Chapter 4, Problem 4.50P

Interpretation Introduction

Interpretation:

Wavelength of the first five lines in Pickering series has to be calculated and also the region in which they occur has to be given.

Explanation of Solution

Ionization energy can be calculated using the equation given by Bohr that is shown below.

    $E_n=\left(-2.1799\text{aJ}\right)\frac{Z^2}{n^2}$

Where,

    $Z$ is the atomic number of the atom.

    $n$ is the integer and the values are $1,2,\mathrm{3.....}$.

It is said that the electron fall to $n=4$ state.  This means the final state $n_f=4$.  Emission line for Pickering series start from 5,6,7… and it ends at 4.  This means the initial state will be $n_i=5,6,\mathrm{7....}$ and $n_f=4$.

Wavenumber of any line that is present in visible spectrum of hydrogen is shown by Balmer and it can be given by using the formula shown below;

    $\overline{\nu }=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

In the above equation, $R_H$ is the Rydberg constant for hydrogen atom and it is equal to $1.097\times {10}^7\text{m}$.  The above equation can be written for the helium atom as shown below;

    $\frac{1}{\lambda }=Z^2{R}_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$        (1)

Wavelength of first five lines of Pickering series:

First line:

Atomic number of helium is $2$.  Photon is emitted when the electron moves from $n_i=5$ to $n_f=5$ state.  Substituting this in equation (1), the wavelength can be calculated as shown below;

    $\begin{array}{c}\frac{1}{\lambda }=Z^2{R}_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ =4\times 1.097\times {10}^7\text{m}\left(\frac{1}{4^2}-\frac{1}{5^2}\right)\\ =\text{0}\text{.09873}\times {10}^7\text{m}\\ \lambda \text{=}\frac{1}{\text{0}\text{.09873}\times {10}^7\text{m}}\\ =10.13\times {10}^{-7}\text{m}\\ \text{=1013}\text{nm}\end{array}$

Second line:

Atomic number of helium is $2$.  Photon is emitted when the electron moves from $n_i=6$ to $n_f=5$ state.  Substituting this in equation (1), the wavelength can be calculated as shown below;

    $\begin{array}{c}\frac{1}{\lambda }=Z^2{R}_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ =4\times 1.097\times {10}^7\text{m}\left(\frac{1}{4^2}-\frac{1}{6^2}\right)\\ =\text{0}\text{.1523}\times {10}^7\text{m}\\ \lambda \text{=}\frac{1}{\text{0}\text{.1523}\times {10}^7\text{m}}\\ =6.566\times {10}^{-7}\text{m}\\ \text{=656}\text{nm}\end{array}$

Third line:

Atomic number of helium is $2$.  Photon is emitted when the electron moves from $n_i=7$ to $n_f=5$ state.  Substituting this in equation (1), the wavelength can be calculated as shown below;

    $\begin{array}{c}\frac{1}{\lambda }=Z^2{R}_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ =4\times 1.097\times {10}^7\text{m}\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\\ =\text{0}\text{.1847}\times {10}^7\text{m}\\ \lambda \text{=}\frac{1}{\text{0}\text{.1847}\times {10}^7\text{m}}\\ =5.4142\times {10}^{-7}\text{m}\\ \text{=541}\text{nm}\end{array}$

Fourth line:

Atomic number of helium is $2$.  Photon is emitted when the electron moves from $n_i=8$ to $n_f=5$ state.  Substituting this in equation (1), the wavelength can be calculated as shown below;

    $\begin{array}{c}\frac{1}{\lambda }=Z^2{R}_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ =4\times 1.097\times {10}^7\text{m}\left(\frac{1}{4^2}-\frac{1}{8^2}\right)\\ =\text{0}\text{.2057}\times {10}^7\text{m}\\ \lambda \text{=}\frac{1}{\text{0}\text{.2057}\times {10}^7\text{m}}\\ =4.861\times {10}^{-7}\text{m}\\ \text{=486}\text{nm}\end{array}$

Fifth line:

Atomic number of helium is $2$.  Photon is emitted when the electron moves from $n_i=9$ to $n_f=5$ state.  Substituting this in equation (1), the wavelength can be calculated as shown below;

    $\begin{array}{c}\frac{1}{\lambda }=Z^2{R}_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\\ =4\times 1.097\times {10}^7\text{m}\left(\frac{1}{4^2}-\frac{1}{9^2}\right)\\ =\text{0}\text{.2203}\times {10}^7\text{m}\\ \lambda \text{=}\frac{1}{\text{0}\text{.2203}\times {10}^7\text{m}}\\ =4.539\times {10}^{-7}\text{m}\\ \text{=454}\text{nm}\end{array}$

The wavelength of the first five lines of Pickering series is summarized as shown below;

$n_i$	$5$	$6$	$7$	$8$	$9$
Wavelength ($\text{nm}$)	$1013$	$656$	$541$	$486$	$454$

From the wavelength, it is found that this series lies in visible region of the electromagnetic spectrum.

In solar astronomy, this series is used n calculation of the properties of stars, age, luminosity, velocity, etc.