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 the transition in a hydrogen atom that emits photons of the lowest frequency, we can follow these steps:

### Step 1: Understand the relationship between energy and frequency
The energy of a photon is given by the equation:
\[ E = h \nu \]
where \( E \) is the energy, \( h \) is Planck's constant, and \( \nu \) (nu) is the frequency of the photon. From this equation, we can see that energy is directly proportional to frequency. Therefore, to find the transition that emits the lowest frequency, we need to find the transition that corresponds to the lowest energy difference between two energy levels.

### Step 2: Identify the energy levels in a hydrogen atom
In a hydrogen atom, the energy levels are quantized and can be expressed using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number (n = 1, 2, 3, ...). As \( n \) increases, the energy levels get closer together, meaning the difference in energy between successive levels decreases.

### Step 3: Determine the transitions and their energy differences
To find the transition with the lowest energy difference, we can consider the transitions between higher energy levels. The energy difference \( \Delta E \) between two levels \( n_i \) and \( n_f \) is given by:
\[ \Delta E = E_{n_f} - E_{n_i} = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2} \]

### Step 4: Calculate the energy differences for possible transitions
Let's consider the transitions between higher energy levels:
1. Transition from \( n = 4 \) to \( n = 3 \)
2. Transition from \( n = 3 \) to \( n = 2 \)
3. Transition from \( n = 2 \) to \( n = 1 \)

Calculating the energy differences:
- For \( n = 4 \) to \( n = 3 \):
\[ \Delta E_{4 \to 3} = E_3 - E_4 = \left(-\frac{13.6}{3^2}\right) - \left(-\frac{13.6}{4^2}\right) \]
\[ = -\frac{13.6}{9} + \frac{13.6}{16} \]

- For \( n = 3 \) to \( n = 2 \):
\[ \Delta E_{3 \to 2} = E_2 - E_3 = \left(-\frac{13.6}{2^2}\right) - \left(-\frac{13.6}{3^2}\right) \]
\[ = -\frac{13.6}{4} + \frac{13.6}{9} \]

- For \( n = 2 \) to \( n = 1 \):
\[ \Delta E_{2 \to 1} = E_1 - E_2 = \left(-\frac{13.6}{1^2}\right) - \left(-\frac{13.6}{2^2}\right) \]
\[ = -13.6 + \frac{13.6}{4} \]

### Step 5: Identify the transition with the minimum energy difference
From the calculations, we can see that the transition from \( n = 4 \) to \( n = 3 \) has the smallest energy difference because the energy levels are closer together at higher values of \( n \). Therefore, this transition will emit photons of the lowest frequency.

### Conclusion
The transition in a hydrogen atom that emits photons of the lowest frequency is:
**Transition from \( n = 4 \) to \( n = 3 \)**.

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