Chapter 12, Problem 165CP

Interpretation Introduction

Interpretation:Wavelength of light required to promote high energy ground state electron to lowest energy excited state is to be calculated.

Concept introduction:Wavefunction is defined as the function of wave that gives information about the properties of wave. For a particle in one dimension box energy is calculated as follows:

  $E=\frac{n^2{h}^2}{8m{L}^2}$

Where,

• $E$ is energy of state.
• $n$ is energy level.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $L$ is dimension of cubical box.

Answer to Problem 165CP

Value of $\lambda$ is $2470\text{ nm}$ .

Explanation of Solution

The expression to calculate energy of a cubical box is as follows:

  $E=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$

Since eight electrons are occupied in allowed energy level and each level occupies two electrons. Therefore, energy for first four energy levels is calculated. The four energy levels are $E_{1,1,1}$ , $E_{1,1,2}$ , $E_{1,2,1}$ and $E_{2,1,1}$ . The energy of last three orbitals is same; therefore, it is a degenerate orbital. The expression to calculate energy of one level is as follows:

  $E_{1,1,1}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$

Where,

• $E_{1,1,1}$ is energy of first level.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $n_x$ , $n_y$ and $n_z$ are energy levels in $x$ , $y$ and $z$ direction respectively.
• $L$ is dimension of cubical box.

Value of $n_x$ is 1.

Value of $n_y$ is 1.

Value of $n_z$ is 1.

Substitute values in above equation.

  $\begin{array}{c}E=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)\\ =\frac{h^2}{8m{L}^2}\left(1+1+1\right)\\ =\frac{3h^2}{8m{L}^2}\end{array}$

Hence, energy of cubical box is $\frac{3h^2}{8m{L}^2}$ .

Since energy of second, third and fourth level is same, therefore, expression used to calculate energy of highest energy level is as follows:

  $E_{1,1,2}=E_{1,2,1}=E_{2,1,1}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$

Where,

• $E_{1,1,2}$ , $E_{1,2,1}$ and $E_{2,1,1}$ .is energy of second, third and fourth level.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $n_x$ , $n_y$ and $n_z$ are energy levels in $x$ , $y$ and $z$ direction respectively.
• $L$ is dimension of cubical box.

Value of $n_x$ is $\sqrt{2}$ .

Value of $n_y$ is $\sqrt{2}$ .

Value of $n_z$ is $\sqrt{2}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{1,1,2}=E_{1,2,1}=E_{2,1,1}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)\\ =\frac{h^2}{8m{L}^2}\left({\left(\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2\right)\\ =\frac{6h^2}{8m{L}^2}\end{array}$

Hence, energy of second, third and fourth level in cubical box of is $\frac{6h^2}{8m{L}^2}$ .

Excitation of electron from highest level to lowest excited state develops next energy levels. Thus, energy of energy levels that are also degenerate is as follows:

  $E_{2,2,1}=E_{1,2,2}=E_{2,1,2}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$

Where,

• $E_{2,2,1}$ , $E_{1,2,2}$ and $E_{2,1,2}$ .is energies of excited level.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $n_x$ , $n_y$ and $n_z$ are energy levels in $x$ , $y$ and $z$ direction respectively.
• $L$ is dimension of cubical box.

Value of $n_x$ is $\sqrt{3}$ .

Value of $n_y$ is $\sqrt{3}$ .

Value of $n_z$ is $\sqrt{3}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{2,2,1}=E_{1,2,2}=E_{2,1,2}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)\\ =\frac{h^2}{8m{L}^2}\left({\left(\sqrt{3}\right)}^2+{\left(\sqrt{3}\right)}^2+{\left(\sqrt{3}\right)}^2\right)\\ =\frac{9h^2}{8m{L}^2}\end{array}$

Hence, energy of excited level in cubical box of is $\frac{9h^2}{8m{L}^2}$ .

The transition of electron from high energy ground state electron to lowest energy excited state is calculated by difference in energy of degenerate levels. The expression for difference in energy is as follows:

  $\begin{array}{c}\Delta E=\frac{9h^2}{8m{L}^2}-\frac{6h^2}{8m{L}^2}\\ =\frac{3h^2}{8m{L}^2}\end{array}$

Where,

• $\Delta E$ is energy change.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $L$ is dimension of cubical box.

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

Value of $m$ is $9.1\times {10}^{-31}\text{ kg}$ .

Value of $L$ is $1.50\text{ nm}$ .

Substitute values in above equation.

  $\begin{array}{c}\Delta E=\frac{3h^2}{8m{L}^2}\\ =\frac{3{\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\left(9.1\times {10}^{-31}\text{ kg}\right)\left(1.50\text{ nm}\right)}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =8.03\times {10}^{-20}\text{ J}\end{array}$

Hence, change in energy is $8.03\times {10}^{-20}\text{ J}$ .

The change in energy for transition is calculated is as follows:

  $\Delta E=\frac{hc}{\lambda }$

Rearrange above equation for $\lambda$ .

  $\lambda =\frac{hc}{\Delta E}$

Value of $c$ is $2.998\times {10}^8\text{ m/s}$ .

Value of $\Delta E$ is $8.03\times {10}^{-20}\text{ J}$ .

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

Substitute values in above equation.

  $\begin{array}{c}\lambda =\frac{hc}{\Delta E}\\ =\frac{\left(6.626\times {10}^{-34}\text{ Js}\right)\left(2.998\times {10}^8\text{ m/s}\right)}{8.03\times {10}^{-20}\text{ J}}\left(\frac{{10}^9\text{ nm}}{1\text{ m}}\right)\\ =2470\text{ nm}\end{array}$

Hence, value of $\lambda$ is $2470\text{ nm}$ .

Conclusion

Value of $\lambda$ is $2470\text{ nm}$ .