12.Atoms
normal

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

A

$\frac{3}{{23}}$

B

$\;\frac{7}{{29}}$

C

$\;\frac{9}{{31}}$

D

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

Solution

The wavelength of different spectral lines of Lyman series is given by

$\frac{1}{{{\lambda _L}}} = R\left[ {\frac{1}{{{1^2}}} – \frac{1}{{{n^2}}}} \right]$ where  $n = 2,3,4, \ldots $

where subscript $L$ refers to Lyman. 

For longest wavelength, $n=2$

$\therefore \quad \frac{1}{\lambda_{L_{\text {longst }}}}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=\frac{3}{4} R$     …. $(i)$

The wavelength of different spectral series of Balmer series is given by

$\frac{1}{{{\lambda _B}}} = R\left[ {\frac{1}{{{2^2}}} – \frac{1}{{{n^2}}}} \right]$ where $n = 3,4,5, \ldots $

where subscript $B$ refers to Balmer.

For longest wavelength, $n=3$

$\therefore \frac{1}{{{\lambda _{{B_{{\text{looget }}}}}}}} = R\left[ {\frac{1}{{{2^2}}} – \frac{1}{{{3^2}}}} \right]$ $ = R\left[ {\frac{1}{4} – \frac{1}{9}} \right] = \frac{{5R}}{{36}}$  …. $(ii)$

Divide $(ii)$ by $(i)$, we get

$\frac{\lambda_{\text {Longest }}}{\lambda_{B_{\text {longest }}}}=\frac{5 R}{36} \times \frac{4}{3 R}=\frac{5}{27}$

Standard 12
Physics

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