The wave number of first emission line in the atomic spectrum of hydrogen in the Balmer series is

Options

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

Answer

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

### Step 1: Understand the Balmer series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The first line in the Balmer series corresponds to the transition from n = 3 to n = 2.

### Step 2: Use the formula for wave number
The wave number (\( \bar{\nu} \)) can be calculated using the formula:
\[ \bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \) or \( 1.097 \times 10^2 \, \text{cm}^{-1} \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Assign values for the first line in the Balmer series
For the first emission line in the Balmer series:
- \( n_1 = 2 \) (the lower energy level),
- \( n_2 = 3 \) (the higher energy level).

### Step 4: Substitute values into the formula
Now, substituting the values into the formula:
\[ \bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the squares:
\[ \bar{\nu} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 5: Find a common denominator and simplify
The common denominator for 4 and 9 is 36:
\[ \bar{\nu} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 6: Final expression for wave number
Thus, the wave number of the first emission line in the Balmer series is:
\[ \bar{\nu} = \frac{5R}{36} \, \text{cm}^{-1} \]

### Conclusion
The final answer is:
\[ \bar{\nu} = \frac{5R}{36} \, \text{cm}^{-1} \]