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Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is
$\frac{3}{{23}}$
$\;\frac{7}{{29}}$
$\;\frac{9}{{31}}$
$\;\frac{5}{{27}}$
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
$\frac{\lambda_{\text {Longest }}}{\lambda_{B_{\text {longest }}}}=\frac{5 R}{36} \times \frac{4}{3 R}=\frac{5}{27}$
Similar Questions
Match List $- I$ (Experiment performed) with List $-II$ (Phenomena discovered/associated) and select the correct option from the options given the lists
List $- I$ | List $- II$ |
$(1)$ Davisson and Genner | $(i)$ Wave nature of electrons |
$(2)$ Millikan's oil drop experiment | $(ii)$ Charge of an electron |
$(3)$ Rutherford experiment | $(iii)$ Quantisation of energy levels |
$(4)$ Franck-Hertz experiment | $(iv)$ Existence of nucleus |