In an ordianary atom, as a first approximation, the motion of the nucleus can be ignored. In a positronium atom a positronreplaces the proton of hydrogen atom. The electron and positron masses are equal and , therefore , the motion of the positron cannot be ignored. One must consider the motion of both electron and positron about their center of mass. A detailed analyasis shows that formulae of Bohr's model apply to positronium atom provided that we replace `m_(e)` by what is known reduced mass is `m_(e)//2`.
When system de-excites from its first excited state to ground state, the wavelngth of radiation is

Options

A. `1217 Å`
B. `2431 Å`
C. `608 Å`
D. none of these

Explanation

To solve the problem of finding the wavelength of radiation emitted when the positronium atom de-excites from its first excited state to the ground state, we can follow these steps:

### Step 1: Understand the System
In a positronium atom, we have an electron and a positron (the antiparticle of the electron). The masses of both are equal, and thus we need to use the concept of reduced mass to analyze the system.

### Step 2: Calculate the Reduced Mass
The reduced mass \( \mu \) for the positronium system is given by:
\[ \mu = \frac{m_e \cdot m_{e^+}}{m_e + m_{e^+}} = \frac{m_e \cdot m_e}{m_e + m_e} = \frac{m_e^2}{2m_e} = \frac{m_e}{2} \]
where \( m_e \) is the mass of the electron (and positron).

### Step 3: Use Bohr's Model Formula
The wavelength of the emitted radiation when the system transitions from an excited state to a lower energy state can be determined using the formula derived from Bohr's model:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant modified for positronium, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy states, respectively.

### Step 4: Determine the Rydberg Constant for Positronium
The Rydberg constant for positronium can be calculated as:
\[ R_{Ps} = \frac{R_H}{\mu/m_e} = \frac{R_H}{1/2} = 2R_H \]
where \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).

### Step 5: Substitute Values for the Transition
For the transition from the first excited state (\( n_2 = 2 \)) to the ground state (\( n_1 = 1 \)):
\[ \frac{1}{\lambda} = 2R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 2R_H \left( 1 - \frac{1}{4} \right) = 2R_H \left( \frac{3}{4} \right) = \frac{3R_H}{2} \]

### Step 6: Calculate the Wavelength
Now, substituting the value of \( R_H \):
\[ \frac{1}{\lambda} = \frac{3 \times 1.097 \times 10^7}{2} \]
\[ \lambda = \frac{2}{3 \times 1.097 \times 10^7} \]
Calculating this gives:
\[ \lambda \approx \frac{2}{3.291 \times 10^7} \approx 6.07 \times 10^{-8} \, \text{m} = 607 \, \text{nm} \]

### Final Answer
The wavelength of the radiation emitted when the positronium atom de-excites from its first excited state to the ground state is approximately **607 nm**.