The most intense line in the Brackett series of the spectrum of atomic hydrogen is the transition :-

Options

A. `n_(2)=ooton_(1)=1`
B. `n_(2)=8ton_(1)=4`
C. `n_(2)=ooton_(1)=4`
D. `n_(2)=4ton_(1)=3`

Answer

To determine the most intense line in the Brackett series of the spectrum of atomic hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Series**: The hydrogen spectrum consists of several series: Lyman, Balmer, Paschen, Brackett, and Pfund. For the Brackett series, the transitions occur from higher energy levels to n=4.

2. **Define n1 and n2**: In the Brackett series, the lower energy level (n1) is 4. The upper energy level (n2) can be any integer greater than 4 (i.e., n2 = 5, 6, 7, ...).

3. **Determine the Most Intense Line**: The intensity of spectral lines is related to the energy difference between the two levels involved in the transition. The energy of the transition can be calculated using the formula:
\[ E = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant.

4. **Maximize Energy**: To find the most intense line, we need to maximize the energy difference. The energy difference is maximized when \( n_2 \) is as large as possible, specifically when \( n_2 \) approaches infinity. This is because:
\[ \frac{1}{n_2^2} \to 0 \text{ as } n_2 \to \infty \]
Therefore, the energy becomes:
\[ E = R_H \left( \frac{1}{4^2} - 0 \right) = R_H \left( \frac{1}{16} \right) \]

5. **Conclusion**: The most intense line in the Brackett series occurs when \( n_2 \) approaches infinity. Thus, the transition that corresponds to the most intense line in the Brackett series is:
\[ n_1 = 4 \text{ and } n_2 = \infty \]

### Final Answer:
The most intense line in the Brackett series of the spectrum of atomic hydrogen is the transition from \( n_2 = \infty \) to \( n_1 = 4 \).

---