Chapter 28, Problem 55P

(a)

To determine

The ground state energy of the electron.

Answer to Problem 55P

The ground state energy of the electron is $0.067\text{ eV}$.

Explanation of Solution

The energy of the photon emitted during transition from the first excited state to the ground state is $0.20\text{ eV}$.

Write the expression for energy of the quantized level in terms of the ground state energy.

$E_n=n^2{E}_1$                                                            (I)

Here, $E_n$ is the energy of the level $n$, $n$ is the quantum number and $E_1$ is the energy of the ground state.

Substituting $2$ for $n$ in (II)

$E_2=2^2{E}_1$                                                           (II)

Here, $E_2$ is the energy of the first excited state.

Write the expression for energy difference between the first excited state and the ground state.

$\Delta E=E_2-E_1$                                                          (III)

Here, $\Delta E$ is the energy difference.

Substituting (II) in (III)

$\Delta E=2^2{E}_1-E_1=3E_1$                                                          (IV)

Rearranging (IV)

$E_1=\frac{\Delta E}{3}$                                                           (V)

Substituting $0.20\text{ eV}$ for $\Delta E$ in (V) to find $E_1$

$\begin{array}{c}{E}_1=\frac{0.20\text{ eV}}{3}\\ =0.067\text{ eV}\end{array}$

Thus, the ground state energy is $0.067\text{ eV}$.

(b)

To determine

The possible energies of the emitted photon if the electron is initially in the third excited state.

Answer to Problem 55P

The possible energies of the emitted photon in the increasing order are $0.20\text{ eV, }0.33\text{ eV, }0.47\text{ eV, }0.53\text{ eV, }0.80\text{ eV and 1}.0\text{ eV}$.

Explanation of Solution

If the initial state of the electron is $4$, The possible transitions are $4\to 3,3\to 2,2\to 1,4\to 2,4\to 1,\text{ and }3\to 1$. The energy differences between these levels are the possible energies of the emitted photon.

Substitute $4\text{}\text{ and 3}$ in (I) to find $E_4\text{ and }{E}_3$  respectively.

$E_4=4^2{E}_1=16E_1$                                                           (VI)

$E_3=3^2{E}_1=9E_1$                                                           (VII)

Subtracting the corresponding energies in the allowed transition to find the energy of the photon and substitute $0.067\text{ eV}$ for $E_1$

For $4\to 3$, $\Delta E=16E_1-9E_1=7E_1=7\times 0.067\text{}\text{ eV}=0.47\text{ eV}$.

For $3\to 2$, $\Delta E=9E_1-4E_1=5E_1=5\times 0.067\text{}\text{ eV}=0.33\text{ eV}$.

For $2\to 1$, $\Delta E=4E_1-E_1=3E_1=3\times 0.067\text{}\text{ eV}=0.20\text{ eV}$.

For $4\to 2$, $\Delta E=16E_1-4E_1=12E_1=12\times 0.067\text{}\text{ eV}=0.80\text{ eV}$.

For $4\to 1$, $\Delta E=16E_1-E_1=15E_1=15\times 0.067\text{}\text{ eV}=1.0\text{ eV}$.

For $3\to 1$, $\Delta E=9E_1-E_1=8E_1=8\times 0.067\text{}\text{ eV}=0.53\text{ eV}$.

Thus, the possible energies of the emitted photon in the increasing order are $0.20\text{ eV, }0.33\text{ eV, }0.47\text{ eV, }0.53\text{ eV, }0.80\text{ eV and 1}.0\text{ eV}$.

(c)

To determine

The plot of the wave function of the electron in the third excited state.

Answer to Problem 55P

The wave function is a sign wave with three nodes.

Explanation of Solution

Write the expression for the wave function of a particle in a box.

${\psi }_n=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)$                                                    (VIII)

Here, ${\psi }_n$ is the wave function of the quantum state $n$, $L$ is the length of the box  and $x$ is the position co-ordinate.

Substitute $4$ for $n$ in (VIII) to find ${\psi }_4$

${\psi }_4=\sqrt{\frac{2}{L}}\sin \left(\frac{4\pi x}{L}\right)$                                                     (VIII)

The plot of the wave function versus position is given below:

(d)

To determine

The effect of increasing the length of the box on its energy.

Answer to Problem 55P

The energy spacing between two levels decreases.

Explanation of Solution

Write the expression for the energy of a particle in a box.

$E_n=\frac{n^2{h}^2}{8m{L}^2}$                                                    (VIII)

Here, $m$ is the mass of the electron and $h$ is the Planck’s constant.

$E_n$ is inversely proportional to the square of the length of the box. Thus, increasing the length decreases the energy of each level and hence the energy spacing also decreases.