Question

What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series ?

• A. $4:1$
• B. $4:3$
• C. $4:9$
• D. $5:9$

Answer 1

Answer: (A)

For a Balmer series
$\frac{1}{{\lambda }_B}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]...\left(i\right)$
where $n=3,4,.....$
By putting $n=\infty$ in equation $(i)$, we obtain the series limit of the Balmer series. This is the shortest wavelength of the Balmer series.
or ${\lambda }_B=\frac{4}{R}...(ii)$
For a Lyman series
$\frac{1}{{\lambda }_L}=R[\frac{1}{1^2}-\frac{1}{n^2}]...(iii)$
where $n=2,3,4,....$
By putting $n=\infty$ in equation $(iii)$, we obtain the series limit of the Lyman series. This is the shortest wavelength of the Lyman series.
or ${\lambda }_L=\frac{1}{R}...(iv)$
Dividing $(ii)$ by $(iv)$, we get
$\frac{{\lambda }_B}{{\lambda }_L}=\frac{4}{1}$