Chapter 34, Problem 41P

(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 and the energy level diagram of the system.

Answer to Problem 41P

The energy of the ground state $\left(n=1\right)$ is $\text{204}\text{.53}\text{MeV}$ and the first two excited states of a neutron in a one-dimensional box are $\text{818 MeV}$ and $\text{1}\text{.84 GeV}$ . The energy level diagram of the system is shown in Figure 1.

Explanation of Solution

Given:

The length of one-dimensional box is $\text{1}\text{.00}\text{fm }\left(\text{about}\text{the}\text{diameter}\text{of}\text{an}\text{atomic}\text{nucleus}\right)$ .

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 stateis calculated as,

  $\begin{array}{c}{E}_1=\frac{h^2}{8m_n{L}^2}\\ =\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.6749\times {10}^{-27}\text{kg}\right){\left(1\times {10}^{-15}\text{m}\right)}^2\left(1.602\times {10}^{-19}\text{J}\right)}\\ =204.53\times {10}^6\text{eV}\end{array}$

Solve further as,

  $\begin{array}{c}{E}_1=204.53\times {10}^6\text{eV}\\ \text{=}\left(204.53\times {10}^6\text{eV}\right)\times \frac{\left({10}^{-6}\text{MeV}\right)}{\left(1\text{eV}\right)}\\ =204.53\text{MeV}\end{array}$

The energy of first excited state is calculated as,

  $\begin{array}{c}{E}_n=n^2{E}_1\\ E_2=2^2\times 204.53\text{MeV}\\ =818\text{MeV}\end{array}$

The energy of second excited state is calculated as,

  $\begin{array}{c}{E}_n=n^2{E}_1\\ E_3=3^2\times 204.53\text{MeV}\\ =1840\text{MeV}\end{array}$

Solve further as,

  $\begin{array}{c}{E}_3=1840\text{MeV}\\ \text{=}\left(1840\text{MeV}\right)\times \frac{\left({10}^{-3}\text{GeV}\right)}{\left(1\text{MeV}\right)}\\ =1.84\text{GeV}\end{array}$

The energy level diagram of the system is shown in Figure 1.

  

Figure 1

Conclusion:

Therefore, the energy of the ground state $\left(n=1\right)$ is $\text{204}\text{.53}\text{MeV}$ and the first two excited states of a neutron in a one-dimensional box are $\text{818 MeV}$ and $\text{1}\text{.84 GeV}$ . The energy level diagram of the system is shown in Figure 1.

(b)

To determine

The wavelength of electromagnetic radiation emitted.

Answer to Problem 41P

The wavelength of electromagnetic radiation emittedis $\text{2}\text{.02}\text{fm}$ .

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,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}\\ =\frac{1240\text{eV}\cdot \text{nm}}{E_2-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{4E_1-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{3\times 204.53\text{MeV}}\end{array}$

Solve further as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{3\times 204.53\text{MeV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{3\times \left(\left(204.53\text{MeV}\right)\times \frac{\left({10}^6\text{eV}\right)}{\left(1\text{MeV}\right)}\right)}\\ =\left(\left(2.02\times {10}^{-6}\text{nm}\right)\times \frac{\left({10}^{-9}\text{fm}\right)}{\left(1\text{nm}\right)}\right)\\ =2.02\text{fm}\end{array}$

Conclusion:

Therefore, the wavelength of electromagnetic radiation emitted is $\text{2}\text{.02}\text{fm}$ .

(c)

To determine

The wavelength of electromagnetic radiation emitted.

Answer to Problem 41P

The wavelength of electromagnetic radiation emitted is $\text{1}\text{.21}\text{fm}$ .

Explanation of Solution

Given:

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

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,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}\\ =\frac{1240\text{eV}\cdot \text{nm}}{E_3-E_2}\\ =\frac{1240\text{eV}\cdot \text{nm}}{9E_1-4E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{5\times 204.53\text{MeV}}\end{array}$

Solve further as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{5\times 204.53\text{MeV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{5\times \left(\left(204.53\text{MeV}\right)\times \frac{\left({10}^6\text{eV}\right)}{\left(1\text{MeV}\right)}\right)}\\ =\left(\left(1.21\times {10}^{-6}\text{nm}\right)\times \frac{\left({10}^{-9}\text{fm}\right)}{\left(1\text{nm}\right)}\right)\\ =1.21\text{fm}\end{array}$

Conclusion:

Therefore, the wavelength of electromagnetic radiation emitted is $\text{1}\text{.21}\text{fm}$ .

(d)

To determine

The wavelength of electromagnetic radiation emitted.

Answer to Problem 41P

The wavelength of electromagnetic radiation emitted is $0.758\text{fm}$ .

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,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{\Delta E}\\ =\frac{1240\text{eV}\cdot \text{nm}}{E_3-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{9E_1-E_1}\\ =\frac{1240\text{eV}\cdot \text{nm}}{8\times 204.53\text{MeV}}\end{array}$

Solve further as,

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{3\times 204.53\text{MeV}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{8\times \left(\left(204.53\text{MeV}\right)\times \frac{\left({10}^6\text{eV}\right)}{\left(1\text{MeV}\right)}\right)}\\ =\left(\left(0.758\times {10}^{-6}\text{nm}\right)\times \frac{\left({10}^{-9}\text{fm}\right)}{\left(1\text{nm}\right)}\right)\\ =0.758\text{fm}\end{array}$

Conclusion:

Therefore, the wavelength of electromagnetic radiation emitted is $0.758\text{fm}$ .