The frequency corresponding to transition `n = 1` to `n = 2` in hydrogen atom is.

Options

A. `15.66 xx 10^10 Hz`
B. `24.66 xx 10^14 Hz`
C. `30.57 xx 10^14 Hz`
D. `40.57 xx 10^24 Hz`

Explanation

To find the frequency corresponding to the transition from \( n = 1 \) to \( n = 2 \) in a hydrogen atom, we can follow these steps:

### Step 1: Use the Rydberg Formula
The Rydberg formula for the wavelength (\( \lambda \)) of the emitted light during an electron transition in a hydrogen atom 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, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \) or \( 109678 \, \text{cm}^{-1} \)
- \( n_1 \) is the lower energy level (1 in this case)
- \( n_2 \) is the higher energy level (2 in this case)

### Step 2: Substitute Values into the Formula
Substituting \( n_1 = 1 \) and \( n_2 = 2 \) into the formula:

\[ \frac{1}{\lambda} = 109678 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating the right side:

\[ \frac{1}{\lambda} = 109678 \left( 1 - \frac{1}{4} \right) = 109678 \left( \frac{3}{4} \right) = 82258.5 \, \text{cm}^{-1} \]

### Step 3: Calculate Wavelength (\( \lambda \))
Now, we can find \( \lambda \):

\[ \lambda = \frac{1}{82258.5} \, \text{cm}^{-1} \approx 1.21567 \times 10^{-5} \, \text{cm} \]

Converting to meters:

\[ \lambda = 1.21567 \times 10^{-5} \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 1.21567 \times 10^{-7} \, \text{m} \]

### Step 4: Calculate Frequency (\( \nu \))
Using the relationship between speed of light (\( c \)), wavelength (\( \lambda \)), and frequency (\( \nu \)):

\[ c = \lambda \cdot \nu \implies \nu = \frac{c}{\lambda} \]

Substituting \( c = 3 \times 10^8 \, \text{m/s} \):

\[ \nu = \frac{3 \times 10^8}{1.21567 \times 10^{-7}} \approx 2.46 \times 10^{15} \, \text{Hz} \]

### Final Answer
The frequency corresponding to the transition from \( n = 1 \) to \( n = 2 \) in the hydrogen atom is approximately:

\[ \nu \approx 2.46 \times 10^{14} \, \text{Hz} \]