Question

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

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

Answer 1

The correct option is A

$4:1$

Step 1: Formula used

The wavelength is given by $\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$ and assume ${\lambda }_L=wavelengthinLymanseries,{\lambda }_B=wavelengthinBalmerseries$

Step 2: Calculation for wavelenght of Lyman and Balmer series

1. For shortest wavelength in Lyman series, take $n_1=1and{n}_2=\infty$,

$$\begin{gathered} \frac{1}{{\lambda }_L}=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right) \\ \frac{1}{{\lambda }_L}=R{Z}^2\to (Eq.1) \end{gathered}$$

2. For shortest wavelength in Balmer series, take $n_1=2and{n}_2=\infty$

$$\begin{gathered} \frac{1}{{\lambda }_B}=R{Z}^2\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right) \\ \frac{1}{{\lambda }_B}=\frac{R{Z}^2}{4}\to (Eq.2) \end{gathered}$$

$Now,\frac{Eq.2}{Eq.1}$

$\frac{{\lambda }_B}{{\lambda }_L}=\frac{4}{1}$

Hence, option A is correct.