The first emission line of Balmer series in H spectrum has the wave number equal to :

Options

A. `(9R)/(400) cm^(-1)`
B. `(7R)/(144) cm^(-1)`
C. `(3R)/(4) cm^(-1)`
D. `(5R)/(36) cm^(-1)`

Explanation

To find the wave number of the first emission line of the Balmer series in the hydrogen spectrum, we can follow these steps:

### Step 1: Understand the Formula
The wave number (ν̅) is given by the formula:
\[ \nu̅ = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant for hydrogen,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 2: Identify the Values of n
For the Balmer series, the lower energy level \( n_1 \) is 2. The first emission line corresponds to \( n_2 = 3 \) (the first transition from n=3 to n=2).

### Step 3: Substitute the Values into the Formula
Substituting \( n_1 = 2 \) and \( n_2 = 3 \) into the formula:
\[ \nu̅ = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
This simplifies to:
\[ \nu̅ = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 4: Calculate the Difference
To calculate \( \frac{1}{4} - \frac{1}{9} \), we need a common denominator:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
Thus,
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 5: Substitute Back into the Wave Number Formula
Now substituting back into the wave number formula:
\[ \nu̅ = R_H \cdot \frac{5}{36} \]

### Step 6: Use the Value of Rydberg Constant
The Rydberg constant \( R_H \) is approximately \( 109677 \, \text{cm}^{-1} \). Thus,
\[ \nu̅ = 109677 \cdot \frac{5}{36} \]

### Step 7: Calculate the Final Value
Calculating the final value:
\[ \nu̅ = \frac{548385}{36} \approx 15232.5 \, \text{cm}^{-1} \]

### Conclusion
The wave number for the first emission line of the Balmer series in the hydrogen spectrum is approximately \( 15232.5 \, \text{cm}^{-1} \).