Chapter 28, Problem 43P

(a)

To determine

Energy level diagram for one-dimensional box.

Answer to Problem 43P

The energy level diagram is given below:

Explanation of Solution

The length of the box is $0.100\text{nm}$.

Write the expression for energy of one dimensional box

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

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

Conclusion

Substitute $6.626\times {10}^{-34}\text{Js}$ for $h$, $1$ for $n$, $0.100\text{nm}$ for $L$ and $9.1\times {10}^{-31}\text{kg}$ for $m$ in (I) to find $E_n$ for the first level.

    $\begin{array}{c}{E}_n=\frac{1^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\text{nm}\right)}^2}\\ =\frac{1^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\times {10}^{-9}\text{m}\right)}^2}\cdot \frac{1\text{ eV}}{1.602\times {10}^{-19}{\text{kgm}}^2{\text{s}}^{-2}}\\ =37.7\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{Js}$ for $h$, $2$ for $n$, $0.100\text{nm}$ for $L$ and $9.1\times {10}^{-31}\text{kg}$ for $m$ in (I) to find $E_n$ for the second level.

    $\begin{array}{c}{E}_n=\frac{2^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\text{nm}\right)}^2}\\ =\frac{2^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\times {10}^{-9}\text{m}\right)}^2}\cdot \frac{1\text{ eV}}{1.602\times {10}^{-19}{\text{kgm}}^2{\text{s}}^{-2}}\\ =151\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{Js}$ for $h$, $3$ for $n$, $0.100\text{nm}$ for $L$ and $9.1\times {10}^{-31}\text{kg}$ for $m$ in (I) to find $E_n$ for the third level

    $\begin{array}{c}{E}_n=\frac{3^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\text{nm}\right)}^2}\\ =\frac{3^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\times {10}^{-9}\text{m}\right)}^2}\cdot \frac{1\text{ eV}}{1.602\times {10}^{-19}{\text{kgm}}^2{\text{s}}^{-2}}\\ =339\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{Js}$ for $h$, $4$ for $n$, $0.100\text{nm}$ for $L$ and $9.1\times {10}^{-31}\text{kg}$ for $m$ in (I) to find $E_n$ for the fourth level

    $\begin{array}{c}{E}_n=\frac{4^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\text{nm}\right)}^2}\\ =\frac{4^2{\left(6.626\times {10}^{-34}{\text{kgm}}^2{\text{s}}^{-1}\right)}^2}{8\left(9.1\times {10}^{-31}\text{kg}\right){\left(0.100\times {10}^{-9}\text{m}\right)}^2}\cdot \frac{1\text{ eV}}{1.602\times {10}^{-19}{\text{kgm}}^2{\text{s}}^{-2}}\\ =603\text{eV}\end{array}$

The energy level diagram is given below:

(b)

To determine

The wavelength of emitted photons during transitions in these four levels.

Answer to Problem 43P

The frequencies of the emitted photon are $\text{4}\text{.71}\text{nm},\text{ 2}\text{.75}\text{nm},\text{2}\text{.20}\text{nm},\text{ 6}\text{.59}\text{nm, 4}\text{.12}\text{nm and 11}\text{.0}\text{nm}$.

Explanation of Solution

The minimum frequency for photoemission corresponds to the cutoff wavelength.

Write the expression for wavelength

    $\lambda =\frac{hc}{\left|E-E^{\prime }\right|}$                                                                                                                (II)

Here, $\lambda$ is the wavelength, $c$ is the speed of light, $E$ is the higher level energy and $E^{\prime }$ is the lower level energy.

Conclusion

For transition, $4\to 3$,

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$, $603\text{eV}$ for $E$ and $339\text{eV}$ for $E^{\prime }$ in (II) to find $\lambda$

    $\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{603\text{eV}-339\text{eV}}\\ =\text{4}\text{.71}\text{nm}\end{array}$

For transition, $4\to 2$,

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$, $603\text{eV}$ for $E$ and $151\text{eV}$ for $E^{\prime }$ in (II) to find $\lambda$

    $\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{603\text{eV}-151\text{eV}}\\ =\text{2}\text{.75}\text{nm}\end{array}$

For transition, $4\to 1$,

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$, $603\text{eV}$ for $E$ and $37.7\text{eV}$ for $E^{\prime }$ in (II) to find $\lambda$

    $\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{603\text{eV}-37.7\text{eV}}\\ =\text{2}\text{.20}\text{nm}\end{array}$

For transition, $3\to 2$,

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$, $339\text{eV}$ for $E$ and $151\text{eV}$ for $E^{\prime }$ in (II) to find $\lambda$

    $\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{339\text{eV}-151\text{eV}}\\ =\text{6}\text{.59}\text{nm}\end{array}$

For transition, $3\to 1$,

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$, $339\text{eV}$ for $E$ and $37.7\text{eV}$ for $E^{\prime }$ in (II) to find $\lambda$

    $\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{339\text{eV}-37.7\text{eV}}\\ =\text{4}\text{.12}\text{nm}\end{array}$

For transition, $2\to 1$,

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$, $151\text{eV}$ for $E$ and $37.7\text{eV}$ for $E^{\prime }$ in (II) to find $\lambda$

    $\begin{array}{c}\lambda =\frac{1240\text{ eV}\cdot \text{nm}}{151\text{eV}-37.7\text{eV}}\\ =\text{11}\text{.0}\text{nm}\end{array}$

Thus, the frequencies of the emitted photon are $\text{4}\text{.71}\text{nm},\text{ 2}\text{.75}\text{nm},\text{2}\text{.20}\text{nm},\text{ 6}\text{.59}\text{nm, 4}\text{.12}\text{nm and 11}\text{.0}\text{nm}$.