Question

The wavelength of the second line in Balmer series of the hydrogen spectrum is 486.4 nm. What is the wavelength of the first line of the Lyman series?

• A. 121.6 nm
• B. 364.8 nm
• C. 729.6 cm
• D. None of these

Answer 1

The correct option is A 121.6 nm

For Balmer series, $\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2})=R(\frac{1}{4}-\frac{1}{n^2});n=3,4,5\cdots$
The second line in balmer series corresponds ot n= 4.
Hence, $\frac{1}{{\lambda }_2}=R(\frac{1}{4}-\frac{1}{16})=\frac{3R}{16}\Rightarrow {\lambda }_2=\frac{16}{3R}$
The wavelength of the first line (n = 2) in Lyman series is given by $\frac{1}{{\lambda }_1}=R(1-\frac{1}{2^2})=R(1-\frac{1}{4})=\frac{3R}{4}\Rightarrow {\lambda }_1=\frac{4}{3R}$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{4}{3R}\times \frac{3R}{16}=\frac{1}{4}\Rightarrow {\lambda }_1=\frac{{\lambda }_2}{4}=\frac{486.4}{4}=121.6nm$