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  4. The ratio of minimum wavelengths of Lyman and Balmer series will be

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Q. The ratio of minimum wavelengths of Lyman and Balmer series will be

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Solution:

For minimum wavelength,
$\lambda \propto n^{2}$
$\therefore \frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\left(\frac{1}{2}\right)^{2}$
$\frac{\lambda_{L}}{\lambda_{B}}=\frac{1}{4}=0.25$