If an electron in a hydrogen atom jumps from the `3rd` orbit to the `2nd` orbit, it emits a photon of wavelength `lambda`. When it jumps form the `4th` orbit to the `3dr` orbit, the corresponding wavelength of the photon will be

Options

A. `(16)/(25)lambda`
B. `(9)/(16)lambda`
C. `(20)/(7)lambda`
D. `(20)/(13)lambda`

Answer

To solve the problem of determining the wavelength of the photon emitted when an electron in a hydrogen atom jumps from the 4th orbit to the 3rd orbit, we can use the Rydberg formula for hydrogen. The Rydberg formula is given by:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( \lambda \) is the wavelength of the emitted photon,
- \( R_H \) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step-by-Step Solution:

1. **Identify the transitions**:
- For the transition from the 3rd orbit to the 2nd orbit, we have \( n_2 = 3 \) and \( n_1 = 2 \).
- For the transition from the 4th orbit to the 3rd orbit, we have \( n_2 = 4 \) and \( n_1 = 3 \).

2. **Calculate the wavelength for the first transition (3rd to 2nd)**:
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_1 = \frac{36}{5 R_H} \]

3. **Calculate the wavelength for the second transition (4th to 3rd)**:
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
\[ = R_H \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = R_H \left( \frac{16 - 9}{144} \right) = R_H \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_2 = \frac{144}{7 R_H} \]

4. **Find the ratio of the wavelengths**:
To find the wavelength for the second transition in terms of the first transition, we can take the ratio:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{144}{7 R_H}}{\frac{36}{5 R_H}} = \frac{144 \times 5}{36 \times 7} = \frac{720}{252} = \frac{120}{42} = \frac{20}{7} \]

5. **Conclusion**:
The wavelength of the photon emitted when the electron jumps from the 4th orbit to the 3rd orbit is:
\[ \lambda_2 = \frac{144}{7 R_H} \]