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  4. Ratio of wavelength for second line of Balmer series and first line of Lymen series in H -atom is -

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Q. Ratio of wavelength for second line of Balmer series and first line of Lymen series in $H$ -atom is -

NTA AbhyasNTA Abhyas 2020

Solution:

$\frac{1}{\left(\lambda \right)_{1}}=R\left(1\right)^{2}\left(\frac{1}{2^{2}} - \frac{1}{4^{2}}\right)\ldots \left(1\right);\left(\lambda \right)_{1}=\frac{16}{3 R}$
$\frac{1}{\left(\lambda \right)_{2}}=R\left(1\right)^{2}\left(\frac{1}{1^{2}} - \frac{1}{2^{2}}\right)\ldots \left(2\right);\left(\lambda \right)_{2}=\frac{4}{3 R}$
$\frac{\lambda _{1}}{\lambda _{2}}=\frac{16}{4}=\frac{4}{1}$