Chapter 26, Problem 100QAP

To determine

(a)

The shortest and longest wavelengths of the Brackett series

Answer to Problem 100QAP

The shortest and longest wavelengths of the Brackett series are $\text{1459}\text{.85 nm}$ and $4050.22\text{ nm}$ respectively.

Explanation of Solution

Given:

For Brackett series $n_f=4$

Formula used:

Rydberg formula is given as

  $\frac{1}{\lambda }=1.09737\times {10}^7\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)mete{r}^{-1}$

Calculation:

For the shortest wavelength of the Brackett series $n_f=4$ and $n_i=\infty$ (continuum)

Putting these values in Eq. (1.1) we get

  $\begin{array}{l}\frac{1}{\lambda }=1.09737\times {10}^7\left(\frac{1}{4^2}-\frac{1}{{\infty }^2}\right)mete{r}^{-1}=0.0685\times {10}^{7}mete{r}^{-1}\\ or,\lambda \text{ = 14}\text{.5985}\times {\text{10}}^{\text{-7}}\text{ meter = 1459}\text{.85 nm}\end{array}$

This emission is in the infrared range.

For the longest wavelength $n_f=4$ and $n_i=5$, so using Eq. 1.1 we get

  $\begin{array}{l}\frac{1}{\lambda }=1.09737\times {10}^7\left(\frac{1}{4^2}-\frac{1}{5^2}\right)mete{r}^{-1}=0.02469\times {10}^{7}mete{r}^{-1}\\ or\text{, }\lambda \text{ = }40.5022\times {10}^{-7}meter=4050.22\text{ nm}\end{array}$

This emission is also in the infrared range.

Conclusion:

Therefore, the shortest and longest wavelengths of the Brackett series are $\text{1459}\text{.85 nm}$ and $4050.22\text{ nm}$ respectively. The emission lines are in the infrared range.

To determine

(b)

The shortest and longest wavelengths of the Pfund series

Answer to Problem 100QAP

Therefore, the shortest and longest wavelengths of the Pfund series are $\text{2283}\text{.10 nm}$ and $7457.12\text{ nm}$ respectively.

Explanation of Solution

Given:

For Pfund series $n_f=5$

Calculation:

For the shortest wavelength of the Pfund series $n_f=5$ and $n_i=\infty$ (continuum)

Putting these values in Eq. (1.1) we get

  $\begin{array}{l}\frac{1}{\lambda }=1.09737\times {10}^7\left(\frac{1}{5^2}-\frac{1}{{\infty }^2}\right)mete{r}^{-1}=0.0438\times {10}^{7}mete{r}^{-1}\\ or,\lambda \text{ = 22}\text{.8310}\times {\text{10}}^{\text{-7}}\text{ meter = 2283}\text{.10 nm}\end{array}$

This emission is in the infrared range.

For the longest wavelength $n_f=5and{n}_i=6$, so using Eq. 1.1 we get

  $\begin{array}{l}\frac{1}{\lambda }=1.09737\times {10}^7\left(\frac{1}{5^2}-\frac{1}{6^2}\right)mete{r}^{-1}=0.01341\times {10}^{7}mete{r}^{-1}\\ or\text{, }\lambda \text{ = }74.5712\times {10}^{-7}meter=7457.12\text{ nm}\end{array}$

This emission is also in the infrared range.

Conclusion:

Therefore, the shortest and longest wavelengths of the Pfund series are $\text{2283}\text{.10 nm}$ and $7457.12\text{ nm}$ respectively. The emission lines are in the infrared range.