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Question

The wavelength of different spectral lines of Lyman series is given by

$\frac{1}{{{\lambda _L}}} = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{n^2}}}}

Vedclass · Accepted Answer

$\;\frac{5}{{27}}$

Vedclass · Answer

The wavelength of different spectral lines of Lyman series is given by

$\frac{1}{{{\lambda _L}}} = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{n^2}}}} ight]$ where  $n = 2,3,4, \ldots $

where subscript $L$ refers to Lyman. 

For longest wavelength, $n=2$

$	herefore \quad \frac{1}{\lambda_{L_{	ext {longst }}}}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}ight]=\frac{3}{4} R$     .... $(i)$

The wavelength of different spectral series of Balmer series is given by

$\frac{1}{{{\lambda _B}}} = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{n^2}}}} ight]$ where $n = 3,4,5, \ldots $

where subscript $B$ refers to Balmer.

For longest wavelength, $n=3$

$	herefore \frac{1}{{{\lambda _{{B_{{	ext{looget }}}}}}}} = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} ight]$ $ = R\left[ {\frac{1}{4} - \frac{1}{9}} ight] = \frac{{5R}}{{36}}$  .... $(ii)$

Divide $(ii)$ by $(i)$, we get

$\frac{\lambda_{	ext {Longest }}}{\lambda_{B_{	ext {longest }}}}=\frac{5 R}{36} 	imes \frac{4}{3 R}=\frac{5}{27}$

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

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