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  4. The shortest wavelengths of Paschen, Balmer and Lyman series are in the ratio

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Q. The shortest wavelengths of Paschen, Balmer and Lyman series are in the ratio

COMEDKCOMEDK 2015Atoms

Solution:

The shortest wavelength of Paschen $(\lambda_P)$ Balmer $(\lambda_B)$ and Lyman $(\lambda_L)$ series are given by
$\frac{1}{\lambda_P} = \frac{R_H}{3^2} ; \frac{1}{\lambda_B} = \frac{R_H}{2^2}$ and $\frac{1}{\lambda_L} = \frac{R_H}{1^2}$
So, $\lambda_p = \frac{9}{R_H} , \lambda_B = \frac{4}{R_H} , \lambda_L = \frac{1}{R_H}$
$\therefore \:\: \lambda_P : \lambda_B : \lambda_L = 9 : 4 : 1$