Positronium consists of an electron and a positron (a particle which has the same mass as an electron, but opposite charge) orbiting round their common centre of mass. Calculate the value of the Rydberg constant `(R_(oo))` for this system

Options

A. `R_(oo)//4`
B. `R_(oo)//2`
C. `2R_(oo)`
D. `R_(oo)`

Explanation

To calculate the value of the Rydberg constant \( R \) for positronium, which consists of an electron and a positron, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
Positronium is a bound state of an electron and a positron. Both particles have the same mass \( m \) but opposite charges.

2. **Define the Reduced Mass**:
The reduced mass \( \mu \) of the system is given by the formula:
\[ \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \]
In this case, both \( m_1 \) (mass of the electron) and \( m_2 \) (mass of the positron) are equal to \( m \):
\[ \mu = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2} \]

3. **Rydberg Constant Relation**:
The Rydberg constant for hydrogen-like systems is given by:
\[ R = R_{\infty} \cdot \frac{\mu}{m_e} \]
where \( R_{\infty} \) is the Rydberg constant for an electron in hydrogen, and \( m_e \) is the mass of the electron.

4. **Substituting the Reduced Mass**:
Since the reduced mass \( \mu \) for positronium is \( \frac{m}{2} \), we substitute this into the equation:
\[ R = R_{\infty} \cdot \frac{\frac{m}{2}}{m} = R_{\infty} \cdot \frac{1}{2} \]

5. **Final Expression**:
Thus, the Rydberg constant for positronium is:
\[ R = \frac{R_{\infty}}{2} \]

### Conclusion:
The value of the Rydberg constant \( R \) for the positronium system is \( \frac{R_{\infty}}{2} \).