Electron in Hydrogen atom first jumps from third excited state to second excited state and then from second excited state to first excited state. The ratio of the wavelengths $\lambda_{1}: \lambda_{2}$ emitted in the two cases respectively is

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Aicharya · Accepted Answer

$\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}ight)=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}ight)=\mathrm{R}\left(\frac{5}{36}ight)$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(1-\frac{1}{2^{2}}ight)=\mathrm{R}\left(\frac{3}{4}ight)$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{3}{4} 	imes \frac{36}{5}=\frac{27}{5}$

Electron in Hydrogen atom first jumps from third excited state to second excited state and then from second…

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