# Classical atomic models

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# Calculate the energy emitted when electrons in one gm. atom of hydrogen undergo transition giving the spectral lines of lowest energy in the visible region of its atomic spectra.

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Solution

## For visible line spectrum i.e., Balmer series and for minimum energy transition :We know that  and                                     [ line has minimum energy]                           $\begin{array}{l}\frac{1}{\lambda }={R}_{\mathrm{H}}\left[\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right]×{Z}^{2}\\ \frac{1}{\lambda }=1.097×{10}^{7}{\mathrm{m}}^{-1}\left[\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right]×{1}^{2}\end{array}$or                       $\frac{1}{\lambda }=1.097×{10}^{7}{\mathrm{m}}^{-1}\left[\frac{1}{4}-\frac{1}{9}\right]$Now                    $E=\frac{hc}{\lambda }=\frac{6.62×{10}^{-34}\mathrm{Js}×3×{10}^{8}{\mathrm{ms}}^{-1}}{6.56×{10}^{-7}\mathrm{m}}$                              $=3.0274×{10}^{-19}\mathrm{J}$  Energy corresponding to $1\mathrm{gm}$ atom of Hydrogen

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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