# Classical atomic models

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# The n2 value of transition in the ${\mathrm{He}}^{+}$ spectrum which has the same wavelength as the first Lyman transition of hydrogen ( to ) ?

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Solution

## At. no. $\left(\mathrm{Z}\right)$ of  . Given  We know that:              $\begin{array}{r}{\overline{v}}_{\mathrm{H}}=\frac{1}{{\lambda }_{\mathrm{H}}}=109678\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right){\mathrm{Z}}^{2}\\ =109678\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)×{1}^{2}{\mathrm{cm}}^{-}\\ =109678×\frac{3}{4}{\mathrm{cm}}^{-}\end{array}$As   $\mathrm{\Delta }{E}_{{\mathrm{He}}^{+}}={\mathrm{Z}}^{2}{\mathrm{E}}_{{\mathrm{H}}^{\mathrm{\prime }}}$ similarly, ${\lambda }_{{\mathrm{He}}^{+}}={Z}_{\mathrm{He}}^{2}×{\lambda }_{\mathrm{H}}$or     ${\lambda }_{{\mathrm{He}}^{+}}=\left(2{\right)}^{2}×\frac{4}{109678×3}\mathrm{cm};$        $\frac{1}{{\lambda }_{{\mathrm{He}}^{+}}}=\frac{109678×3}{16}{\mathrm{cm}}^{-}$For ${\mathrm{He}}^{+},$ let  Thus we have :             ${\overline{\mathrm{\mu }}}_{{\mathrm{He}}^{+}}=\frac{1}{{\lambda }_{{\mathrm{He}}^{+}}}=109678\left(\frac{1}{{2}^{2}}-\frac{1}{{n}_{2}^{2}}\right){\mathrm{cm}}^{-}$or $\frac{109678×3}{16}{\mathrm{cm}}^{-}=109678\left(\frac{1}{4}-\frac{1}{{n}_{2}^{2}}\right){\mathrm{cm}}^{-};$Hence               ${\mathrm{n}}_{2}={\left(16\right)}^{1/2}=4.$

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