Electromagnetic radiation (photon) with highest wavelength result when an electron in the hydrogen atom falls from" `n=6` to :

Options

A. (a) `n=1`
B. (b) `n=2`
C. (c) `n=3`
D. (d) `n=5`

Answer

To determine the transition of an electron in a hydrogen atom that results in the highest wavelength of electromagnetic radiation (photon) when falling from \( n = 6 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Wavelength and Energy Relationship**:
The wavelength (\( \lambda \)) of the emitted photon is inversely related to the energy difference between the two energy levels. The formula relating these quantities is given by the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Identify the Transition**:
Since the electron is falling from \( n = 6 \), we need to find which lower energy level \( n_1 \) will yield the maximum wavelength (minimum energy difference). The possible values for \( n_1 \) are 1, 2, 3, 4, and 5.

3. **Calculate the Wavelength for Each Transition**:
We will calculate \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \) for each possible \( n_1 \):
- For \( n_1 = 1 \):
\[ \frac{1}{1^2} - \frac{1}{6^2} = 1 - \frac{1}{36} = \frac{36 - 1}{36} = \frac{35}{36} \]
- For \( n_1 = 2 \):
\[ \frac{1}{2^2} - \frac{1}{6^2} = \frac{1}{4} - \frac{1}{36} = \frac{9 - 1}{36} = \frac{8}{36} = \frac{2}{9} \]
- For \( n_1 = 3 \):
\[ \frac{1}{3^2} - \frac{1}{6^2} = \frac{1}{9} - \frac{1}{36} = \frac{4 - 1}{36} = \frac{3}{36} = \frac{1}{12} \]
- For \( n_1 = 4 \):
\[ \frac{1}{4^2} - \frac{1}{6^2} = \frac{1}{16} - \frac{1}{36} = \frac{9 - 4}{144} = \frac{5}{144} \]
- For \( n_1 = 5 \):
\[ \frac{1}{5^2} - \frac{1}{6^2} = \frac{1}{25} - \frac{1}{36} = \frac{36 - 25}{900} = \frac{11}{900} \]

4. **Determine the Maximum Wavelength**:
The maximum wavelength corresponds to the minimum value of \( \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \). From our calculations:
- \( n_1 = 1 \): \( \frac{35}{36} \)
- \( n_1 = 2 \): \( \frac{2}{9} \)
- \( n_1 = 3 \): \( \frac{1}{12} \)
- \( n_1 = 4 \): \( \frac{5}{144} \)
- \( n_1 = 5 \): \( \frac{11}{900} \)

The smallest value occurs when \( n_1 = 5 \).

5. **Conclusion**:
Therefore, the transition that results in the highest wavelength when an electron falls from \( n = 6 \) is to \( n = 5 \).

### Final Answer:
The electron in the hydrogen atom falls from \( n = 6 \) to \( n = 5 \) to produce the photon with the highest wavelength.