Chapter 12, Problem 164CP

(a)

Interpretation Introduction

Interpretation:Energies of lowest levels of cubic box 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 164CP

Energy of three lowest levelsis $\frac{3h^2}{8m{L}^2}$ , $\frac{6h^2}{8m{L}^2}$ and $\frac{9h^2}{8m{L}^2}$ .

Explanation of Solution

Energy for three lowest levels of a cubical box are $E_{111}$ , $E_{112}$ and $E_{122}$ . The expression to calculate energy of a cubical box, $E_{111}$ is as follows:

  $E_{111}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$

Where,

• $E_{111}$ is energy of first lowest level of cubical box.
• $h$ isplanck’s constant.
• $m$ is mass of box.
• $n_x$ , $n_y$ and $n_z$ are energy levels in $x$ , $y$ and $z$ direction respectively.
• $L$ isdimension 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 first lowest level of cubical box is $\frac{3h^2}{8m{L}^2}$ .

The expression to calculate energy of a cubical box, $E_{112}$ is as follows:

  $E_{112}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$

Where,

• $E_{112}$ is energy of second lowest level of cubical box.
• $h$ isplanck’s constant.
• $m$ is mass of box.
• $n_x$ , $n_y$ 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 2.

Substitute values in above equation.

  $\begin{array}{c}{E}_{112}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)\\ =\frac{h^2}{8m{L}^2}\left(1^2+1^2+2^2\right)\\ =\frac{3h^2}{8m{L}^2}\left(1+1+4\right)\\ =\frac{6h^2}{8m{L}^2}\end{array}$

Hence, energy of first lowest level of cubical box is $\frac{6h^2}{8m{L}^2}$ .

The expression to calculate energy of a cubical box, $E_{122}$ is as follows:

  $E_{122}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$

Where,

• $E_{122}$ is energy of third lowest level of cubical box.
• $h$ isplanck’s constant.
• $m$ is mass of box.
• $n_x$ , $n_y$ 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 2.

Value of $n_z$ is 1.

Substitute values in above equation.

  $\begin{array}{c}{E}_{122}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)\\ =\frac{h^2}{8m{L}^2}\left(1+2^2+2^2\right)\\ =\frac{9h^2}{8m{L}^2}\end{array}$

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

(b)

Interpretation Introduction

Interpretation:Degenracy that corresponds to quantum number that has values 1 or 2 in a rectangular box should be calculated. Also, if box changes the effect of degeneracy is to be determined.

Concept introduction:Degenrate levels are those that has same energy but different wavefunctions. Wavefunction is defined as the function of wave that gives information about the properties of wave.

Explanation of Solution

For a quantum number 1 or 2, degeneracies is as follows:

  $\begin{array}{cc}\text{Energy state}& \text{Degeneracy}\\ E_{111}& \text{single state}\left(\text{no degeneracy}\right)\\ E_{112}& 3\left(\begin{array}{c}{E}_{112}\\ E_{121}\\ E_{211}\end{array}\right)\\ E_{122}& 3\left(\begin{array}{c}{E}_{122}\\ E_{212}\\ E_{221}\end{array}\right)\\ E_{222}& \text{single state}\end{array}$

If the box is changed that is $L_x\ne L_y\ne L_z$ , then degeneracy will be as follows:

  $\begin{array}{cc}\text{Energy state}& \text{Levels}\\ E_{111}& \text{single state}\left(\text{no degeneracy}\right)\\ E_{112}& 3\left(\text{no degeneracy}\right)\\ E_{122}& 3\left(\text{no degeneracy}\right)\\ E_{222}& \text{single state}\end{array}$