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Question

${\left( {\frac{{{\lambda _{Lyman}}}}{{{\lambda _{Balmer}}}}} \right)_{\max }} = \frac{{\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)}}{{\lef

Vedclass · Accepted Answer

$\frac {5}{27}$

Vedclass · Answer

${\left( {\frac{{{\lambda _{Lyman}}}}{{{\lambda _{Balmer}}}}} \right)_{\max }} = \frac{{\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)}}{{\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)}} = \frac{{5/36}}{{3/4}} = \frac{5}{{27}}$

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

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