• # question_answer If the series limit wavelength of the Lyman series for hydrogen atom is $912\overset{\text{o}}{\mathop{\text{A}}}\,\text{,}$ then the series limit wave- length for the Balmer series for the hydrogen atom is: A) $912\,\overset{\text{o}}{\mathop{\text{A}}}\,$                  B) $912\,\times 2\overset{\text{o}}{\mathop{\text{A}}}\,$C) $912\,\times 4\overset{\text{o}}{\mathop{\text{A}}}\,$                  D)  $\frac{912}{2}\overset{\text{o}}{\mathop{\text{A}}}\,$

For series limit of Balmer series. ${{n}_{i}}=\infty ,{{n}_{f}}=2$ $\frac{l}{{{\lambda }_{b}}}=R\left[ \frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} \right]$or $\frac{1}{{{\lambda }_{b}}}=R\left[ \frac{1}{4}-\frac{1}{{{\infty }^{2}}} \right]=\frac{R}{4}$ $\Rightarrow$${{\lambda }_{b}}=\frac{4}{R}.$ where${{\lambda }_{b}}$ is the series limit of the Balmer series. Limit of Lymen series. $\frac{1}{{{\lambda }_{\ell }}}=R\left[ 1-\frac{1}{\infty } \right]$ Where ${{\lambda }_{\ell }}$is the series limit of lymen series. $\therefore$${{\lambda }_{\ell }}=\frac{1}{R}$ So, $\frac{{{\lambda }_{b}}}{{{\lambda }_{\ell }}}=4\Rightarrow {{\lambda }_{0}}={{\lambda }_{\ell }}\times 4=912\times 4\overset{o}{\mathop{A}}\,$             Hence, the correction option is [c].