Question

What is the shortest wavelength in the Paschen series of spectral lines?

Answer 1

Step 1: Formula for the calculation of wavelength

$\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Here, $\lambda$ is the wavelength

$R$ is Rydberg constant,

$n_1$=1,2........ , and $n_2$= $n_1+1$,.$n_1+2$……

The first five series of lines that correspond to $n_1$are known as the Lyman, Balmer, Paschen, Bracket, and Pfund series, respectively.

Step 2: Calculating the shortest wavelength for paschen series

For the Paschen series,

$n_1=3$

$n_2=\infty$

$R=109,677{\text{cm}}^{-1}$.

$$\begin{gathered} \frac{1}{\lambda }=109,677{\text{cm}}^{-1}\left[\frac{1}{3^2}-\frac{1}{\infty }\right] \\ \frac{1}{\lambda }\text{=}\frac{109,677{\text{cm}}^{-1}}{9} \\ \lambda =\frac{9}{109,677{\text{cm}}^{-1}} \\ \lambda =8.205\times {10}^{-5}\text{cm} \end{gathered}$$

Hence, the shortest wavelength is $8.205\times {10}^{-5}\text{cm}$.