Chapter 34, Problem 42P

(a)

To determine

The energy of the ground state $\left(n=1\right)$ and the first two excited states of a neutron in a one-dimensional box.

Answer to Problem 42P

The energy of the ground state $\left(n=1\right)$ is $\text{5}\text{.1133}\text{meV}$ and the first two excited states of a neutron in a one-dimensional box are $\text{20}\text{.5 meV}$ and $\text{46}\text{.0 meV}$ .

Explanation of Solution

Given:

The length of one-dimensional box is $\text{0}\text{.200}\text{nm }\left(\text{about}\text{the}\text{diameter}\text{of}\text{a}{H}_2\text{nucleus}\right)$ .

Formula used:

The expression for energy of ground state is given by,

  $E_1=\frac{h^2}{8m_n{L}^2}$

The expression for energy of nth states is given by,

  $E_n=n^2{E}_1$

Calculation:

The energy of ground state is calculated as,

  $\begin{array}{c}{E}_1=\frac{h^2}{8m_n{L}^2}\\ =\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.6749\times {10}^{-27}\text{kg}\right){\left(0.200\text{nm}\right)}^2\left(1.602\times {10}^{-19}\text{J}\right)}\\ =\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.6749\times {10}^{-27}\text{kg}\right){\left(\left(0.200\text{nm}\right)\times \frac{\left({10}^{-9}\text{m}\right)}{\left(1\text{nm}\right)}\right)}^2\left(1.602\times {10}^{-19}\text{J}\right)}\\ =5.1133\times {10}^{-3}\text{eV}\end{array}$

Solve further as,

  $\begin{array}{c}{E}_1=5.1133\times {10}^{-3}\text{eV}\\ \text{=}\left(5.1133\times {10}^{-3}\text{eV}\right)\times \frac{\left({10}^{-6}\text{meV}\right)}{\left(1\text{eV}\right)}\\ =5.1133\text{meV}\end{array}$

The energy of first excited state is calculated as,

  $\begin{array}{c}{E}_n=n^2{E}_1\\ E_2=2^2\times 5.1133\text{meV}\\ =20.5\text{meV}\end{array}$

The energy of second excited state is calculated as,

  $\begin{array}{c}{E}_n=n^2{E}_1\\ E_3=3^2\times 5.1133\text{meV}\\ =46.0\text{meV}\end{array}$

Conclusion:

Therefore, the energy of the ground state $\left(n=1\right)$ is $\text{5}\text{.1133}\text{meV}$ and the first two excited states of a neutron in a one-dimensional box are $\text{20}\text{.5 meV}$ and $\text{46}\text{.0 meV}$ .

(b)

To determine

The wavelength of electromagnetic radiation emitted.

Answer to Problem 42P

The wavelength of electromagnetic radiation emitted is $80.8\text{μm}$ .

Explanation of Solution

Given:

The neutron makes transition from $n=2$ to $n=1$

Formula used:

The expression for wavelength of electromagnetic radiation emitted is given by,

  $\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}$

Calculation:

The wavelength of electromagnetic radiation emitted is calculated as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}\\ =\frac{1240\text{eV}\cdot \text{nm}}{E_2-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{4E_1-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{3\times 5.1133\text{meV}}\end{array}$

Solve further as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{3\times 5.1133\text{meV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{3\times \left(\left(5.1133\text{meV}\right)\times \frac{\left({10}^6\text{eV}\right)}{\left(1\text{meV}\right)}\right)}\\ =\left(\left(80.8\times {10}^3\text{nm}\right)\times \frac{\left({10}^{-3}\text{μm}\right)}{\left(1\text{nm}\right)}\right)\\ =80.8\text{μm}\end{array}$

Conclusion:

Therefore, the wavelength of electromagnetic radiation emitted is $80.8\text{μm}$ .

(c)

To determine

The wavelength of electromagnetic radiation emitted.

Answer to Problem 42P

The wavelength of electromagnetic radiation emitted is $48.5\text{μm}$ .

Explanation of Solution

Given:

The neutron makes transition from $n=3$ to $n=2$ .

Formula used:

The expression for wavelength of electromagnetic radiation emitted is given by,

  $\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}$

Calculation:

The wavelength of electromagnetic radiation emitted is calculated as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}\\ =\frac{1240\text{eV}\cdot \text{nm}}{E_3-E_2}\\ =\frac{1240\text{eV}\cdot \text{nm}}{9E_1-4E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{5\times 5.1133\text{meV}}\end{array}$

Solve further as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{5\times 5.1133\text{meV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{5\times \left(\left(5.1133\text{meV}\right)\times \frac{\left({10}^6\text{eV}\right)}{\left(1\text{meV}\right)}\right)}\\ =\left(\left(48.5\times {10}^3\text{nm}\right)\times \frac{\left({10}^{-3}\text{μm}\right)}{\left(1\text{nm}\right)}\right)\\ =48.5\text{μm}\end{array}$

Conclusion:

Therefore, the wavelength of electromagnetic radiation emitted is $48.5\text{μm}$ .

(d)

To determine

The wavelength of electromagnetic radiation emitted.

Answer to Problem 42P

The wavelength of electromagnetic radiation emitted is $30.3\text{μm}$ .

Explanation of Solution

Given:

The neutron makes transition from $n=3$ to $n=1$

Formula used:

The expression for wavelength of electromagnetic radiation emitted is given by,

  $\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}$

Calculation:

The wavelength of electromagnetic radiation emitted is calculated as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}\\ =\frac{1240\text{eV}\cdot \text{nm}}{E_3-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{9E_1-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{8\times 5.1133\text{meV}}\end{array}$

Solve further as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{8\times 5.1133\text{meV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{8\times \left(\left(5.1133\text{meV}\right)\times \frac{\left({10}^6\text{eV}\right)}{\left(1\text{meV}\right)}\right)}\\ =\left(\left(30.3\times {10}^3\text{nm}\right)\times \frac{\left({10}^{-3}\text{μm}\right)}{\left(1\text{nm}\right)}\right)\\ =30.3\text{μm}\end{array}$

Conclusion:

Therefore, the wavelength of electromagnetic radiation emitted is $30.3\text{μm}$ .