Question

If ${\lambda }_1$ and ${\lambda }_2$ are the wavelength of the first numbers of the Lyman and Paschen series respectively. Then ${\lambda }_1:{\lambda }_2$ ____________

• A. $1:30$
• B. $1:3$
• C. $7:50$
• D. $7:108$

Answer 1

The correct option is D $7:108$

For Lyman series $\frac{1}{{\lambda }_1}=R[\frac{1}{1^2}-\frac{1}{n^2}]$
$\frac{1}{{\lambda }_1}=R[\frac{1}{1}-\frac{1}{2^2}]$ [for the first number, $n=2$]
$\frac{1}{{\lambda }_1}=R\left[\frac{4-1}{4}\right]$
${\lambda }_1=\frac{4}{3R}$ ....(i)
For Paschen series
$\frac{1}{{\lambda }_2}=R[\frac{1}{3^2}-\frac{1}{4^2}]$ [for $n=4$]
${\lambda }_2=\frac{144}{7R}$ ....(ii)
So, ${\lambda }_1:{\lambda }_2=\frac{4}{3R}\times \frac{7R}{144}=7:108$