Question

the wavelength of the first member of balmer series in the hydrogen spectrum is 6563A.calculate the first member of lyman series in the same spectrum

Answer 1

Dear Student ,
here in this question the wavelength of the spectral lines in Hydrogen atom are given by ,$\frac{1}{\lambda }=\frac{1}{R}\left(\frac{1}{{n_f}^2}-\frac{1}{{n_i}^2}\right)$ where R is the Rydberg constant .
Now for the first member of the Balmer series ,$n_f=2and{n}_i=3.$
Therefore ,

$\frac{1}{{\lambda }_1}=\frac{1}{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5R}{36}$ .......(1)
Now for the first member of the Lyman series $n_f=1and{n}_i=2.$
Therefore,
$$\begin{gathered} \frac{1}{{\lambda }_1\text{'}}=\frac{1}{R}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3R}{4}.........\left(2\right) \\ Nowfromequations1and2wecanwritethat, \\ \frac{{\lambda }_1}{{\lambda }_1\text{'}}=\frac{5R}{36}\times \frac{4}{3R}=\frac{5}{27} \\ \therefore {\lambda }_1\text{'}=\frac{5}{27}\times {\lambda }_1=\frac{5}{27}\times 6563=1215\stackrel{0}{A} \end{gathered}$$
Regards