The transition of electron in H atom that will emit maximum energy is

Options

A. `n_(3) rarr n_(2)`
B. `n_(4) rarr n_(3)`
C. `n_(5) rarr n_(4)`
D. `n_(6) rarr n_(5)`

Explanation

To determine the transition of an electron in a hydrogen atom that will emit maximum energy, we can follow these steps:

### Step 1: Understand Energy Levels in Hydrogen Atom
In a hydrogen atom, electrons occupy specific energy levels, denoted by the principal quantum number \( n \). The energy levels are quantized and can be represented as:
- \( n = 1 \) (ground state)
- \( n = 2 \)
- \( n = 3 \)
- \( n = 4 \)
- \( n = 5 \)
- \( n = 6 \)

### Step 2: Identify the Formula for Energy Difference
The energy associated with each level can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
To find the energy emitted during a transition from a higher energy level \( n_i \) to a lower energy level \( n_f \), we use:
\[ \Delta E = E_f - E_i = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2} \]
This can be simplified to:
\[ \Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

### Step 3: Calculate Energy Differences for Given Transitions
We need to calculate the energy differences for the possible transitions:
1. **Transition from \( n = 3 \) to \( n = 2 \)**:
\[ \Delta E_{3 \to 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{9 - 4}{36} \right) = 13.6 \left( \frac{5}{36} \right) \approx 1.89 \, \text{eV} \]

2. **Transition from \( n = 4 \) to \( n = 3 \)**:
\[ \Delta E_{4 \to 3} = 13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \left( \frac{16 - 9}{144} \right) = 13.6 \left( \frac{7}{144} \right) \approx 0.66 \, \text{eV} \]

3. **Transition from \( n = 5 \) to \( n = 4 \)**:
\[ \Delta E_{5 \to 4} = 13.6 \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = 13.6 \left( \frac{1}{16} - \frac{1}{25} \right) = 13.6 \left( \frac{25 - 16}{400} \right) = 13.6 \left( \frac{9}{400} \right) \approx 0.31 \, \text{eV} \]

4. **Transition from \( n = 6 \) to \( n = 5 \)**:
\[ \Delta E_{6 \to 5} = 13.6 \left( \frac{1}{5^2} - \frac{1}{6^2} \right) = 13.6 \left( \frac{1}{25} - \frac{1}{36} \right) = 13.6 \left( \frac{36 - 25}{900} \right) = 13.6 \left( \frac{11}{900} \right) \approx 0.17 \, \text{eV} \]

### Step 4: Compare Energy Differences
Now, we compare the energy differences calculated:
- \( \Delta E_{3 \to 2} \approx 1.89 \, \text{eV} \)
- \( \Delta E_{4 \to 3} \approx 0.66 \, \text{eV} \)
- \( \Delta E_{5 \to 4} \approx 0.31 \, \text{eV} \)
- \( \Delta E_{6 \to 5} \approx 0.17 \, \text{eV} \)

### Conclusion
The transition that emits the maximum energy is from \( n = 3 \) to \( n = 2 \).

### Final Answer
The transition of electron in H atom that will emit maximum energy is **from \( n = 3 \) to \( n = 2 \)**.

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