Which of the transitions in hydrogen atom emits a photon of lowest frequecny `(n =` quantum number)?

Options

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

Explanation

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

### Step 1: Understand the relationship between energy and frequency
The energy of a photon is related to its frequency by the equation:
\[ E = h \cdot f \]
where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is the frequency. Therefore, a lower energy corresponds to a lower frequency.

### Step 2: Calculate the energy levels of the hydrogen atom
The energy levels of a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number (1, 2, 3, ...).

- For \( n = 1 \):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{13.6}{3^2} \approx -1.51 \, \text{eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{13.6}{4^2} = -0.85 \, \text{eV} \]

### Step 3: Calculate the energy differences for each transition
Now, we will calculate the energy differences for the possible transitions:

1. **Transition from \( n = 2 \) to \( n = 1 \)**:
\[ \Delta E_{2 \to 1} = E_1 - E_2 = (-13.6) - (-3.4) = -10.2 \, \text{eV} \]

2. **Transition from \( n = 3 \) to \( n = 2 \)**:
\[ \Delta E_{3 \to 2} = E_2 - E_3 = (-3.4) - (-1.51) \approx -1.89 \, \text{eV} \]

3. **Transition from \( n = 4 \) to \( n = 1 \)**:
\[ \Delta E_{4 \to 1} = E_1 - E_4 = (-13.6) - (-0.85) = -12.75 \, \text{eV} \]

4. **Transition from \( n = 4 \) to \( n = 3 \)**:
\[ \Delta E_{4 \to 3} = E_3 - E_4 = (-1.51) - (-0.85) \approx -0.66 \, \text{eV} \]

### Step 4: Identify the transition with the lowest energy difference
Now we compare the energy differences calculated:

- \( \Delta E_{2 \to 1} = -10.2 \, \text{eV} \)
- \( \Delta E_{3 \to 2} \approx -1.89 \, \text{eV} \)
- \( \Delta E_{4 \to 1} = -12.75 \, \text{eV} \)
- \( \Delta E_{4 \to 3} \approx -0.66 \, \text{eV} \)

The transition with the lowest energy difference is from \( n = 4 \) to \( n = 3 \) with an energy difference of approximately \( -0.66 \, \text{eV} \).

### Step 5: Conclusion
Since the transition from \( n = 4 \) to \( n = 3 \) has the lowest energy difference, it emits a photon of the lowest frequency.

**Answer: The transition from \( n = 4 \) to \( n = 3 \) emits a photon of the lowest frequency.**

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