Chapter 12, Problem 39E

a)

Interpretation Introduction

Interpretation: The reason behind closer energy levels with higher states should be explained with mathematical concept.

Concept introduction: Salient characteristics of Bohr’s model are as follows:

Electron revolved in certain discrete circular orbits that were associated with fixed angular momentum.

Such discrete orbits are termed non-radiating or stationary orbits. The quantized energy for each of such discrete orbit can be obtained from the expression as follows:

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

Where

• $R_{\text{H}}$ is constant that has value of $2.178\times {10}^{-18}\text{ J}$ .
• $Z$ represent atomic number.
• $n$ denotes particular stationary state.
• $E_n$ denotes energy associated with particular stationary state.

Energy absorption or emission only occurred when electron jumped from one energy sate to another hence each such probable unique transition gave rise to specific lines in emission spectrum of hydrogen atom. This was based on the relation as follows:

  $\Delta E=h\cdot \Delta v$

Where

• $\Delta v$ is difference in frequency of two stationary states involved in transition.
• $\Delta E$ denotes energy absorbed or emitted in transition.

Explanation of Solution

The quantized energy for each of such discrete orbit can be obtained from the expression as follows:

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

This expression suggests $E_n$ is inversely related to square of $n$ . So higher the value of particular stationary state more is the energy associated with it. Hence for consecutive energy levels, the energy difference is almost minimal. So they begin to appear as continuum.

(b)

Interpretation Introduction

Interpretation: The colors depicted for transition between various energy levels should be verified.

Concept introduction: Salient characteristics of Bohr’s model are as follows:

Electron revolved in certain discrete circular orbits that were associated with fixed angular momentum.

Such discrete orbits are termed non-radiating or stationary orbits. The quantized energy for each of such discrete orbit can be obtained from the expression as follows:

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

Where

• $R_{\text{H}}$ is constant that has value of $2.178\times {10}^{-18}\text{ J}$ .
• $Z$ represent atomic number.
• $n$ denotes particular stationary state.
• $E_n$ denotes energy associated with particular stationary state.

Hydrogen emission spectrum that shows characteristics lines could be accounted by the assumption made in Bohr’s model. Since energy absorption or emission only occurred when electron jumped from one energy sate to another hence each such probable unique transition gave rise to specific lines in emission spectrum of hydrogen atom. This was based on the relation as follows:

  $\Delta E=h\cdot \Delta v$

Where

• $\Delta v$ is difference in frequency of two stationary states involved in transition.
• $\Delta E$ denotes energy absorbed or emitted in transition.

Explanation of Solution

The quantized energy for each of such discrete orbit can be obtained from the expression as follows:

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

For $n=3$

Value of $R_{\text{H}}$ is $2.178\times {10}^{-18}\text{ J}$ .

Value of $Z$ for hydrogen atom is 1.

Substitute the values in above formula.

  $\begin{array}{c}{E}_3=\left(-R_{\text{H}}\right)\frac{Z^2}{n^2}\\ =\left(-2.178\times {10}^{-18}\text{ J}\right)\frac{1}{{\left(3\right)}^2}\end{array}$

For $n=2$

Substitute the values in above formula.

  $\begin{array}{c}{E}_2=\left(-R_{\text{H}}\right)\frac{Z^2}{n^2}\\ =\left(-2.178\times {10}^{-18}\text{ J}\right)\frac{1}{{\left(2\right)}^2}\end{array}$

Difference in theses energies gives energy for red color transition is calculated as follows:

  $\begin{array}{c}{E}_3-E_2=\left(-2.178\times {10}^{-18}\text{ J}\right)\left[\frac{1}{{\left(3\right)}^2}-\frac{1}{{\left(2\right)}^2}\right]\\ =3.025\times {10}^{-19}\text{ J}\end{array}$

Similarly, energy for green color transition as follows:

  $\begin{array}{c}{E}_4-E_2=\left(-2.178\times {10}^{-18}\text{ J}\right)\left[\frac{1}{{\left(4\right)}^2}-\frac{1}{{\left(2\right)}^2}\right]\\ =4.083\times {10}^{-19}\text{ J}\end{array}$

Likewise, energy for blue color transition as follows:

  $\begin{array}{c}{E}_5-E_2=\left(-2.178\times {10}^{-18}\text{ J}\right)\left[\frac{1}{{\left(5\right)}^2}-\frac{1}{{\left(2\right)}^2}\right]\\ =5.445\times {10}^{-19}\text{ J}\end{array}$

Formula to compute wavelength from energy difference is given as follow:

  $\text{λ}=h\cdot \frac{c}{\Delta E}$

Value of $\Delta E$ is $3.025\times {10}^{-19}\text{ J}$ .

Value of $h$ is $6.626\times {10}^{-34}\text{ J s}$ .

Value of $c$ is $3\times {10}^8\text{ m/s}$ .

Substitute the value in above equation to calculate wavelength for red color.

  $\begin{array}{c}\text{λ}=h\cdot \frac{c}{\Delta E}\\ =\left(6.626\times {10}^{-34}\text{ J s}\right)\frac{\left(3\times {10}^8\text{ m/s}\right)}{\left(3.025\times {10}^{-19}\text{ J}\right)}\\ =\left(6.571\times {10}^{-7}\text{ m}\right)\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =657.1\text{ nm}\end{array}$

Corresponding wavelength is calculated as follows;

Value of $\Delta E$ is $3.025\times {10}^{-19}\text{ J}$ .

Substitute the value in above equation to calculate wavelength for green color.

  $\begin{array}{c}\text{λ}=h\cdot \frac{c}{\Delta E}\\ =\left(6.626\times {10}^{-34}\text{ J s}\right)\frac{\left(3\times {10}^8\text{ m/s}\right)}{\left(4.083\times {10}^{-19}\text{ J}\right)}\\ =\left(4.868\times {10}^{-7}\text{ m}\right)\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =486.8\text{ nm}\end{array}$

Value of $\Delta E$ is $5.445\times {10}^{-19}\text{ J}$ .

Substitute the value in above equation to calculate wavelength for blue color.

  $\begin{array}{c}\text{λ}=h\cdot \frac{c}{\Delta E}\\ =\left(6.626\times {10}^{-34}\text{ J s}\right)\frac{\left(3\times {10}^8\text{ m/s}\right)}{\left(5.445\times {10}^{-19}\text{ J}\right)}\\ =\left(1.6528\times {10}^{-7}\text{ m}\right)\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =165.28\text{ nm}\end{array}$

Since the wavelength for red, green and blue light also correspond to wavelength values of $657.1\text{ nm}$

  $486.8\text{ nm}$ and $165.28\text{ nm}$ therefore the colored lines depicted are correct.