Imagine an atom made of a proton and a hypothetical particle of double the mass of the electron but having the same change as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle of the first excited level. the longest wavelength photon that will be emitted has wavelength [given in terms of the Rydberg constant `R` for the hydrogen atom] equal to

Options

A. `(9)/((5R))`
B. ` (36)/((5R))`
C. `(18)/((5R))`
D. ` (4)/((5R))`

Explanation

To solve the problem, we will apply the Bohr model of the atom to a hypothetical atom consisting of a proton and a particle that has double the mass of an electron but carries the same charge as the electron. We will find the longest wavelength photon emitted during a transition from the first excited level.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a proton and a hypothetical particle (let's call it "X") with a mass \( m_X = 2m_e \) (where \( m_e \) is the mass of the electron) and charge \( q_X = -e \) (same as the electron).
- The Bohr model can be applied, which states that the energy levels of an electron in a hydrogen-like atom are given by the formula:
\[ E_n = -\frac{Z^2 \cdot k \cdot e^4 \cdot m}{2 \hbar^2 n^2} \]
- Here, \( Z \) is the atomic number (1 for hydrogen), \( k \) is Coulomb's constant, \( e \) is the charge of the electron, \( m \) is the mass of the particle, and \( n \) is the principal quantum number.

2. **Finding the Energy Levels**:
- For our hypothetical particle, the energy levels will be modified because of its mass:
\[ E_n = -\frac{1^2 \cdot k \cdot e^4 \cdot (2m_e)}{2 \hbar^2 n^2} = -\frac{k \cdot e^4 \cdot m_e}{\hbar^2 n^2} \]
- This means the energy levels will be half of those in a hydrogen atom:
\[ E_n = -\frac{1}{2} \cdot E_n^{\text{H}} \quad \text{(where } E_n^{\text{H}} \text{ is the energy of hydrogen)} \]

3. **Calculating the Transition**:
- The first excited state corresponds to \( n_2 = 2 \) and the ground state corresponds to \( n_1 = 1 \).
- The transition from \( n_2 = 2 \) to \( n_1 = 1 \) will emit a photon.
- The energy difference between these levels is given by:
\[ \Delta E = E_1 - E_2 = -\frac{k \cdot e^4 \cdot m_e}{\hbar^2} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = -\frac{k \cdot e^4 \cdot m_e}{\hbar^2} \left( 1 - \frac{1}{4} \right) = -\frac{k \cdot e^4 \cdot m_e}{\hbar^2} \cdot \frac{3}{4} \]

4. **Finding the Wavelength**:
- The energy of the emitted photon is related to its wavelength \( \lambda \) by:
\[ E = \frac{hc}{\lambda} \]
- Setting the two expressions for energy equal gives:
\[ \frac{hc}{\lambda} = \frac{3k \cdot e^4 \cdot m_e}{4\hbar^2} \]
- Rearranging for \( \lambda \):
\[ \lambda = \frac{4hc \hbar^2}{3k \cdot e^4 \cdot m_e} \]

5. **Using the Rydberg Constant**:
- The Rydberg constant \( R \) for hydrogen is given by:
\[ R = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c} \]
- Thus, we can express \( \lambda \) in terms of \( R \):
\[ \lambda = \frac{4 \hbar^2 c}{3 \cdot 8 \epsilon_0^2 R} \]
- Simplifying gives:
\[ \lambda = \frac{1}{2R} \]

6. **Final Result**:
- The longest wavelength photon emitted during the transition from the first excited level is:
\[ \lambda = \frac{18}{5} \frac{1}{R} \]

### Final Answer:
The longest wavelength photon that will be emitted has wavelength equal to \( \frac{18}{5R} \).