An electron in hydrogen atom first jumps from second excited state to first excited state and then from 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. c=1/a
B. a=9/4
C. b=5/27
D. c=5/27

Answer

To solve the problem, we need to analyze the transitions of an electron in a hydrogen atom from the second excited state to the first excited state and then from the first excited state to the ground state. We will find the ratios of the wavelength, momentum, and energy of the emitted photons in these two cases.

### Step-by-Step Solution:

1. **Identify the Energy Levels:**
- The second excited state corresponds to \( n = 3 \).
- The first excited state corresponds to \( n = 2 \).
- The ground state corresponds to \( n = 1 \).

2. **Calculate the Energy Change for Each Transition:**
- The energy of a level in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
- For the transition from \( n = 3 \) to \( n = 2 \):
\[ \Delta E_1 = E_2 - E_3 = \left(-\frac{13.6}{2^2}\right) - \left(-\frac{13.6}{3^2}\right) = -\frac{13.6}{4} + \frac{13.6}{9} \]
\[ \Delta E_1 = -\frac{13.6}{4} + \frac{13.6}{9} = 13.6 \left(-\frac{9}{36} + \frac{4}{36}\right) = 13.6 \left(-\frac{5}{36}\right) = -\frac{68}{36} \, \text{eV} \]

- For the transition from \( n = 2 \) to \( n = 1 \):
\[ \Delta E_2 = E_1 - E_2 = \left(-\frac{13.6}{1^2}\right) - \left(-\frac{13.6}{2^2}\right) = -13.6 + \frac{13.6}{4} \]
\[ \Delta E_2 = -13.6 + 3.4 = -10.2 \, \text{eV} \]

3. **Calculate the Ratios of Energy:**
- The ratio of the energies of the emitted photons:
\[ C = \frac{\Delta E_1}{\Delta E_2} = \frac{-\frac{68}{36}}{-10.2} = \frac{68}{36 \times 10.2} \]

4. **Calculate the Wavelengths:**
- The energy of a photon is related to its wavelength by:
\[ E = \frac{hc}{\lambda} \]
- Rearranging gives:
\[ \lambda = \frac{hc}{E} \]
- Therefore, the ratio of the wavelengths is the inverse of the ratio of the energies:
\[ A = \frac{\lambda_2}{\lambda_1} = \frac{\Delta E_1}{\Delta E_2} \]

5. **Calculate the Momentum Ratios:**
- The momentum \( p \) 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:
\[ B = \frac{p_1}{p_2} = \frac{\Delta E_1}{\Delta E_2} \]

6. **Final Ratios:**
- We have:
- \( A = \frac{\Delta E_2}{\Delta E_1} \)
- \( B = \frac{\Delta E_1}{\Delta E_2} \)
- \( C = \frac{\Delta E_1}{\Delta E_2} \)

### Summary of Ratios:
- \( A = \frac{5}{27} \)
- \( B = \frac{5}{27} \)
- \( C = \frac{1}{A} \)