The wave number of the first emission line in the Balmer series of H - Spectrum is `(n)/(36)R`. The value of 'n' is. (R = Rydberg constant):

Answer

To solve the problem, we need to find the value of 'n' in the equation for the wave number of the first emission line in the Balmer series of the hydrogen spectrum, given as \( \frac{n}{36}R \).

### Step-by-Step Solution:

1. **Understand the Balmer Series**:
The Balmer series corresponds to transitions of electrons in hydrogen from higher energy levels (n2) to the second energy level (n1 = 2). The first emission line corresponds to the transition from n2 = 3 to n1 = 2.

2. **Use the Formula for Wave Number**:
The wave number (\( \bar{v} \)) for the hydrogen spectrum can be calculated using the formula:
\[ \bar{v} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (which is 1 for hydrogen), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Substituting Values**:
For the first emission line in the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \)
- \( Z = 1 \)

Substitute these values into the wave number formula:
\[ \bar{v} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
This simplifies to:
\[ \bar{v} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

4. **Finding a Common Denominator**:
To subtract the fractions, find a common denominator, which is 36:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
Therefore:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

5. **Final Expression for Wave Number**:
Now substituting back into the wave number equation:
\[ \bar{v} = R \cdot \frac{5}{36} \]

6. **Comparing with Given Expression**:
The problem states that the wave number is given as \( \frac{n}{36}R \). By comparing both expressions, we have:
\[ \frac{n}{36}R = R \cdot \frac{5}{36} \]
Since \( R \) and \( 36 \) are common in both sides, we can simplify:
\[ n = 5 \]

### Conclusion:
The value of \( n \) is **5**.