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  4. The ratio of the maximum to the minimum wavelengths in Balmer series of H spectrum is

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Q. The ratio of the maximum to the minimum wavelengths in Balmer series of H spectrum is

COMEDKCOMEDK 2008Atoms

Solution:

Wavelength of Balmer series is given by
$\frac{ 1 }{\lambda} = R \left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right) n =3, 4,5, ...... \infty$
For minimum wavelength , $ n = \infty \frac{1}{\lambda_{min}} = R \left(\frac{1}{4} - \frac{1}{\infty}\right) = \frac{R}{4}$
or $\lambda_{min} = \frac{4}{R} \, \, \, \, \, \, \, $.......(i)
For maximum wavelength, n = 3
$\frac{1}{\lambda_{max}} = R \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36} $
$\lambda_{max}= \frac{36}{5R} \, \, \, \, \, \, \, \, $ .......(ii)
$ \therefore \, \, \frac{\lambda_{max}}{\lambda_{min} } = \frac{36}{5R} \times\frac{R}{4}= \frac{9}{5} $