The first line in the Balmer series in the H atom will have the frequency

Options

A. `4.57 xx 10^(14)s^(-1)`
B. `3.29 xx 10^(15) s^(-1)`
C. `8.22 xx 10^(15) s^(-1)`
D. `8.05 xx 10^(13) s^(-1)`

Explanation

To find the frequency of the first line in the Balmer series of the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Balmer Series**:
The Balmer series corresponds to electronic transitions in the hydrogen atom where the electron falls to the n=2 energy level from higher energy levels (n=3, 4, 5,...).

2. **Identify the Transition**:
The first line in the Balmer series corresponds to the transition from n=3 to n=2. Here, n1 = 2 and n2 = 3.

3. **Use the Rydberg Formula**:
The Rydberg formula for the frequency of emitted light is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

4. **Convert to Frequency**:
We know that frequency \( \nu \) is related to wavelength \( \lambda \) by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Therefore, we can express the frequency in terms of the Rydberg formula:
\[ \nu = R \cdot c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

5. **Substitute Values**:
- Rydberg constant \( R = 1.097 \times 10^7 \, \text{m}^{-1} \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)
- \( n_1 = 2 \), \( n_2 = 3 \)

Now substituting these values into the equation:
\[ \nu = 1.097 \times 10^7 \cdot 3 \times 10^8 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

6. **Calculate the Terms**:
- Calculate \( \frac{1}{2^2} = \frac{1}{4} = 0.25 \)
- Calculate \( \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \)
- Now, calculate \( 0.25 - 0.1111 = 0.1389 \)

7. **Final Calculation**:
\[ \nu = 1.097 \times 10^7 \cdot 3 \times 10^8 \cdot 0.1389 \]
\[ \nu \approx 4.57 \times 10^{14} \, \text{s}^{-1} \]

### Final Answer:
The frequency of the first line in the Balmer series for the hydrogen atom is approximately \( 4.57 \times 10^{14} \, \text{s}^{-1} \).

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