Question

Positronium consists of an electron and a positron (a particle that has the same mass as an electron, but opposite charge) orbiting around their common center of mass. Calculate the value of the Rydberg constant for this system.

Answer 1

Step 1 Rydberg constant for any atom:

In the Bohr model of an atom, for the calculation of the Rydberg constant, the mass of an atom is considered to be very large when it is compared to the electron,

The Rydberg constant is ${\mathrm{R}}_{\mathrm{H}}=1.097\times {10}^{-7}{\mathrm{m}}^{-1}$

The Rydberg constant for any atom is given as $R_m=R_H\left(1-\frac{m_e}{M}\right)$ …..(1)

Where, ${\mathrm{m}}_{\mathrm{e}}$ is the mass of the electron and $\mathrm{M}$ is the total mass of an atom.

Step 2 Rydberg constant for Positronium:

Positronium consists of an electron and a positron that has the same mass as an electron but has the opposite charge

Here,

$$\begin{gathered} \mathrm{M}={\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{e}} \\ \mathrm{M}=2{\mathrm{m}}_{\mathrm{e}} \end{gathered}$$

Substituting this in the equation (1)

$$\begin{gathered} {\mathrm{R}}_{\mathrm{m}}={\mathrm{R}}_{\mathrm{H}}\left(1-\frac{{\mathrm{m}}_{\mathrm{e}}}{2{\mathrm{m}}_{\mathrm{e}}}\right) \\ {\mathrm{R}}_{\mathrm{m}}={\mathrm{R}}_{\mathrm{H}}\left(1-\frac{1}{2}\right) \\ {\mathrm{R}}_{\mathrm{m}}={\mathrm{R}}_{\mathrm{H}}\times \frac{1}{2} \end{gathered}$$

Therefore, the value of the Rydberg constant for Positronium is ${\mathrm{R}}_{\mathrm{M}}=\frac{{\mathrm{R}}_{\mathrm{H}}}{2}$.