The frequency of the `H_(beta)`-line of the Balmer series for hydrogen is

Options

A. `6.17xx10^(14)`
B. `6.12xx10^(12)`
C. `6.15xx10^(12)`
D. `6.18xx10^(13)`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Balmer Series**: The Balmer series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (n > 2) to the second energy level (n = 2). The H-beta line specifically corresponds to the transition from n = 4 to n = 2.

2. **Use the Rydberg Formula**: The Rydberg formula for the wavelength of the 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 (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)), \( n_1 \) is the lower energy level (2 for Balmer series), and \( n_2 \) is the higher energy level (4 for H-beta).

3. **Substitute Values into the Formula**: For the H-beta line:
\[ n_1 = 2, \quad n_2 = 4 \]
Therefore,
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) \]

4. **Calculate the Terms**: Calculate \( \frac{1}{4} - \frac{1}{16} \):
\[ \frac{1}{4} = \frac{4}{16}, \quad \text{thus} \quad \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]

5. **Substitute Back into the Formula**: Now substitute back:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{16} \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1875 = 2.060625 \times 10^6 \, \text{m}^{-1} \]

6. **Calculate Wavelength (\( \lambda \))**: Taking the reciprocal gives:
\[ \lambda = \frac{1}{2.060625 \times 10^6} \approx 4.85 \times 10^{-7} \, \text{m} \]

7. **Convert Wavelength to Frequency**: Use the relationship between speed of light (\( c \)), wavelength (\( \lambda \)), and frequency (\( f \)):
\[ c = f \lambda \implies f = \frac{c}{\lambda} \]
where \( c \approx 3 \times 10^8 \, \text{m/s} \):
\[ f = \frac{3 \times 10^8}{4.85 \times 10^{-7}} \approx 6.19 \times 10^{14} \, \text{Hz} \]

### Final Answer:
The frequency of the H-beta line of the Balmer series for hydrogen is approximately \( 6.19 \times 10^{14} \, \text{Hz} \).