Consider an atom made up of a protons and a hypothetical particle of triple the mass of electron but having same charge as electron. Apply bohr model and consider all possible transition of this hypothetical from second excited state to lower states.The possible wavelengths emitted is (are) (given in term of the Rydberg constant R for the hydrogen atom)

Options

A. `(8)/(5R)`
B. `(3)/(8R)`
C. `(4)/(9R)`
D. `(12)/(5R)`

Answer

To solve the problem, we will apply the Bohr model to an atom consisting of a proton and a hypothetical particle with a mass three times that of an electron but with the same charge as an electron. We will calculate the possible wavelengths emitted when this particle transitions from the second excited state (n=3) to lower energy states.

### Step-by-Step Solution:

1. **Understanding the System**:
- The hypothetical particle has a mass \( m' = 3m_e \) (where \( m_e \) is the mass of the electron) and a charge \( q' = -e \) (same as the electron).
- The Rydberg formula for the hydrogen atom is modified due to the mass difference.

2. **Energy Levels**:
- The energy levels in the Bohr model are given by:
\[ E_n = -\frac{m' e^4 Z^2}{2 \hbar^2 n^2} \]
- Here, \( m' = 3m_e \), so the energy levels become:
\[ E_n = -\frac{3m_e e^4 Z^2}{2 \hbar^2 n^2} \]

3. **Rydberg Constant**:
- The Rydberg constant \( R \) for hydrogen is given by:
\[ R = \frac{m_e e^4}{8 \epsilon_0^2 h^3} \]
- For our hypothetical particle, the Rydberg constant \( R' \) becomes:
\[ R' = \frac{3m_e e^4}{8 \epsilon_0^2 h^3} = 3R \]

4. **Possible Transitions**:
- The second excited state corresponds to \( n=3 \). Possible transitions to lower states are:
- \( n=3 \) to \( n=2 \)
- \( n=3 \) to \( n=1 \)
- \( n=2 \) to \( n=1 \)

5. **Calculating Wavelengths**:
- Using the formula for the wavelength emitted during a transition:
\[ \frac{1}{\lambda} = R' \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

- **Transition from \( n=3 \) to \( n=2 \)**:
\[ \frac{1}{\lambda_1} = 3R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 3R \left( \frac{1}{4} - \frac{1}{9} \right) = 3R \left( \frac{9 - 4}{36} \right) = 3R \left( \frac{5}{36} \right) = \frac{15R}{36} = \frac{5R}{12} \]
\[ \lambda_1 = \frac{12}{5R} \]

- **Transition from \( n=3 \) to \( n=1 \)**:
\[ \frac{1}{\lambda_2} = 3R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 3R \left( 1 - \frac{1}{9} \right) = 3R \left( \frac{8}{9} \right) = \frac{24R}{9} = \frac{8R}{3} \]
\[ \lambda_2 = \frac{3}{8R} \]

- **Transition from \( n=2 \) to \( n=1 \)**:
\[ \frac{1}{\lambda_3} = 3R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 3R \left( 1 - \frac{1}{4} \right) = 3R \left( \frac{3}{4} \right) = \frac{9R}{4} \]
\[ \lambda_3 = \frac{4}{9R} \]

6. **Final Results**:
- The possible wavelengths emitted are:
- \( \lambda_1 = \frac{12}{5R} \)
- \( \lambda_2 = \frac{3}{8R} \)
- \( \lambda_3 = \frac{4}{9R} \)