• # question_answer For any given series of spectral lines of atomic hydrogen, let $\Delta \text{\bar{v}=}{{\text{\bar{v}}}_{\max }}-{{\text{\bar{v}}}_{\min }}$ be the difference in maximum and minimum frequencies in $c{{m}^{-1}}.$The ratio $\Delta {{\text{\bar{v}}}_{Lyman}}/\Delta {{\text{\bar{v}}}_{Balmer}}$is : [JEE Main 9-4-2019 Morning] A) 27 : 5              B) 4 : 1C) 5 : 4                D) 9 : 4

For Lyman ${{\text{\bar{v}}}_{\max }}={{R}_{H}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)={{R}_{H}}$ ${{\text{\bar{v}}}_{\min }}={{R}_{H}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3}{4}{{R}_{H}}$ $\Delta {{\text{\bar{v}}}_{Lyman}}=\frac{{{R}_{H}}}{4}$ For Balmer ${{\text{\bar{v}}}_{\max }}={{R}_{H}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=\frac{{{R}_{H}}}{4}$ ${{\text{\bar{v}}}_{\min }}={{R}_{H}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5}{36}{{R}_{H}}$ $\Delta {{\text{\bar{v}}}_{Balmer}}=\frac{{{R}_{H}}}{4}-\frac{5{{R}_{H}}}{36}=\frac{4{{R}_{H}}}{36}=\frac{{{R}_{H}}}{9}$ $\frac{\Delta {{{\text{\bar{v}}}}_{Lyman}}}{\Delta {{{\text{\bar{v}}}}_{Balmer}}}=\frac{{}^{{{R}_{H}}}/{}_{4}}{{}^{{{R}_{H}}}/{}_{9}}=\frac{9}{4}$ $\therefore$ Ans. is