Let `F_1` be the frequency of second line of Lyman series and `F_2` be the frequency of first line of Balmer series then frequency of first line of Lyman series is given by

Options

A. `F_(1) - F_(2)`
B. `F_(1) + F_(2)`
C. `F_(2) - F_(1)`
D. `(F_(1)F_(2))/(F_(1) + F_(2))`

Answer

To find the frequency of the first line of the Lyman series, we will follow these steps:

### Step 1: Identify the Frequencies
Let \( F_1 \) be the frequency of the second line of the Lyman series and \( F_2 \) be the frequency of the first line of the Balmer series.

### Step 2: Determine \( F_1 \)
The second line of the Lyman series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 1 \). We can use the Rydberg formula for hydrogen:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting \( n_1 = 1 \) and \( n_2 = 3 \):

\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]

Thus,

\[ \lambda_1 = \frac{9}{8R} \]

Now, the frequency \( F_1 \) is given by:

\[ F_1 = \frac{c}{\lambda_1} = \frac{c \cdot 8R}{9} \]

### Step 3: Determine \( F_2 \)
The first line of the Balmer series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \). Again using the Rydberg formula:

\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ \frac{1}{\lambda_2} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

Thus,

\[ \lambda_2 = \frac{36}{5R} \]

Now, the frequency \( F_2 \) is given by:

\[ F_2 = \frac{c}{\lambda_2} = \frac{c \cdot 5R}{36} \]

### Step 4: Determine the Frequency of the First Line of Lyman Series
The first line of the Lyman series corresponds to the transition from \( n_2 = 2 \) to \( n_1 = 1 \):

\[ \frac{1}{\lambda_3} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

Thus,

\[ \lambda_3 = \frac{4}{3R} \]

Now, the frequency \( F_3 \) is given by:

\[ F_3 = \frac{c}{\lambda_3} = \frac{c \cdot 3R}{4} \]

### Step 5: Relate \( F_1 \), \( F_2 \), and \( F_3 \)
From our calculations:

\[ F_1 = \frac{8cR}{9}, \quad F_2 = \frac{5cR}{36}, \quad F_3 = \frac{3cR}{4} \]

Now, we need to find the relationship between \( F_1 \), \( F_2 \), and \( F_3 \):

\[ F_1 - F_2 = \frac{8cR}{9} - \frac{5cR}{36} \]

Finding a common denominator (36):

\[ F_1 - F_2 = \frac{32cR}{36} - \frac{5cR}{36} = \frac{27cR}{36} = \frac{3cR}{4} \]

Thus, we find:

\[ F_1 - F_2 = F_3 \]

### Final Answer
The frequency of the first line of the Lyman series is given by:

\[ F_3 = F_1 - F_2 \]