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  3. Chemistry
  4. For any given series of spectral lines of atomic hydrogen, let Δ V = Vmax - Vmin be the difference in maximum and minimum frequencies in cm-1. The ratio Δ barV textLyman / Δ barV textBalmer is :

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

JEE MainJEE Main 2019Structure of Atom

Solution:

For Lyman
$\bar{V_{max}} \, R_H \bigg( \frac{1}{1^2} - \frac{1}{\infty^2}\bigg) \, = \, R_H$
$\bar{V_{max}} \, R_H \bigg( \frac{1}{1^2} - \frac{1}{2^2}\bigg) \, = \frac{3}{4}\, R_H$
$\Delta \, \bar{V_{Lyman}} \, = \, \frac{R_H}{4}$
For Balmer
$\bar{v_{max}} = R_H \bigg(\frac{1}{2^2} - \frac{1}{\infty^2}\bigg) \, = \, \frac{R_H}{4}$
$\bar{v}_{min} \, = \, R_H \bigg(\frac{1}{2^2} - \frac{1}{3^2}\bigg) \, = \, \frac{5}{36} R_H$
$\Delta \bar{V_{Balmer}} \, = \, \frac{R_H}{4} - \frac{5R_H}{36} = \frac{4R_H}{36} = \frac{RH}{9}$ |
$\frac{\Delta \bar{v_{Lyman}}}{\Delta \bar{v}_{Balmer}} \, = \, \frac{R_H/4}{R_H/9} \, = \, \frac{9}{4}$