The wavelength for n=3 to n=2 transition of the hydrogen atom is 656.3 nm. What are the wavelength for this same transition in (a) positronium, which consists of an electron and a positron (b) singly ionized helium

Explanation

To find the wavelengths for the n=3 to n=2 transition in positronium and singly ionized helium, we can follow these steps:

### Step 1: Understand the Transition and Given Wavelength
We know that the wavelength for the n=3 to n=2 transition in hydrogen is given as 656.3 nm.

### Step 2: Wavelength in Positronium
Positronium is a bound state of an electron and a positron. The mass of the positron is equal to the mass of the electron, so we can calculate the reduced mass (\( \mu \)) of positronium as follows:

\[ \mu = \frac{m_e \cdot m_p}{m_e + m_p} = \frac{m_e \cdot m_e}{m_e + m_e} = \frac{m_e}{2} \]

Since the reduced mass is half of the electron's mass, the energy levels will be affected. The energy is directly proportional to the reduced mass, which means that the wavelength is inversely proportional to the reduced mass.

Thus, if the reduced mass is halved, the wavelength will double:

\[ \lambda_{positronium} = 2 \times \lambda_{hydrogen} = 2 \times 656.3 \, \text{nm} = 1312.6 \, \text{nm} \]

### Step 3: Wavelength in Singly Ionized Helium
For singly ionized helium (He\(^+\)), the atomic number \( Z \) is 2. The wavelength for the transition can be calculated using the formula:

\[ \lambda_{He^+} = \frac{\lambda_{H}}{Z^2} \]

Substituting the values:

\[ \lambda_{He^+} = \frac{656.3 \, \text{nm}}{2^2} = \frac{656.3 \, \text{nm}}{4} = 164.075 \, \text{nm} \]

### Final Results
- Wavelength in positronium: **1312.6 nm**
- Wavelength in singly ionized helium: **164.075 nm**