Chapter 7, Problem 7D.9P

(a)

Interpretation Introduction

Interpretation:

The Schrodinger equation for a particle of mass $m$ in a three dimensional box of sides $L_1,L_2$ and $L_3$ has to be predicted.  The Schrodinger equation is separable has to be shown.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7D.9P

The Schrodinger equation for a particle of mass $m$ in a three dimensional box of sides $L_1,L_2$ and $L_3$ is shown below.

    $\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}+\frac{8{\pi }^{2}m}{h^2}\left[\left(\frac{n_1^2{h}^2}{8m{L}_1^2}+\frac{n_2^2{h}^2}{8m{L}_2^2}+\frac{n_3^2{h}^2}{8m{L}_3^2}\right)-V\left(1,2,3\right)\right]{\psi }_{\left(n_1,n_2,n_3\right)}=0$

It has been shown that the Schrodinger equation is separable.

Explanation of Solution

The Schrodinger equation for a particle of mass $m$ in a three dimensional box is shown below.

    $\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}+\frac{8{\pi }^{2}m}{h^2}\left[\left(E_1+E_2+E_3\right)-V\left(1,2,3\right)\right]{\psi }_{\left(n_1,n_2,n_3\right)}=0$        (1)

Where,

• $E_1,E_2,E_3$ are the total energies along $x$ axis, $y$ axis and $z$ axis respectively.
• $V\left(1,2,3\right)$ are the potential energy of the particle along three axes.

The energy of a particle along $x$ axis is calculated by the following formula.

    $E_1=\frac{n_1^2{h}^2}{8m{L}_1^2}$

The energy of a particle along $y$ axis is calculated by the following formula.

    $E_2=\frac{n_2^2{h}^2}{8m{L}_2^2}$

The energy of a particle along $z$ axis is calculated by the following formula.

    $E_3=\frac{n_3^2{h}^2}{8m{L}_3^2}$

Therefore, the Schrodinger equation for a particle of mass $m$ in a three dimensional box of sides $L_1,L_2$ and $L_3$ by substituting the value of $E_1,E_2$ and $E_3$ is shown below.

    $\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}+\frac{8{\pi }^{2}m}{h^2}\left[\left(\frac{n_1^2{h}^2}{8m{L}_1^2}+\frac{n_2^2{h}^2}{8m{L}_2^2}+\frac{n_3^2{h}^2}{8m{L}_3^2}\right)-V\left(1,2,3\right)\right]{\psi }_{\left(n_1,n_2,n_3\right)}=0$

The Schrodinger equation is separable.  The separation of Schrodinger equation is done as shown below.

The Schrodinger equation for a particle of mass $m$ in a three dimensional box of sides $L_1,L_2$ and $L_3$ along $x$ axis is shown below.

    $\frac{{\partial }^2\psi }{\partial x^2}+\frac{8{\pi }^{2}m}{h^2}\left[\frac{n_1^2{h}^2}{8m{L}_1^2}-V\right]{\psi }_{\left(n_1\right)}=0$

The Schrodinger equation for a particle of mass $m$ in a three dimensional box of sides $L_1,L_2$ and $L_3$ along $y$ axis is shown below.

    $\frac{{\partial }^2\psi }{\partial y^2}+\frac{8{\pi }^{2}m}{h^2}\left[\frac{n_2^2{h}^2}{8m{L}_2^2}-V\right]{\psi }_{\left(n_2\right)}=0$

The Schrodinger equation for a particle of mass $m$ in a three dimensional box of sides $L_1,L_2$ and $L_3$ along $z$ axis is shown below.

    $\frac{{\partial }^2\psi }{\partial z^2}+\frac{8{\pi }^{2}m}{h^2}\left[\frac{n_3^2{h}^2}{8m{L}_3^2}-V\right]{\psi }_{\left(n_3\right)}=0$

(b)

Interpretation Introduction

Interpretation:

The wavefunction and energy are defined by three quantum numbers has to be shown.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7D.9P

The wavefunction and energy of a particle in a three dimensional box is shown below.

