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 solve the problem of finding the wavelengths for the n=3 to n=2 transition in positronium and singly ionized helium, we will follow these steps:

### Step 1: Understand the Transition in Hydrogen
We are given that the wavelength for the transition from n=3 to n=2 in hydrogen is 656.3 nm. This can be expressed using the Rydberg formula for hydrogen:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
where \( R_H \) is the Rydberg constant for hydrogen, \( n_1 = 2 \), and \( n_2 = 3 \).

### Step 2: Calculate the Wavelength for Positronium
Positronium is a bound state of an electron and a positron. The effective mass of the system is half the mass of an electron:
\[ m_{positronium} = \frac{m_e}{2} \]
The Rydberg constant for positronium \( R_P \) can be derived from the Rydberg constant for hydrogen:
\[ R_P = \frac{R_H}{2} \]
Now, we can use the Rydberg formula for positronium:
\[ \frac{1}{\lambda_P} = R_P \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting \( R_P \):
\[ \frac{1}{\lambda_P} = \frac{R_H}{2} \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the difference:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus:
\[ \frac{1}{\lambda_P} = \frac{R_H}{2} \cdot \frac{5}{36} \]
Now substituting \( R_H = \frac{1}{656.3 \, \text{nm}} \):
\[ \lambda_P = \frac{656.3 \, \text{nm}}{2} \cdot \frac{36}{5} = \frac{656.3 \cdot 36}{10} \, \text{nm} \]
Calculating this gives:
\[ \lambda_P = 1312.6 \, \text{nm} \]

### Step 3: Calculate the Wavelength for Singly Ionized Helium
For singly ionized helium (He\(^+\)), the atomic number \( Z = 2 \). The Rydberg formula for helium will be:
\[ \frac{1}{\lambda_{He}} = R_H Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting \( Z = 2 \):
\[ \frac{1}{\lambda_{He}} = R_H \cdot 4 \cdot \frac{5}{36} \]
Thus:
\[ \lambda_{He} = \frac{656.3 \, \text{nm}}{4} = 164.075 \, \text{nm} \]

### Final Answers
- Wavelength for positronium: \( \lambda_P = 1312.6 \, \text{nm} \)
- Wavelength for singly ionized helium: \( \lambda_{He} = 164.075 \, \text{nm} \)