The frequency for a series limit of Balmer and paschen serial respectively are `f_1and f_3` if the frequency of the first line of Balmer series is then the relation between `f _1,f_2and f_3` is

Options

A. `f_1-f_2=f_3`
B. `f_1+f_3=f_2`
C. `f_1+f_2=f_3`
D. `f_2-f_3=2f_1`

Answer

To solve the problem, we need to establish the relationship between the frequencies \( f_1 \), \( f_2 \), and \( f_3 \) for the Balmer and Paschen series in the hydrogen spectrum.

### Step-by-Step Solution:

1. **Understanding the Series**:
- The Balmer series corresponds to transitions where the electron falls to the second energy level (\( n_1 = 2 \)).
- The Paschen series corresponds to transitions where the electron falls to the third energy level (\( n_1 = 3 \)).
- The series limits correspond to transitions from \( n_2 = \infty \) to \( n_1 \).

2. **Frequency for Series Limit of Balmer Series**:
- For the Balmer series, the series limit frequency \( f_1 \) can be expressed using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 2 \) and \( n_2 = \infty \).
- Therefore, we have:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4} \]
- The frequency \( f_1 \) is then given by:
\[ f_1 = c \cdot \frac{1}{\lambda} = c \cdot \frac{R}{4} \]

3. **Frequency for Series Limit of Paschen Series**:
- For the Paschen series, the series limit frequency \( f_3 \) can be expressed similarly:
\[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \]
- The frequency \( f_3 \) is then given by:
\[ f_3 = c \cdot \frac{1}{\lambda} = c \cdot \frac{R}{9} \]

4. **Frequency of the First Line of the Balmer Series**:
- The first line of the Balmer series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Finding a common denominator (36):
\[ \frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} \]
- Thus, the frequency \( f_2 \) is given by:
\[ f_2 = c \cdot \frac{1}{\lambda} = c \cdot \frac{5R}{36} \]

5. **Establishing the Relationship**:
- Now we have:
\[ f_1 = c \cdot \frac{R}{4}, \quad f_2 = c \cdot \frac{5R}{36}, \quad f_3 = c \cdot \frac{R}{9} \]
- To find the relationship between \( f_1 \), \( f_2 \), and \( f_3 \), we can express \( f_2 \) in terms of \( f_1 \) and \( f_3 \):
- Rearranging gives:
\[ f_3 = f_1 - f_2 \]

### Final Relation:
\[ f_3 = f_1 - f_2 \]