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 $
Click to See Answer :
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 $