Calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom.

Options

A. `3.7xx10^(14)Hz`
B. `9.1xx10^(15)Hz`
C. `10.23xx10^(14)Hz`
D. `29.7xx10^(15)Hz`

Explanation

To calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom, we can follow these steps:

### Step 1: Understand the Paschen Series
The Paschen series corresponds to transitions of electrons from higher energy levels (n ≥ 4) to the n = 3 energy level in a hydrogen atom.

### Step 2: Identify the Transition for Highest Frequency
The highest frequency photon in the Paschen series is emitted when an electron transitions from the highest possible energy level (n = ∞) to n = 3.

### Step 3: Use the Rydberg Formula
The frequency of the emitted photon can be calculated using the Rydberg formula for hydrogen:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant (\( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( n_1 \) is the lower energy level (3 for the Paschen series)
- \( n_2 \) is the upper energy level (∞ for the highest frequency)

### Step 4: Substitute Values into the Rydberg Formula
Substituting \( n_1 = 3 \) and \( n_2 = \infty \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{9} \right) \]
\[ \frac{1}{\lambda} = \frac{R_H}{9} \]

### Step 5: Calculate the Wavelength
Now, substituting the value of \( R_H \):
\[ \frac{1}{\lambda} = \frac{1.097 \times 10^7}{9} \approx 1.219 \times 10^6 \, \text{m}^{-1} \]
Thus, the wavelength \( \lambda \) is:
\[ \lambda = \frac{1}{1.219 \times 10^6} \approx 8.2 \times 10^{-7} \, \text{m} = 820 \, \text{nm} \]

### Step 6: Calculate the Frequency
Now, we can calculate the frequency \( f \) using the speed of light \( c \):
\[ f = \frac{c}{\lambda} \]
where \( c = 3 \times 10^8 \, \text{m/s} \):
\[ f = \frac{3 \times 10^8}{8.2 \times 10^{-7}} \approx 3.66 \times 10^{14} \, \text{Hz} \]

### Conclusion
The highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom is approximately \( 3.66 \times 10^{14} \, \text{Hz} \).

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