# For any given series of spectral lines of atomic hydrogen,

Question:

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

1. $4: 1$

2.  $9: 4$

3. $5: 4$

4. $27: 5$

Correct Option: , 2

Solution:

$\bar{v} \propto \Delta \mathrm{E}$

For H-atom

$\bar{v}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$

For Lyman series,

$\bar{v}(\max )=13.6\left(1-\frac{1}{\infty}\right)$

$\bar{v}(\mathrm{~min})=13.6\left(1-\frac{1}{4}\right)$

$\therefore \bar{v}_{\max }-\bar{v}_{\min }=13.6\left(\frac{1}{4}\right)$

For Balmer series,

$\bar{v}(\max )=13.6\left(\frac{1}{4}-\frac{1}{\infty}\right)$

$\bar{v}(\min )=13.6\left(\frac{1}{4}-\frac{1}{9}\right)$

$\therefore \bar{v}_{\max }-\bar{v}_{\min }=13.6\left(\frac{1}{9}\right)$

$\frac{\Delta \bar{v}_{\text {Lyman }}}{\Delta \bar{v}_{\text {Balmer }}}=\frac{9}{4}$