Atomic spectra

Vitthal, thanks for posting the solution. Nice problem. The original post as follows.

Using Rydberg formula for hydrogen-like element

$\frac{1}{\lambda} = RZ^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ where:

• $R = 1.097373 \times 10^7 \space m^{-1}$ is the Rydberg constant
• $Z$ is the atomic number of the element

And putting $Z=2$ , $n_1$ the first line, $n_2 \to \infty$ for the series limit, and $\dfrac{1}{\lambda} = 2745100 \space cm^{-1} = 2745100 \space m^{-1}$ , we have:

$\begin{aligned} \frac{1}{\lambda} & = RZ^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \\ 2745100 & = 10973730 \times 2^2 \frac{1}{n_1^2} \\ n_1^2 & = 16.002 \\ \Rightarrow n_1 & = 4 \end{aligned}$

Since the first line $n_1 = 4$ , the lower energy level is $n' = n_1 - 1 = 3$ which is the $\boxed{\text{Paschen}}$ series ( see here ).