Which of the following links lines of the H-atom spectrum belongs to the Balmer series ?

Options

A. `1025Å`
B. `1218Å`
C. `4861Å`
D. `18751Å`

Answer

To determine which line of the hydrogen atom spectrum belongs to the Balmer series, we need to understand the characteristics of the Balmer series and apply the Rydberg formula.

### Step-by-Step Solution:

1. **Understand the Balmer Series**:
The Balmer series corresponds to electronic transitions in a hydrogen atom where the electron falls to the n=2 energy level from higher energy levels (n=3, 4, 5, ...).

2. **Rydberg Formula**:
The wavelength (λ) of the emitted light during these transitions can be calculated using 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, \( n_1 = 2 \) for the Balmer series, and \( n_2 \) can be any integer greater than 2.

3. **Set \( n_1 \) and Calculate for Different \( n_2 \)**:
For the Balmer series, set \( n_1 = 2 \) and calculate for \( n_2 = 3, 4, 5, ... \):
- For \( n_2 = 3 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
- For \( n_2 = 4 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) \]
- Continue this for higher values of \( n_2 \).

4. **Calculate Specific Wavelengths**:
Using the Rydberg constant, we can calculate specific wavelengths for transitions:
- For \( n_2 = 3 \):
\[ \lambda_{3 \to 2} \approx 6561 \, \text{Å} \]
- For \( n_2 = 4 \):
\[ \lambda_{4 \to 2} \approx 4861 \, \text{Å} \]
- For \( n_2 = 5 \):
\[ \lambda_{5 \to 2} \approx 4340 \, \text{Å} \]
- For \( n_2 = 6 \):
\[ \lambda_{6 \to 2} \approx 4102 \, \text{Å} \]

5. **Identify the Wavelengths**:
The wavelengths calculated above correspond to the visible spectrum of the hydrogen atom. The line that falls within the visible range and corresponds to the Balmer series is:
- \( 4861 \, \text{Å} \) (for the transition from \( n=4 \) to \( n=2 \)).

6. **Conclusion**:
From the calculations, the correct answer is the wavelength corresponding to the transition from \( n=4 \) to \( n=2 \), which is \( 4861 \, \text{Å} \).

### Final Answer:
The line of the hydrogen atom spectrum that belongs to the Balmer series is **4861 Å**.