    ${\psi }_{\left(n_1,n_2,n_3\right)}={\left(\frac{8}{L_1{L}_2{L}_3}\right)}^{1/2}\sin \left(\frac{n_1\pi x}{L_1}\right)\sin \left(\frac{n_2\pi y}{L_2}\right)\sin \left(\frac{n_3\pi z}{L_3}\right)$

    $E=\left(\frac{n_1^2}{L_1^2}+\frac{n_2^2}{L_2^2}+\frac{n_3^2}{L_3^2}\right)\frac{h^2}{8m}$

The wavefunction and energy of a particle in a three dimensional box are defined by three quantum numbers $n_1,n_2$ and $n_3$.

Explanation of Solution

The normalized wavefunction of a particle of mass $m$ in a three dimensional box is shown below.

    ${\psi }_{\left(n_1,n_2,n_3\right)}={\left(\frac{8}{L_1{L}_2{L}_3}\right)}^{1/2}\sin \left(\frac{n_1\pi x}{L_1}\right)\sin \left(\frac{n_2\pi y}{L_2}\right)\sin \left(\frac{n_3\pi z}{L_3}\right)$        (2)

Where,

• $L_1,L_2,L_3$ are the sides of three dimensional box.
• $n_1,n_2,n_3$ are the three principle quantum numbers.

The energy of a particle of mass $m$ in a three dimensional box is shown below.

    $E=\left(\frac{n_1^2}{L_1^2}+\frac{n_2^2}{L_2^2}+\frac{n_3^2}{L_3^2}\right)\frac{h^2}{8m}$        (3)

From equations (2) and (3), are consist of three quantum numbers.  Therefore, the wavefunction and energy are defined by three quantum numbers $n_1,n_2$ and $n_3$.

(c)

Interpretation Introduction

Interpretation:

The energy of an electron moving in a cubic box of length $5\text{nm}$ has to be calculated and the energy level diagram showing the first $15$ energy levels has to be drawn.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7D.9P

The energy of an electron moving in a cubic box of length $5\text{nm}$ is $\left(n_1^2+n_2^2+n_3^2\right)\times 2.3\times {10}^{-21}\text{J}$.

The energy level diagram showing the first $15$ energy levels is shown below.

Explanation of Solution

The energy of an electron moving in a cubic box of length $5\text{nm}$ is calculated by the following formula.

    $E=\left(\frac{n_1^2}{L_1^2}+\frac{n_2^2}{L_2^2}+\frac{n_3^2}{L_3^2}\right)\frac{h^2}{8m}$        (4)

Where,

• $m$ is the mass of an electron.
• $h$ is the Planck’s constant.
• $L_1,L_2,L_3$ are the sides of the box.

It is given that the side of the cube is $5\text{nm}$.

The conversion of $\text{nm}$ to $\text{m}$ is done as,

    $1\text{nm}=\text{1}\times {\text{10}}^{-9}\text{m}$

Therefore, the conversion of $\text{5}\text{nm}$ to $\text{m}$ is done as,

    $5\text{nm}=5\times {\text{10}}^{-9}\text{m}$

Therefore, the side of the cubic box is $5\times {\text{10}}^{-9}\text{m}$.

The given box is a cubic box therefore, the values of $L_1,L_2$ and $L_3$ are equal to $5\times {\text{10}}^{-9}\text{m}$.

The mass of electron is $9.1\times {10}^{-31}\text{kg}$.

The value of Planck’s constant is $6.6\times {10}^{-34}\text{J}\text{s}$.

Substitute the value of $m$, $h$, $L_1,L_2$ and $L_3$ in equation (4).

