Question

The shortest wavelength for the Lyman series is $912A^{\circ }$. Find the shortest wavelength for Paschen and Brackett series in the Hydrogen atom.

Answer 1

Step 1: Electron Transition for Hydrogen atom:

The electron Transition for the Hydrogen atom is as follows,

Step 2: Finding the shortest wavelength

Using the formula for the shortest wavelength $\lambda$, $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1}-\frac{1}{n_2})$

Where $R=1.097\times {10}^7$, $Z$ is the atomic number and $n$ is the level.

For the shortest wavelength in the Paschen series, the two levels showing transition should be maximum. $n=\infty$

$$\begin{gathered} \frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{3^2}-\frac{1}{{\infty }^2}) \\ \lambda =8.20\times {10}^{-7}m \\ \lambda =8200A^{\circ } \end{gathered}$$

For the shortest wavelength in the Brackett series, the two levels showing transition should be maximum. $n=\infty$

$$\begin{gathered} \frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{4^2}-\frac{1}{{\infty }^2}) \\ \lambda =15.58\times {10}^{-7}m \\ \lambda =14580A^{\circ } \end{gathered}$$

Hence, for Paschen and Brackett series the shortest wavelength is $8200A^{\circ }$ and $14580A^{\circ }$.