The positronium consisting of an electron and a positron corresponding binding energy and wavelength of first line in Lyman series of such a system.

Explanation

To find the binding energy and the wavelength of the first line in the Lyman series of positronium (which consists of an electron and a positron), we can follow these steps:

### Step 1: Determine the Reduced Mass
The reduced mass (\( \mu \)) of the positronium system can be calculated using the formula:
\[ \mu = \frac{m_e \cdot m_p}{m_e + m_p} \]
where \( m_e \) is the mass of the electron and \( m_p \) is the mass of the positron. Since the mass of the positron is equal to the mass of the electron, we have:
\[ \mu = \frac{m_e \cdot m_e}{m_e + m_e} = \frac{m_e}{2} \]

### Step 2: Calculate the Binding Energy
The binding energy (\( E_1 \)) for the first energy level of positronium can be derived from the hydrogen-like atom formula:
\[ E_n = -\frac{\mu e^4}{2 \hbar^2 n^2} \]
For positronium, at the first level (\( n = 1 \)):
\[ E_1 = -\frac{\frac{m_e}{2} e^4}{2 \hbar^2} = -\frac{1}{2} \cdot 13.6 \, \text{eV} = -6.8 \, \text{eV} \]
Thus, the binding energy of positronium is \( 6.8 \, \text{eV} \).

### Step 3: Calculate the Wavelength of the First Line in the Lyman Series
The wavelength (\( \lambda \)) of the first line in the Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \). The Rydberg formula for the wavelength is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant for positronium, which is half of that for hydrogen:
\[ R = \frac{1}{2} \cdot R_H = \frac{1}{2} \cdot 1.097 \times 10^7 \, \text{m}^{-1} = 5.485 \times 10^6 \, \text{m}^{-1} \]
For the transition \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Substituting the value of \( R \):
\[ \frac{1}{\lambda} = 5.485 \times 10^6 \cdot \frac{3}{4} = 4.114 \times 10^6 \, \text{m}^{-1} \]
Thus, the wavelength \( \lambda \) is:
\[ \lambda = \frac{1}{4.114 \times 10^6} \approx 2.43 \times 10^{-7} \, \text{m} = 2430 \, \text{Å} \]

### Final Answers
- **Binding Energy**: \( 6.8 \, \text{eV} \)
- **Wavelength of the First Line in Lyman Series**: \( 2430 \, \text{Å} \)