Chapter 28, Problem 55A

To determine

To calculate: Wavelength of $E_2$ to $E_1$ transition describing the orbital radii and energy changes at each level.

Answer to Problem 55A

Wavelength of $E_2$ to $E_1$ transitionof thepositronium atom = $\underset{\_}{\text{244 nm}}$

Explanation of Solution

Given:

The mass of the electron is replaced by one half of its mass in the Bohr model.

Formula used:

The electron orbital radius in hydrogen is given by,

  $r=\frac{h^2{n}^2}{4{\pi }^{2}Km{q}^2}$

Where, $h$ is the Planck’s constant, $K$ is a constant from Coulomb’s Law and has a value of $9.0\times {10}^9{\text{ Nm}}^2/{\text{C}}^2$ , $m$ is the mass of electron and $q$ is the charge of an electron.

Energy of an atom is given by,

  $E=\frac{-2\pi K^{2}m{q}^4}{h^2}$

Where, $h$ is the Planck’s constant, $K$ is a constant from Coulomb’s Law and has a value of $9.0\times {10}^9{\text{ Nm}}^2/{\text{C}}^2$ , $m$ is the mass of electron and $q$ is the charge of an electron.

Energy of a hydrogen is given by,

  $E_n=-13.6\text{ eV}\times \frac{1}{n^2}$

Where $n$ is principal quantum number.

Energy of every photon is given by,

  $E=\frac{hc}{\lambda }$

Where $h$ is Planck’s constant, $c$ is speed of light and $\lambda$ is wavelength.

Calculation:

The positronium atom’s orbital radius is calculated as:

  $\begin{array}{l}{r}_p=\frac{h^2{n}^2}{4{\pi }^{2}Km{q}^2}\\ \text{ =}\frac{h^2{n}^2}{4{\pi }^{2}K(\frac{m}{2})q^2}\\ \text{ =2}\cdot \frac{h^2{n}^2}{4{\pi }^{2}Km{q}^2}\\ \text{ =2r}\end{array}$

Hence the radii will be twice as large.

Energy level of the positronium can be calculated a:

  $\begin{array}{l}{E}_p=\frac{-2\pi K^{2}m{q}^4}{h^2}\\ \text{ =}\frac{-2\pi K^2(\frac{m}{2})q^4}{h^2}\\ \text{ =}\frac{-\pi K^{2}m{q}^4}{h^2}\\ \text{ =}\frac{1}{2}E\end{array}$

Hence the energy level will be half as large.

Energy emitted from $E_2$ to $E_1$ transition is calculated as:

  $E=\frac{hc}{\lambda }$

Since energy of the positronium atom is half as large,

  $\begin{array}{l}{E}_p=\frac{1}{2}E\\ \frac{hc}{{\lambda }_p}=\frac{1}{2}\cdot \frac{hc}{\lambda }\\ {\lambda }_p=2\lambda \end{array}$

Energy change from $E_2$ to $E_1$ transition of hydrogen atom is calculated as:

  $\begin{array}{l}E=E_2-E_1\\ \text{ =}-\text{13}\text{.6 eV}\times \frac{1}{{(2)}^2}-(-13.6\text{ eV}\times \frac{1}{1^2})\\ \text{ =}-\text{13}\text{.6 eV}\left(\frac{1}{4}-\frac{1}{1}\right)\\ \text{ =}-\text{13}\text{.6 eV}\times (-0.75)\\ \text{ =10}\text{.2 eV}\\ \end{array}$

Wavelength of hydrogen atom is calculated as:

  $\begin{array}{l}\lambda \text{=}\frac{hc}{E}\\ =\frac{1240\text{ eV-nm}}{10.2\text{ eV}}\text{ [hc=1240 eV-nm]}\\ =121.5\text{ nm}\\ \simeq 122\text{ nm}\end{array}$

Since the wavelength of the positronium atom is twice as large as the hydrogen atom.

  $\begin{array}{l}{\lambda }_p=2\lambda \\ \text{ =2}\cdot \text{122}\\ \text{ =244 nm}\end{array}$

Conclusion:

Wavelength of $E_2$ to $E_1$ transitionof thepositronium atom = $\underset{\_}{\text{244 nm}}$