# Classical atomic models

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# For deuterium  the value of Rydberg constant is . (This value reflects a refinement of simple Bohr theory, wherein the orbital radii and Rydberg constant do not depend on electron mass but on the so called reduced mass. The latter mass, in turn, varies slightly with the mass of the nucleus). Find the shortest wavelength in the absorption spectrum of deuterium

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Solution

## (i) $\mathrm{R}$ for  The shortest wavelength transition corresponds to highest frequency and hence to the highest energy  So, the transitionwill be from ground state (lowest energy state) for which  to the highest state for which Hence,      $\overline{\mathrm{u}}=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right){\mathrm{cm}}^{-}$                  $=109707×{1}^{2}\left(\frac{1}{{1}^{2}}-\frac{1}{\left(\mathrm{\infty }{\right)}^{2}}\right){\mathrm{cm}}^{-}$or          $\frac{1}{\lambda }=109707\left(1-0\right){\mathrm{cm}}^{-}=109707{\mathrm{cm}}^{-}$or                     Ans.

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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