In which transition of a hydrogen atom, photons of lowest frequency are emitted ?

Options

A. n=4 to n=3
B. n=4 to n=2
C. n=2 to n=1
D. n=3 to n=1

Answer

To determine which transition of a hydrogen atom emits photons of the lowest frequency, we can follow these steps:

### Step 1: Understand the Energy Transition Formula
The energy of a photon emitted during a transition between two energy levels in a hydrogen atom can be calculated using the formula:
\[ E = -13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.

### Step 2: Identify the Transitions
We need to analyze the following transitions:
1. \( n_4 \) to \( n_3 \)
2. \( n_4 \) to \( n_2 \)
3. \( n_3 \) to \( n_1 \)
4. \( n_2 \) to \( n_1 \)

### Step 3: Calculate Energy for Each Transition
1. **Transition from \( n_4 \) to \( n_3 \)**:
\[ E_{4 \to 3} = -13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = -13.6 \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = -13.6 \left( \frac{16 - 9}{144} \right) = -13.6 \left( \frac{7}{144} \right) \approx -0.66 \, \text{eV} \]

2. **Transition from \( n_4 \) to \( n_2 \)**:
\[ E_{4 \to 2} = -13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = -13.6 \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ = -13.6 \left( \frac{4 - 1}{16} \right) = -13.6 \left( \frac{3}{16} \right) \approx -2.55 \, \text{eV} \]

3. **Transition from \( n_3 \) to \( n_1 \)**:
\[ E_{3 \to 1} = -13.6 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = -13.6 \left( 1 - \frac{1}{9} \right) \]
\[ = -13.6 \left( \frac{8}{9} \right) \approx -12.09 \, \text{eV} \]

4. **Transition from \( n_2 \) to \( n_1 \)**:
\[ E_{2 \to 1} = -13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = -13.6 \left( 1 - \frac{1}{4} \right) \]
\[ = -13.6 \left( \frac{3}{4} \right) \approx -10.2 \, \text{eV} \]

### Step 4: Determine Frequency
The frequency of the emitted photon can be calculated using the relation:
\[ E = h \nu \quad \Rightarrow \quad \nu = \frac{E}{h} \]
Since \( h \) is a constant, the frequency is directly proportional to the energy. Thus, the transition with the lowest energy will correspond to the lowest frequency.

### Step 5: Compare Energies
Now we compare the energies calculated:
- \( E_{4 \to 3} \approx -0.66 \, \text{eV} \)
- \( E_{4 \to 2} \approx -2.55 \, \text{eV} \)
- \( E_{3 \to 1} \approx -12.09 \, \text{eV} \)
- \( E_{2 \to 1} \approx -10.2 \, \text{eV} \)

From these values, the transition with the lowest energy (and therefore the lowest frequency) is:
- **Transition from \( n_4 \) to \( n_3 \)**.

### Conclusion
The transition in a hydrogen atom that emits photons of the lowest frequency is from \( n_4 \) to \( n_3 \).

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