Chapter 28, Problem 72P

(a)

To determine

The length of the one-dimensional box.

Answer to Problem 72P

The length of the box is $6.1\text{ nm}$.

Explanation of Solution

The ground state energy of the electron in a box is $0.010\text{ eV}$.

Write the expression for energy of the quantized level.

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

Here, $E_n$ is the energy of the level $n$, $n$ is the quantum number, $h$ is the Planck’s constant, $m$ is the mass of the electron and $L$ is the length of the box.

Write the expression for ground state energy for a particle in box.

$E_1=\frac{h^2}{8m{L}^2}$                                                           (II)

Here, $E_1$ is the ground state energy.

Rearranging (I)

$L=\sqrt{\frac{h^2}{8m{E}_1}}$                                                            (III)

Substituting $9.109\times {10}^{-31}\text{ kg}$ for $m$, $0.010\text{ eV}$ for $E_1$ and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (II) to find $L$.

 $\begin{array}{c}L=\sqrt{\frac{{\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\times 9.109\times {10}^{-31}\text{ kg}\times \text{}0.010\text{ eV}}}\\ =\sqrt{\frac{{\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\times 9.109\times {10}^{-31}\text{ kg}\times \text{}0.010\text{ eV}}\times \frac{1\text{ eV}}{1.602\times {10}^{-19}\text{J}}}\\ =6.1\times {10}^{-9}\text{}\text{m}\\ \simeq 6.1\text{nm}\end{array}$

Thus, the length of the box is $6.1\text{ nm}$.

(b)

To determine

The plot of wave functions of the first three state.

Explanation of Solution

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

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

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 $1,2\text{ and 3}$ for $n$ and $6.1\text{ nm}$ for $L$  in (VIII) to find the fourth excited state wave function, ${\psi }_5$.

${\psi }_1=\sqrt{\frac{2}{6.1\text{ nm}}}\sin \left(\frac{\pi x}{6.1\text{ nm}}\right)$                                                    (V)

Here, ${\psi }_1$ is the ground state wave function.

${\psi }_2=\sqrt{\frac{2}{6.1\text{ nm}}}\sin \left(\frac{2\pi x}{6.1\text{ nm}}\right)$                                                    (VI)

Here, ${\psi }_2$ is the ground state wave function.

${\psi }_3=\sqrt{\frac{2}{6.1\text{ nm}}}\sin \left(\frac{3\pi x}{6.1\text{ nm}}\right)$                                                    (VII)

Here, ${\psi }_3$ is the ground state wave function.

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

(c)

To determine

The wavelength of the electron in the third state.

Answer to Problem 72P

The wave length of the electron in the second excited state is $4.1\text{}\text{nm}$.

Explanation of Solution

Substituting $3$ for $n$ in (III)

$E_3=\frac{3^2{h}^2}{8m{L}^2}$                                                           (IV)

Here, $E_3$ is the energy of the second excited state.

Write the expression for de Broglie wavelength of an electron.

$\lambda =\frac{h}{\sqrt{2mK}}$                                                                     (V)

Here, $\lambda$ is the de Broglie wavelength of the electron and $K$ is the kinetic energy of the electron.

The particle in a box model renders only the kinetic energy since the potential is zero at the box and is infinite everywhere else. Thus, consider, $K=E_3$.

Substituting $E_3$ for $K$  in (V)

$\lambda =\frac{h}{\sqrt{2mK}}\sqrt{\frac{8m{L}^2}{9h^2}}$                                                                     (VI)

Rearranging

$\lambda =\frac{2L}{3}$                                                                   (VII)

Substituting $6.1\text{nm}$ for $L$ in (VII) to find $\lambda$.

$\begin{array}{c}\lambda =\frac{2\times 6.1\text{ nm}}{3}\\ =4.1\text{ nm}\end{array}$

Thus, the wave length of the electron in the second excited state is $4.1\text{}\text{nm}$.

(d)

To determine

The wavelength of the emitted photon during the downward transition of the electron.

