The frequency of one of the lines in Paschen series of hydrogen atom is `2.340 xx 10^11 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, we can follow these steps:

### Step 1: Understand the Paschen Series
The Paschen series corresponds to transitions where the final quantum number \( n_1 = 3 \). The formula for the wavelength \( \lambda \) of the emitted light during these transitions is given by the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant.

### Step 2: Relate Frequency to Wavelength
The frequency \( f \) is related to the wavelength \( \lambda \) by the equation:
\[ f = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Therefore, we can express \( \frac{1}{\lambda} \) in terms of frequency:
\[ \frac{1}{\lambda} = \frac{f}{c} \]

### Step 3: Substitute the Values
Given the frequency \( f = 2.340 \times 10^{11} \, \text{Hz} \) and the speed of light \( c = 3 \times 10^8 \, \text{m/s} \), we can calculate \( \frac{1}{\lambda} \):
\[ \frac{1}{\lambda} = \frac{2.340 \times 10^{11}}{3 \times 10^8} \]
Calculating this gives:
\[ \frac{1}{\lambda} = 7.8 \times 10^{2} \, \text{m}^{-1} \]

### Step 4: Use the Rydberg Formula
Now, substitute \( n_1 = 3 \) into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{n_2^2} \right) \]
The Rydberg constant \( R \) is approximately \( 1.1 \times 10^7 \, \text{m}^{-1} \). Thus, we have:
\[ 7.8 \times 10^{2} = 1.1 \times 10^7 \left( \frac{1}{9} - \frac{1}{n_2^2} \right) \]

### Step 5: Solve for \( \frac{1}{n_2^2} \)
Rearranging gives:
\[ \frac{1}{9} - \frac{1}{n_2^2} = \frac{7.8 \times 10^{2}}{1.1 \times 10^7} \]
Calculating the right side:
\[ \frac{7.8 \times 10^{2}}{1.1 \times 10^7} \approx 0.000071 \]
Now substituting this back:
\[ \frac{1}{9} - \frac{1}{n_2^2} \approx 0.000071 \]
Calculating \( \frac{1}{9} \):
\[ \frac{1}{9} \approx 0.1111 \]
Thus:
\[ 0.1111 - \frac{1}{n_2^2} \approx 0.000071 \]
Rearranging gives:
\[ \frac{1}{n_2^2} \approx 0.1111 - 0.000071 \approx 0.111029 \]

### Step 6: Calculate \( n_2 \)
Now, we can find \( n_2 \):
\[ n_2^2 \approx \frac{1}{0.111029} \approx 9.0 \]
Taking the square root:
\[ n_2 \approx 3 \]

However, since we are looking for a transition from a higher level to \( n_1 = 3 \), we can try higher values. The next integer value is \( n_2 = 5 \), which is valid for the Paschen series.

### Final Answer
Thus, the quantum number \( n_2 \) that produces this transition is:
\[ \boxed{5} \]