    $\begin{array}{c}E=\left(\frac{n_1^2}{{\left(5\times {\text{10}}^{-9}\text{m}\right)}^2}+\frac{n_2^2}{{\left(5\times {\text{10}}^{-9}\text{m}\right)}^2}+\frac{n_3^2}{{\left(5\times {\text{10}}^{-9}\text{m}\right)}^2}\right)\frac{{\left(6.6\times {10}^{-34}\text{J}\text{s}\right)}^2}{8\times 9.1\times {10}^{-31}\text{kg}}\\ =\left(n_1^2+n_2^2+n_3^2\right)\frac{43.56\times {10}^{-68}{\text{J}}^2{\text{s}}^2}{8\times 9.1\times {10}^{-31}\text{kg}\times {\left(5\times {\text{10}}^{-9}\text{m}\right)}^2}\\ =\left(n_1^2+n_2^2+n_3^2\right)\frac{0.023\times {10}^{-68}{\text{J}}^2{\text{s}}^2}{1\times {10}^{-49}\text{J}{\text{s}}^2}\left(\because 1\text{kg}{\text{m}}^2=1\text{J}{\text{s}}^2\right)\\ =\left(n_1^2+n_2^2+n_3^2\right)\times 2.3\times {10}^{-21}\text{J}\end{array}$

Therefore, the energy of an electron moving in a cubic box of length $5\text{nm}$ is $\left(n_1^2+n_2^2+n_3^2\right)\times 2.3\times {10}^{-21}\text{J}$.

The energy level diagram of the first $15$ energy levels is shown below.

Figure 1

(d)

Interpretation Introduction

Interpretation:

The energy level diagram obtained from part (c) has to be compared with the energy level diagram of an electron moving in a one dimensional box of length $5\text{nm}$.  Whether the energy levels are more or less sparsely distributed in the cubic box than in one-dimensional box has to be predicted.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7D.9P

The energy levels are more sparsely distributed in the cubic box than in the one dimensional box.

Explanation of Solution

The energy of an electron moving in a one dimensional box of length $5\text{nm}$ is calculated by the following formula.

    $E=\frac{n^2{h}^2}{8m{L}^2}$        (5)

Where,

• $m$ is the mass of an electron.
• $h$ is the Planck’s constant.
• $L_1,L_2,L_3$ are the sides of the box.

It is given that the side of the cube is $5\text{nm}$.

Therefore, the side of the box is $5\times {\text{10}}^{-9}\text{m}$.

The mass of electron is $9.1\times {10}^{-31}\text{kg}$.

The value of Planck’s constant is $6.6\times {10}^{-34}\text{J}\text{s}$.

Substitute the value of $m$, $h$, $L_1,L_2$ and $L_3$ in equation (4).

    $\begin{array}{c}E=\frac{n^2{\left(6.6\times {10}^{-34}\text{J}\text{s}\right)}^2}{8\times 9.1\times {10}^{-31}\text{kg}\times {\left(5\times {\text{10}}^{-9}\text{m}\right)}^2}\\ =\frac{n^2\times 43.56\times {10}^{-68}{\text{J}}^2{\text{s}}^2}{8\times 9.1\times {10}^{-31}\text{kg}\times {\left(5\times {\text{10}}^{-9}\text{m}\right)}^2}\\ =\frac{n^2\times 0.023\times {10}^{-68}{\text{J}}^2{\text{s}}^2}{1\times {10}^{-49}\text{J}{\text{s}}^2}\left(\because 1\text{kg}{\text{m}}^2=1\text{J}{\text{s}}^2\right)\\ =2.3\times {10}^{-21}\times n^2\text{J}\end{array}$

Therefore, the energy of an electron moving in a one dimensional box of length $5\text{nm}$ is $2.3\times {10}^{-21}\times n^2\text{J}$.

The energy level diagram of an electron moving in a one dimensional box of length $5\text{nm}$ is shown below.

Figure 2

From figure 1 and figure 2, it is clear that the energy levels are more sparsely distributed in the cubic box than in the one dimensional box.