Chapter 1, Problem 1A.16E

Interpretation Introduction

Interpretation:

The values of $n$ for the initial and final energy levels of the electron during the emission of energy that leads to the spectral line observed at $434nm$ have to be determined.

Concept Introduction:

The relationship between wavelength, frequency and speed of light is given below.

  $\lambda \times \nu =c$

Where, $\lambda$ is the wavelength, $\nu$ is the frequency of radiation and $c$ is the speed of light.

Rydberg equation:

  $\begin{array}{l}\nu =R\left\{\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right\}\\ \\ Where,n_1=1,2,\mathrm{....},n_2=n_1+1,n_1+2,\mathrm{....}\end{array}$

R is the Rydberg constant.  Its value is $3.29\times {10}^{15}Hz.$

Answer to Problem 1A.16E

The values of $n$ for the beginning and ending energy levels of the electron are $2and5$ respectively.

Explanation of Solution

The values of initial and final energy levels can be calculated by using Rydberg formula.

  $\begin{array}{c}\nu =R\left\{\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right\}\\ \\ \lambda =\frac{c}{\nu }=\frac{c}{\left(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right)R}\\ \\ \left(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right)=\frac{c}{\lambda R}=\frac{3\times {10}^{8}m/s}{\left(434\times {10}^{-9}m\right)\left(3.29\times {10}^{15}{s}^{-1}\right)}=0.21\\ \\ \frac{n_2{}^2-n_1{}^2}{{\left(n_1.n_2\right)}^2}=\frac{21}{100}\end{array}$

There are two equations possible here.

  $\begin{array}{l}{n}_2{}^2-n_1{}^2=\mathrm{21........}\left(1\right)\\ \\ {\left(n_1.n_2\right)}^2=100\\ \\ n_1.n_2=10\\ \\ n_1=\frac{10}{n_2}\mathrm{......}\left(2\right)\end{array}$

Now plugging equation $\left(2\right)$ in equation $\left(1\right)$, the following result can be obtained.

  $\begin{array}{l}{n}_2{}^2-n_1{}^2=21\\ \\ n_2{}^2-{\left(\frac{10}{n_2}\right)}^2=21\\ \\ n_2{}^2-\frac{100}{n_2{}^2}=21\\ \\ n_2{}^4-21n_2{}^2-100=\mathrm{0.....}\left(3\right)\end{array}$

The roots of the above quadratic equation can be obtained as given below.

  $\begin{array}{l}Let\text{'}stake{n}_2{}^2=x,then{x}^2-21x-100=0\\ \\ x=\frac{-\left(-21\right)\pm \sqrt{{\left(-21\right)}^2-\mathrm{4.1.}\left(-100\right)}}{2.1}\\ \\ x=\frac{21\pm \sqrt{841}}{2}\\ \Rightarrow x_1=\frac{21+29}{2}=25\\ \\ \Rightarrow x_2=\frac{21-29}{2}=-4\left(Cannotbeaccepted\right)\\ \\ So,x=25\Rightarrow n_2{}^2=25\Rightarrow n_2=5\\ \\ Equation\left(2\right)\Rightarrow n_1=\frac{10}{n_2}=\frac{10}{5}=2\end{array}$

Therefore, the values of $n$ for the beginning and ending energy levels of the electron are $2and5$ respectively.