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

Explanation

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:

1. **Understanding the transitions**:
- The electron transitions from the 3rd orbit (n=3) to the 2nd orbit (n=2) and emits a photon of wavelength λ.
- We need to find the wavelength when the electron transitions from the 4th orbit (n=4) to the 3rd orbit (n=3).

2. **Using the Rydberg formula**:
The Rydberg formula for the wavelength of emitted light during an electron transition is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)), and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.

3. **Calculating the wavelength for the first transition (3rd to 2nd)**:
For the transition from n=3 to n=2:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36} \]

4. **Calculating the wavelength for the second transition (4th to 3rd)**:
For the transition from n=4 to n=3:
\[ \frac{1}{\lambda'} = R \cdot 1^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
\[ \frac{1}{\lambda'} = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = R \cdot \frac{7}{144} \]

5. **Finding the relationship between the two wavelengths**:
We can set up the ratio of the two wavelengths:
\[ \frac{\frac{1}{\lambda}}{\frac{1}{\lambda'}} = \frac{R \cdot \frac{5}{36}}{R \cdot \frac{7}{144}} = \frac{5}{36} \cdot \frac{144}{7} = \frac{5 \cdot 4}{7} = \frac{20}{7} \]
Thus,
\[ \frac{\lambda'}{\lambda} = \frac{7}{20} \]
Therefore,
\[ \lambda' = \frac{7}{20} \lambda \]

6. **Final result**:
The wavelength of the photon emitted when the electron jumps from the 4th orbit to the 3rd orbit is:
\[ \lambda' = \frac{7}{20} \lambda \]