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Q. The ratio of longest wavelength lines in the Balmer and Paschen series of hydrogen spectrum is

AP EAMCETAP EAMCET 2015

Solution:

The wavelength of Balmer series is given by
$\frac{1}{\lambda_{B}}=R\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5}{36} R$ .......(i)
The wavelength for paschen series is given by
$\frac{1}{\lambda_{p}}=R\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]=\frac{7}{144} R$ ......(ii)
On dividing Eq. (ii) by Eq (i), we get
$\frac{\lambda_{B}}{\lambda_{p}}=\frac{7}{20}$