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Minimum energy released in balmer series (H) will be corresponding to following wavelength

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\\\text{Wavelength of photon emitted in Balmer series of Hydrogen atom } \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right) \\ \text{where, } \mathrm{n}=3,4,5, \ldots \ldots \\ \text{For minimum wavelength }n=\infty

\large \begin{array}{l} \text { So, } \frac{1}{\lambda_{\min }}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{R}{4} \\\\ \Rightarrow \lambda_{\min }=\frac{4}{R}=\frac{4}{1.097 \times 10^{7}}=364.6 \mathrm{nm} \end{array}

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