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Home » Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms: [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 .
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
\begin{array}{|l|c|c|} \hline \text { Substance } & \text { Atomic Mass (u) } & \begin{array}{l} \text { Density }\left(10^{3}\right. \\ \left.\mathrm{Kg} \mathrm{m}^{-3}\right) \end{array} \\ \hline \text { Carbon (diamond) } & 12.01 & 2.22 \\ \hline \text { Gold } & 197.00 & 19.32 \\ \hline \text { Nitrogen (liquid) } & 14.01 & 1.00 \\ \hline \text { Lithium } & 6.94 & 0.53 \\ \hline \text { Fluorine (liquid) } & 19.00 & 1.14 \\ \hline \end{array}
[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 \AA].
CBSE | Class 11 | Kinetic Theory | NCERT | Physics
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
\begin{array}{|l|c|c|} \hline \text { Substance } & \text { Atomic Mass (u) } & \begin{array}{l} \text { Density }\left(10^{3}\right. \\ \left.\mathrm{Kg} \mathrm{m}^{-3}\right) \end{array} \\ \hline \text { Carbon (diamond) } & 12.01 & 2.22 \\ \hline \text { Gold } & 197.00 & 19.32 \\ \hline \text { Nitrogen (liquid) } & 14.01 & 1.00 \\ \hline \text { Lithium } & 6.94 & 0.53 \\ \hline \text { Fluorine (liquid) } & 19.00 & 1.14 \\ \hline \end{array}
[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 \AA].

If r is the radius of the atom then the volume of each atom will be (4 / 3) \pi r^{3}

Volume of all the substance will be =(4 / 3) \pi r^{3} \times N=M / \rho

M= atomic mass of the substance

\rho= density of the substance

We know, One mole of the substance has 6.023 \times 10^{23} atoms

r=\left(3 M / 4 \pi \rho \times 6.023 \times 10^{23}\right)^{1 / 3}

For carbon, M=12.01 \times 10^{-3} \mathrm{~kg} and \rho=2.22 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}

R=\left(3 \times 12.01 \times 10^{-3} / 4 \times 3.14 \times 2.22 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}

=\left(36.03 \times 10^{-3} / 167.94 \times 10^{26}\right)^{1 / 3}

1.29 \times 10^{-10} \mathrm{~m}=1.29 \AA

For gold, M=197 \times 10^{-3} \mathrm{~kg} and \rho=19.32 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}

R=\left(3 \times 197 \times 10^{-3} / 4 \times 3.14 \times 19.32 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}

=1.59 \times 10^{-10} \mathrm{~m}=1.59 \AA

For lithium, M=6.94 \times 10^{-3} \mathrm{~kg} and \rho=0.53 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}

R=\left(3 \times 6.94 \times 10^{-3} / 4 \times 3.14 \times 0.53 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}

=1.73 \times 10^{-10} \mathrm{~m}=1.73 \AA

For nitrogen (liquid), M=14.01 \times 10^{-3} \mathrm{~kg} and \rho=1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}

R=\left(3 \times 14.01 \times 10^{-3} / 4 \times 3.14 \times 1.00 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}

=1.77 \times 10^{-10} \mathrm{~m}=1.77 \AA

For fluorine (liquid), M=19.00 \times 10^{-3} \mathrm{~kg} and \rho=1.14 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}

R=\left(3 \times 19 \times 10^{-3} / 4 \times 3.14 \times 1.14 \times 10^{3} \times 6.023 \times 10^{23}\right)^{1 / 3}

=1.88 \times 10^{-10} \mathrm{~m}=1.88 \AA

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