Atoms and Nuclei

From formula, $\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$ For Balmer series, $n_{1}=2$ $\therefore$ For longest wavelength, transition occurs from $n=3$ to $n=2$. $\therefore \frac{1}{\lambda_{\max}} = R\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]$ for shortest wavelength transition occurs from $n=\infty$ to $n=2$ $\therefore \frac{1}{\lambda_{\min}} = R\left[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right]$ $\therefore \frac{\lambda_{\text{longest}}}{\lambda_{\text{shortest}}} = \frac{9}{5}$ For Paschen series, $n_{1}=3$ $\lambda_{\text{longest}}$ of Balmer $= \frac{36}{5R}$ and $\lambda_{\text{shortest}}$ of Paschen $= \frac{9}{R}$ Hence there is no overlap between the wavelength ranges of Balamer and Paschen series. For Lyman series, $n_{1}=1$ $\frac{1}{\lambda}=R\left[\frac{1}{1}-\frac{1}{m^{2}}\right]$ Also $\frac{1}{\lambda_{0}}=R$ $\therefore \frac{1}{\lambda} = \frac{1}{\lambda_{0}}\left[1-\frac{1}{m^{2}}\right] \Rightarrow \lambda = \frac{\lambda_{0}}{1-\frac{1}{m^{2}}}$ $\lambda_{\text{longest}}$ of Lyman $= \frac{4}{3R}$ and $\lambda_{\text{shortest}}$ of Balmer $= \frac{4}{R}$ Hence that wavelenght ranges of Lyman and Balmer series do not overlap.