Answer to Problem 72P

The possible wavelengths of the emitted photons are $15.5\text{ μm, 25 μm and}\text{41 μm}$.

Explanation of Solution

The wavelength of the photon is $15.5\text{ μm}$.

Write the expression for energy of a photon.

$E=\frac{hc}{\lambda }$                                                                    (IX)

Here, $E$ is the energy of the photon, $\lambda$ is the wavelength of the photon  and $c$ is the speed of light.

Write the expression for energy of the quantized levels in terms of the ground state energy

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

Subtracting $E_1$ from (X)

 $E_n-E_1=n^2{E}_1-E_1$                                                                     (XI)

Here, $E_n-E_1$ is the energy absorbed by the electron from the photon to undergo an upward transition. Thus, $E_n-E_1$ is same as $E$ and hence (X) becomes

$E=\left(n^2-1\right)E_1$                                                                     (XII)

Rearranging (XII)

$n=\sqrt{\frac{E}{E_1}+1}$                                                                 (XIII)

Substituting $15.5\text{ μm}$ for $\lambda$, $1240\text{ eV}$ for $hc$ in (IX) to find $E$

$\begin{array}{c}E=\frac{1240\text{eV}\cdot \text{nm}}{15.5\text{ μm}}\\ =0.0800\text{ eV}\end{array}$

Substitute $0.0800\text{ eV}$ for $E$ and  $0.010\text{ eV}$ for $E_1$ in (XII) to find $n$

$\begin{array}{c}n=\sqrt{\frac{0.0800\text{ eV}}{0.010\text{ eV}}+1}\\ =3\end{array}$

Thus when the electron absorbs the photon, it is excited to the third state.

The possible downward transitions are $3\to 2,2\to 1\text{and }3\to 1$.

Write the expression for energy of the photon during a downward transition.

$\lambda =\frac{hc}{E_{n^{\prime }}-E_n}$                                                               (XIV)

Here, $E_{n^{\prime }}$ is the energy in the higher state and $E_n$ is the energy in the lower state.

Substituting $2\text{and 3}$ for $n$ in (X) to find $E_2\text{}\text{and }{E}_3$ respectively

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

$E_3=3^2{E}_1$                                                          (XVI)

Substituting $2$ for $n^{\prime }$ and $1$ for $n$ in (XIV) to find $\lambda$

$\lambda =\frac{hc}{E_2-E_1}$                                                     (XVII)

Substituting $1240\text{ eV}$ for $hc$, $2^2{E}_1$ for $E_2$ and $0.010\text{ eV}$ for $E_1$ in (XVII) to find $\lambda$

$\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{4\left(0.010\text{ eV}\right)-0.010\text{ eV}}\\ =41\times {10}^{-6}\text{ m}\\ =\text{41 μm}\end{array}$

For, $2\to 1$ transition, the wavelength of the photon is $\text{41 μm}$.

Substituting $3$ for $n^{\prime }$ and $2$ for $n$ in (XIV) to find $\lambda$

$\lambda =\frac{hc}{E_3-E_2}$                                                    (XVIII)

Substituting $1240\text{ eV}$ for $hc$, $3^2{E}_1$ for $E_3$, $2^2{E}_1$ for $E_2$ and $0.010\text{ eV}$ for $E_1$ in (XVIII) to find $\lambda$

$\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{9\left(0.010\text{ eV}\right)-4\left(0.010\text{ eV}\right)}\\ =25\times {10}^{-6}\text{ m}\\ =2\text{5}\text{}\text{μm}\end{array}$

For, $3\to 2$ transition, the wavelength of the photon is $2\text{5}\text{}\text{μm}$.

For, $3\to 1$ transition, the emitted photon has the same energy as the absorbed photon. Thus, the wavelength of the photon is $\text{15}\text{.5}\text{}\text{μm}$.

Thus, the possible wavelengths of the emitted photons are $15.5\text{ μm, 25 μm and}\text{41 μm}$.