Q: The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Å. The wavelength of the first line is

(a) $ \displaystyle \frac{27}{20}\times 4861 A^o $

(b) $ \displaystyle \frac{20}{27}\times 4861 A^o $

(c) $ \displaystyle 20 \times 4861 A^o $

(d) $ \displaystyle 4861 A^o $

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Sol:$ \displaystyle \frac{1}{\lambda_2} = R (\frac{1}{2^2}-\frac{1}{4^2})$

$ \displaystyle \frac{1}{\lambda_2} = R (\frac{1}{4}-\frac{1}{16})$

$ \displaystyle \frac{1}{\lambda_2} = R (\frac{3}{16}) $ …(i)

$ \displaystyle \frac{1}{\lambda_1} = R (\frac{1}{2^2}-\frac{1}{3^2})$

$\displaystyle \frac{1}{\lambda_2} = R (\frac{1}{4}-\frac{1}{9})$

$ \displaystyle \frac{1}{\lambda_2} = R (\frac{5}{36}) $ …(ii)

On dividing (i) by (ii)

$ \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{3R/16}{5R/36} $

$ \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{27}{20} $

$ \displaystyle \lambda_1 = \frac{27}{20}\times \lambda_2 $

$ \displaystyle \lambda_1 = \frac{27}{20}\times 4861 A^o $