Chapter 31, Problem 67P

(a)

To determine

The maximum wavelength of radiation.

Answer to Problem 67P

The maximum wavelength of radiation is $588\text{nm}$ .

Explanation of Solution

Given:

The energy at ground state is, $E_0=0$ .

The first excited of atoms is $E_1=2.11\text{eV}$ .

The second excited of atoms is $E_2=3.20\text{eV}$ .

The third excited of atoms is $E_3=4.35\text{eV}$ .

Formula used:

The maximum wavelength of radiation is given as,

  $\lambda =\frac{h\times c}{\Delta E}$

Here, $h$ is the Planck’s constant and its value is $6.26\times {10}^{-34}\text{J}\cdot \text{s}$ , $c$ is the speed of light and its value $3\times {10}^8\text{m}/\text{s}$ and $\Delta E$ is the difference in the energy state.

Calculation:

The maximum wavelength of radiation from state 2 to ground state can be calculated as,

  $\begin{array}{c}{\lambda }_{2-0}=\frac{h\times c}{E_2-E_0}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8\text{m}/\text{s}}{\left(3.20\text{eV}\times \left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)-0\right)}\\ =388\times {10}^{-9}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =388\text{nm}\end{array}$

The wavelength of radiation from state 2 to state 1 can be calculated as,

  $\begin{array}{c}{\lambda }_{2-1}=\frac{h\times c}{E_2-E_1}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8\text{m}/\text{s}}{\left(3.20\text{eV}\times \left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)-2.11\text{eV}\times \left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)\right)}\\ =1140\times {10}^{-9}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =1140\text{nm}\end{array}$

The maximum wavelength of radiation from state 1 to ground state can be calculated as,

  $\begin{array}{c}{\lambda }_{1-0}=\frac{h\times c}{E_1-E_0}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8\text{m}/\text{s}}{\left(2.11\text{eV}\times \left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)-0\right)}\\ =588\times {10}^{-9}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =588\text{nm}\end{array}$

Conclusion:

Therefore, the maximum wavelength of radiation for resonance absorption is $588\text{nm}$ .

(b)

To determine

The wavelength at excitation state of $4.35\text{eV}$ and the wavelength of radiation at excited states.

Answer to Problem 67P

The wavelength at excitation state of $4.35\text{eV}$ is $285\text{nm}$ , the wavelengths of resonance fluorescence at excited state are $554\text{nm}$ and $558\text{nm}$

Explanation of Solution

Calculation:

The wavelength of radiation from excited state 3 to state 1 can be calculated as,

  $\begin{array}{c}{\lambda }_{3-1}=\frac{h\times c}{E_3-E_1}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8\text{m}/\text{s}}{\left(4.35\text{eV}\times \left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)-2.11\text{eV}\times \left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)\right)}\\ =554\times {10}^{-9}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =554\text{nm}\end{array}$

The wavelength of radiation from excited state 1 to ground state can be calculated as,

  $\begin{array}{c}{\lambda }_{1-0}=\frac{h\times c}{E_1-E_0}\\ =\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8\text{m}/\text{s}}{\left(2.11\text{eV}\times \left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)-0\right)}\\ =558\times {10}^{-9}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =558\text{nm}\end{array}$

Conclusion:

Therefore, the wavelength at excitation state of $4.35\text{eV}$ is $285\text{nm}$ , the wavelengths of resonance fluorescence at excited state are $554\text{nm}$ and $558\text{nm}$