 1.14 \times {10^3}\;kg\;{m^{ - 3}}$
$\therefore r = {(\dfrac{{3 \times 19 \times {{10}^{ - 3}}}}{{4\pi \times 1.14 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$
$ \Rightarrow r = 1.88\;{A^ \circ }$
Hence the radius of the liquid fluorine atom is $1.88\;{A^ \circ }$.

Note: For the above question the proper conversions should be made correctly . Also for a unit cell with z number of atoms, M mass the density is calculated according to the formula
$\rho = \dfrac{{ZM}}{{{N_A}{a^3}}}$
Where Z is the number of atoms, M is the molecular mass , ${N_A}$ is the Avogadro number and a is the edge length of the unit cell.