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 in which transition of a hydrogen atom photons of the lowest frequency are emitted, we can follow these steps:

### Step 1: Understand the relationship between energy and frequency
The energy of a photon emitted during a transition between two energy levels in a hydrogen atom is given by the formula:
\[ \Delta E = E_0 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \(E_0\) is the energy of the ground state (13.6 eV for hydrogen), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.

### Step 2: Identify the transitions
We need to evaluate the energy differences for the following transitions:
1. \(n_2 = 4\) to \(n_1 = 3\)
2. \(n_2 = 4\) to \(n_1 = 2\)
3. \(n_2 = 2\) to \(n_1 = 1\)
4. \(n_2 = 3\) to \(n_1 = 1\)

### Step 3: Calculate the energy for each transition
1. **Transition from \(n=4\) to \(n=3\)**:
\[ \Delta E = E_0 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = 13.6 \left( \frac{16 - 9}{144} \right) = 13.6 \left( \frac{7}{144} \right) = \frac{95.2}{144} \approx 0.66 \text{ eV} \]

2. **Transition from \(n=4\) to \(n=2\)**:
\[ \Delta E = E_0 \left( 1 - \frac{1}{16} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \left( \frac{3}{16} \right) = \frac{40.8}{16} = 2.55 \text{ eV} \]

3. **Transition from \(n=2\) to \(n=1\)**:
\[ \Delta E = E_0 \left( 1 - \frac{1}{4} \right) = 13.6 \left( \frac{3}{4} \right) = 10.2 \text{ eV} \]

4. **Transition from \(n=3\) to \(n=1\)**:
\[ \Delta E = E_0 \left( 1 - \frac{1}{9} \right) = 13.6 \left( \frac{8}{9} \right) \approx 12.09 \text{ eV} \]

### Step 4: Compare the energies
Now we compare the energies calculated:
- \(n=4\) to \(n=3\): \(0.66 \text{ eV}\)
- \(n=4\) to \(n=2\): \(2.55 \text{ eV}\)
- \(n=2\) to \(n=1\): \(10.2 \text{ eV}\)
- \(n=3\) to \(n=1\): \(12.09 \text{ eV}\)

The transition with the lowest energy is from \(n=4\) to \(n=3\).

### Step 5: Conclusion
Since the energy of the photon is lowest for the transition \(n=4\) to \(n=3\), the frequency of the emitted photon will also be the lowest. Thus, the answer is:
**The transition from \(n=4\) to \(n=3\) emits photons of the lowest frequency.**

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