Thank you for reporting, we will resolve it shortly
Q.
Taking Rydberg's constant $R_{ H }=1.097 \times 10^{7} m$ first and second wavelength of Balmer series in hydrogen spectrm, is :
Punjab PMETPunjab PMET 2004Atoms
Solution:
The wavelength of the lines in Balmer series is represented by,
$\frac{1}{\lambda}=R_{H}\left[\frac{1}{2^{2}}-\frac{1}{n_{2}^{2}}\right]$
For first wavelength
$\frac{1}{\lambda_{1}} =1.097 \times 10^{7}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]$
$=1.524 \times 10^{6} m$
or $\lambda_{1} =\frac{1}{1.524 \times 10^{6}}$
$=6.562 \times 10^{-7}$
or $\lambda_{1} =6562\, \mathring{A}$
For second wavelength
$\frac{1}{\lambda_{1}} =1.097 \times 10^{-7}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]$
$=2.056 \times 10^{6}$
or $\lambda_{2} =\frac{1}{2.056 \times 10^{6}}$
or $\lambda_{2} =4.863-7 \times 10 \times 10^{-10}$
or $\lambda_{2} =4863\, \mathring{A}$