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If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is
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$\frac{36}{5 R}$
As we know $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\quad \frac{1}{\lambda}=\mathrm{R}\left(\frac{9-4}{36}\right)=\frac{5 \mathrm{R}}{36}$ $\therefore \quad \lambda=\frac{36}{5 \mathrm{R}}$
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