In a hydrogen atom, an electron jumps from the third orbit to the first orbit. Find out the frequency and wavelength of the spectral line.

Explanation

To find the frequency and wavelength of the spectral line corresponding to an electron transition from the third orbit to the first orbit in a hydrogen atom, we can follow these steps:

### Step 1: Identify the Rydberg Formula
The Rydberg formula for the wavelength of light emitted 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:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant (\( R = 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level (1 in this case),
- \( n_2 \) is the higher energy level (3 in this case).

### Step 2: Substitute Values into the Rydberg Formula
Substituting the values into the formula:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

Calculating the right side:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{9} \right) = 1.097 \times 10^7 \left( \frac{8}{9} \right) \]

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.8889 \approx 0.9749 \times 10^7 \, \text{m}^{-1} \]

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

\[ \lambda = \frac{1}{0.9749 \times 10^7} \approx 1.0254 \times 10^{-7} \, \text{m} \]

To convert this into angstroms (1 angstrom = \( 10^{-10} \) m):

\[ \lambda \approx 1.0254 \times 10^{-7} \, \text{m} = 1025.4 \, \text{angstrom} \]

### Step 4: Calculate Frequency
Now, we can find the frequency \( \nu \) using the speed of light \( c \):

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

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

\[ \nu = \frac{3 \times 10^8}{1.0254 \times 10^{-7}} \approx 2.9257 \times 10^{15} \, \text{s}^{-1} \]

### Final Results
- Wavelength \( \lambda \approx 1025.4 \, \text{angstrom} \)
- Frequency \( \nu \approx 2.9257 \times 10^{15} \, \text{s}^{-1} \)