The frequency of `H_(beta)` line of lyman seris of hydrogen is

Options

A. `2.92 xx10^(15)` Hz
B. `3.07 xx10^(15)` Hz
C. `1.02 xx10^(7)` Hz
D. `9.76 xx10^(6)` Hz

Answer

To find the frequency of the H-beta line of the Lyman series of hydrogen, we can follow these steps:

### Step 1: Identify the relevant formula
The frequency (ν) can be calculated using the formula:
\[ \nu = \frac{c}{\lambda} \]
where:
- \( c \) is the speed of light (approximately \( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda \) is the wavelength.

### Step 2: Determine the wavelength using the Rydberg formula
The Rydberg formula for hydrogen is:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (approximately \( 1.1 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 \) is the lower energy level (for the Lyman series, \( n_1 = 1 \))
- \( n_2 \) is the higher energy level (for the H-beta line, \( n_2 = 3 \))

### Step 3: Substitute the values into the Rydberg formula
Substituting the values into the formula:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{9} \right) \]
\[ \frac{1}{\lambda} = R \left( \frac{8}{9} \right) \]

### Step 4: Calculate the wavelength
Now substituting the value of \( R \):
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \cdot \frac{8}{9} \]
\[ \lambda = \frac{9}{8 \cdot 1.1 \times 10^7} \]

Calculating \( \lambda \):
\[ \lambda = \frac{9}{8.8 \times 10^7} \]
\[ \lambda \approx 1.02 \times 10^{-7} \, \text{m} \]

### Step 5: Calculate the frequency
Now we can calculate the frequency using the wavelength:
\[ \nu = \frac{c}{\lambda} \]
\[ \nu = \frac{3 \times 10^8}{1.02 \times 10^{-7}} \]
\[ \nu \approx 2.94 \times 10^{15} \, \text{Hz} \]

### Conclusion
The frequency of the H-beta line of the Lyman series of hydrogen is approximately \( 2.94 \times 10^{15} \, \text{Hz} \).