For any given series of spectral lines of atomic hydrogen, let $Δ \bar{v} = {\bar{v}}_{max} - {\bar{v}}_{min}$ be the difference in maximum and minimum frequencies in cm^-1. The ratio $Δ {\bar{v}}_{Lyman} / Δ {\bar{v}}_{Balmer}$ is

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27 : 5

5 : 4

9 : 4

4 : 1

answer is D.

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Detailed Solution

For Lyman
$\begin{array}{l} {\bar{v}}_{max} = R_{H} (\frac{1}{1^{2}} - \frac{1}{\infty^{2}}) = R_{H} \\ {\bar{v}}_{min} = R_{H} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{3}{4} R_{H} \\ Δ {\bar{v}}_{Lyman} = \frac{R_{H}}{4} \end{array}$
For Balmer
$\begin{array}{l} {\bar{v}}_{max} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4} \\ {\bar{v}}_{min} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5}{36} R_{H} \\ Δ {\bar{v}}_{Balmer} = \frac{R_{H}}{4} - \frac{5 R_{H}}{36} = \frac{4 R_{H}}{36} = \frac{R_{H}}{9} \\ \frac{Δ {\bar{v}}_{Lyman}}{Δ {\bar{v}}_{Balmer}} = \frac{R_{H} / 4}{R_{H} / 9} = \frac{9}{4} . \end{array}$