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.

Options

A. 6
B. 5
C. 4
D. 3

Explanation

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

### Step 1: Calculate the Wavelength \( \lambda \)
We know that the speed of light \( c \) is related to frequency \( \nu \) and wavelength \( \lambda \) by the equation:
\[ c = \nu \lambda \]
Rearranging this gives:
\[ \lambda = \frac{c}{\nu} \]
Substituting the values:
- \( c = 3.0 \times 10^8 \, \text{m/s} \)
- \( \nu = 2.340 \times 10^{14} \, \text{Hz} \)

Calculating \( \lambda \):
\[ \lambda = \frac{3.0 \times 10^8}{2.340 \times 10^{14}} \approx 1.288 \times 10^{-6} \, \text{m} \]

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelengths of spectral lines in hydrogen 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, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

For the Paschen series, \( n_1 = 3 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{n_2^2} \right) \]

### Step 3: Substitute Known Values
Using \( R = 1.097 \times 10^7 \, \text{m}^{-1} \) and \( \lambda = 1.288 \times 10^{-6} \, \text{m} \):
\[ \frac{1}{\lambda} = \frac{1}{1.288 \times 10^{-6}} \approx 776,636 \, \text{m}^{-1} \]
Substituting this into the Rydberg formula:
\[ 776636 = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{n_2^2} \right) \]

### Step 4: Solve for \( n_2 \)
Rearranging the equation:
\[ \frac{1}{9} - \frac{1}{n_2^2} = \frac{776636}{1.097 \times 10^7} \]
Calculating the right side:
\[ \frac{776636}{1.097 \times 10^7} \approx 0.0707 \]
Now, substituting back:
\[ \frac{1}{n_2^2} = \frac{1}{9} - 0.0707 \]
Calculating \( \frac{1}{9} \):
\[ \frac{1}{9} \approx 0.1111 \]
Thus:
\[ \frac{1}{n_2^2} \approx 0.1111 - 0.0707 \approx 0.0404 \]
Taking the reciprocal gives:
\[ n_2^2 \approx \frac{1}{0.0404} \approx 24.75 \]
Taking the square root:
\[ n_2 \approx 5 \]

### Final Answer
The quantum number \( n_2 \) that produces this transition is \( n_2 = 5 \).

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