In hydrogen atomic spectrum , a series limit is found at `12186.3 cm^(-1)`. Then it belong to :

Options

A. Lyman series
B. Balmer series
C. Paschen series
D. Bracket series

Explanation

To determine which series the series limit of 12186.3 cm^(-1) belongs to in the hydrogen atomic spectrum, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Series Limit**:
The series limit refers to the highest energy transition in a series, which occurs when an electron transitions from a higher energy level (n2) to a lower energy level (n1). In this case, the series limit corresponds to n2 approaching infinity.

2. **Using the Rydberg Formula**:
The Rydberg formula for the hydrogen spectrum is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant (approximately \( 109677 \, \text{cm}^{-1} \)), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Setting Up the Equation**:
For the series limit, we set \( n_2 = \infty \). Thus, the equation simplifies to:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - 0 \right) \]
This means:
\[ \frac{1}{\lambda} = \frac{R_H}{n_1^2} \]

4. **Substituting the Given Value**:
We know that \( \frac{1}{\lambda} = 12186.3 \, \text{cm}^{-1} \). Now substitute this into the equation:
\[ 12186.3 = \frac{109677}{n_1^2} \]

5. **Solving for \( n_1 \)**:
Rearranging gives:
\[ n_1^2 = \frac{109677}{12186.3} \]
Calculating this:
\[ n_1^2 \approx 9.0 \quad \Rightarrow \quad n_1 \approx 3 \]

6. **Identifying the Series**:
Now that we have \( n_1 = 3 \), we can identify the series:
- Lyman series: \( n_1 = 1 \)
- Balmer series: \( n_1 = 2 \)
- Paschen series: \( n_1 = 3 \)
- Brackett series: \( n_1 = 4 \)

Since \( n_1 = 3 \), this corresponds to the Paschen series.

### Conclusion:
The series limit at \( 12186.3 \, \text{cm}^{-1} \) belongs to the **Paschen series**.