The frequency of one of the lines in Paschen series of hydrogen atom is `2.340 xx 10^14 Hz`. The quantum number `n_2` Which produces this transition is.

Explanation

To find the quantum number \( n_2 \) that produces the transition in the Paschen series of the hydrogen atom given the frequency \( f = 2.340 \times 10^{14} \, \text{Hz} \), we can follow these steps:

### Step 1: Understand the Paschen Series
The Paschen series corresponds to transitions where the electron falls to the \( n_1 = 3 \) energy level. The general formula for the wavelengths in the hydrogen spectrum is given by:

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

where:
- \( R \) is the Rydberg constant (\( R = 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (for Paschen series, \( n_1 = 3 \)),
- \( n_2 \) is the higher energy level.

### Step 2: Relate Frequency to Wavelength
We know that the speed of light \( c \) is related to wavelength \( \lambda \) and frequency \( f \) by the equation:

\[ c = \lambda f \]

Thus, we can express \( \lambda \) as:

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

### Step 3: Substitute Values
Substituting the known values:
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)
- Frequency \( f = 2.340 \times 10^{14} \, \text{Hz} \)

Calculating \( \lambda \):

\[ \lambda = \frac{3 \times 10^8}{2.340 \times 10^{14}} = 1.282 \times 10^{-6} \, \text{m} \]

### Step 4: Use the Rydberg Formula
Now we can substitute \( \lambda \) back into the Rydberg formula:

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

Calculating \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{1}{1.282 \times 10^{-6}} \approx 7.8 \times 10^5 \, \text{m}^{-1} \]

### Step 5: Set Up the Equation
Now we can set up the equation:

\[ 7.8 \times 10^5 = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{n_2^2} \right) \]

### Step 6: Solve for \( n_2 \)
Substituting \( n_1 = 3 \):

\[ 7.8 \times 10^5 = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{n_2^2} \right) \]

Calculating \( \frac{1}{9} \):

\[ \frac{1}{9} \approx 0.1111 \]

Now substituting this back into the equation:

\[ 7.8 \times 10^5 = 1.097 \times 10^7 \left( 0.1111 - \frac{1}{n_2^2} \right) \]

Rearranging gives:

\[ \frac{7.8 \times 10^5}{1.097 \times 10^7} = 0.1111 - \frac{1}{n_2^2} \]

Calculating the left side:

\[ 0.0711 \approx 0.1111 - \frac{1}{n_2^2} \]

Now solving for \( \frac{1}{n_2^2} \):

\[ \frac{1}{n_2^2} = 0.1111 - 0.0711 = 0.04 \]

Taking the reciprocal:

\[ n_2^2 = \frac{1}{0.04} = 25 \]

Thus, \( n_2 = 5 \).

### Final Answer
The quantum number \( n_2 \) which produces this transition is:

\[ \boxed{5} \]