# Classical atomic models

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# How will you show that radius of first Bohr orbit $\left({\mathrm{a}}_{0}\right)$ is $0.529{\mathrm{A}}^{\mathrm{o}}$ with the help of permittivity of free space instead of coloumb' s law constant ?    $\left({\epsilon }_{0}=8.854×{10}^{-12}{\mathrm{C}}^{2}/{\mathrm{Nm}}^{2},h=6.626×{10}^{-34}\mathrm{Js}\right).$

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Solution

## We know, ${a}_{0}=\frac{{\epsilon }_{0}{h}^{2}}{\pi m{e}^{2}}$ where $\mathrm{m}$ and $\mathrm{e}$ are the mass and charge on the electron.            Hence : ${a}_{0}=\frac{8.854×{10}^{-12}{\mathrm{C}}^{2}}{\mathrm{N}.{\mathrm{m}}^{2}}$                         $\begin{array}{l}×\frac{{\left(6.626×{10}^{-34}\mathrm{Js}\right)}^{2}}{3.14×9.109×{10}^{-31}\mathrm{kg}×{\left(1.602×{10}^{-19}\mathrm{C}\right)}^{2}}\\ =\frac{5.29×{10}^{-11}{\mathrm{J}}^{2}{\mathrm{s}}^{2}}{{\mathrm{Nm}}^{2}×\mathrm{kg}}=\frac{5.29×{10}^{-11}{\mathrm{J}}^{2}{\mathrm{s}}^{2}}{\mathrm{Nm}×\mathrm{m}×\mathrm{kg}}\\ =\frac{5.29×{10}^{-11}{\mathrm{J}}^{2}{\mathrm{s}}^{2}}{\mathrm{J}×\mathrm{m}×\mathrm{kg}}=\frac{5.29×{10}^{-11}{\mathrm{Js}}^{2}}{\mathrm{m}×\mathrm{kg}}\\ =\frac{5.29×{10}^{-11}{\mathrm{kgm}}^{2}{\mathrm{s}}^{-2}×{\mathrm{s}}^{2}}{\mathrm{m}×\mathrm{kg}}\end{array}$                          $\begin{array}{r}=5.29×{10}^{-11}\mathrm{m}×\frac{{10}^{10}{\mathrm{A}}^{\mathrm{o}}}{1\mathrm{m}}=0.529{\mathrm{A}}^{\mathrm{o}}\\ \left[\because 1\mathrm{Nm}=\mathrm{J};\mathrm{J}={\mathrm{kgm}}^{2}{\mathrm{s}}^{-2}\right]\end{array}$

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The ground state energy of hydrogen atom is -13.6 eV. The Energy of second exited state of ${\mathrm{He}}^{+}$ ion in eV is