In the emission spectrum of `H-` atom from energy level `'n'` to ground state in one more step, no line belonging to the Brackett series is observed. The wave number of lines belonging to Balmer series may be

Options

A. `(8R)/(9),(5R)/(36)`
B. `(3R)/(16),(8R)/(9)`
C. `(5R)/(36),(3R)/(16)`
D. `(3R)/(4),(3R)/(16)`

Explanation

To solve the question regarding the emission spectrum of the H- atom and the wave numbers of lines belonging to the Balmer series, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Series**:
- The Balmer series corresponds to transitions where the final energy level (n1) is 2. The transitions can occur from n = 3, 4, 5, etc., to n = 2.

2. **Formula for Wave Number**:
- The wave number (ṽ) is given by the formula:
\[ ṽ = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, R is the Rydberg constant, \( n_1 \) is the lower energy level (2 for Balmer series), and \( n_2 \) is the higher energy level (3, 4, etc.).

3. **Calculate Wave Number for Transition from n=3 to n=2**:
- For the transition from n=3 to n=2:
\[ ṽ = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Finding a common denominator (36):
\[ ṽ = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

4. **Calculate Wave Number for Transition from n=4 to n=2**:
- For the transition from n=4 to n=2:
\[ ṽ = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
- Finding a common denominator (16):
\[ ṽ = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]

5. **Conclusion**:
- The wave numbers for the transitions in the Balmer series from n=3 to n=2 and from n=4 to n=2 are:
- From n=3 to n=2: \( ṽ = \frac{5R}{36} \)
- From n=4 to n=2: \( ṽ = \frac{3R}{16} \)

### Final Answer:
The wave numbers of lines belonging to the Balmer series may be \( \frac{5R}{36} \) and \( \frac{3R}{16} \).