An electron in hydrogen atom first jumps form second excited state to first excited state and then form first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively, Then

Options

A. `a=9/4`
B. `b=5/27`
C. `c=5/27`
D. `c=1/a`

Explanation

To solve the problem step by step, we need to analyze the transitions of the electron in a hydrogen atom and calculate the ratios of the wavelength, momentum, and energy of the emitted photons during these transitions.

### Step 1: Identify the energy levels
The electron transitions are as follows:
1. From the second excited state (n = 3) to the first excited state (n = 2).
2. From the first excited state (n = 2) to the ground state (n = 1).

### Step 2: Calculate the energy of the transitions
The energy of the photon emitted during a transition from a higher energy level (n_i) to a lower energy level (n_f) in a hydrogen atom can be calculated using the formula:
\[ E = -13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV} \]

#### For the first transition (n = 3 to n = 2):
\[ E_1 = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = -13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the fractions:
\[ E_1 = -13.6 \left( \frac{9 - 4}{36} \right) = -13.6 \left( \frac{5}{36} \right) = -\frac{68}{36} \text{ eV} = -1.89 \text{ eV} \]

#### For the second transition (n = 2 to n = 1):
\[ E_2 = -13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = -13.6 \left( 1 - \frac{1}{4} \right) = -13.6 \left( \frac{3}{4} \right) = -10.2 \text{ eV} \]

### Step 3: Calculate the ratios of energy
The ratio of the energies of the emitted photons is:
\[ C = \frac{E_1}{E_2} = \frac{-1.89}{-10.2} = \frac{1.89}{10.2} = \frac{5}{27} \]

### Step 4: Calculate the momentum of the photons
The momentum of a photon is given by:
\[ P = \frac{E}{c} \]
Thus, the ratio of the momenta is the same as the ratio of the energies:
\[ \frac{P_1}{P_2} = \frac{E_1}{E_2} = C = \frac{5}{27} \]

### Step 5: Calculate the wavelengths of the emitted photons
The wavelength of a photon is related to its energy by:
\[ \lambda = \frac{hc}{E} \]
Thus, the ratio of the wavelengths is:
\[ A = \frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1} = \frac{-10.2}{-1.89} = \frac{10.2}{1.89} = \frac{27}{5} \]

### Step 6: Summarize the ratios
- \( A = \frac{27}{5} \)
- \( B = \frac{5}{27} \)
- \( C = \frac{5}{27} \)

### Conclusion
The ratios of the wavelength, momentum, and energy of the photons emitted during the transitions are:
- \( A = \frac{27}{5} \)
- \( B = \frac{5}{27} \)
- \( C = \frac{5}{27} \)