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

Options

A. `n_(3) rarr n_(2)`
B. `n_(4) rarr n_(3)`
C. `n_(2) 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 need to analyze the energy levels of the hydrogen atom and the transitions between them.

### Step-by-Step Solution:

1. **Understanding Energy Levels**:
In a hydrogen atom, electrons occupy specific energy levels denoted by quantum numbers (n). The energy of these levels is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number (n = 1, 2, 3, ...).

2. **Energy Transition Calculation**:
The energy emitted during a transition from a higher energy level (n2) to a lower energy level (n1) can be calculated using the formula:
\[ \Delta E = E_{n1} - E_{n2} = -\frac{13.6 \, \text{eV}}{n_1^2} + \frac{13.6 \, \text{eV}}{n_2^2} \]
This can be rearranged to:
\[ \Delta E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

3. **Identifying Maximum Energy Transition**:
To find the transition that emits maximum energy, we need to maximize the difference \( \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \). The largest energy difference occurs when the electron transitions from the highest possible energy level to the lowest energy level (ground state).

4. **Choosing the Transition**:
The maximum energy transition in a hydrogen atom is from \( n = \infty \) (where the electron is free) to \( n = 1 \) (the ground state). This transition can be represented as:
\[ \Delta E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = 13.6 \, \text{eV} \]

5. **Conclusion**:
Therefore, the transition of an electron in a hydrogen atom that will emit maximum energy is from \( n = \infty \) to \( n = 1 \).