In which transition minimum energy is emitted :-

Options

A. `oorarr1`
B. `2rarr1`
C. `3rarr2`
D. `nrarr(n-1)(nge4)`

Explanation

To determine in which transition minimum energy is emitted, we will calculate the energy released for each transition using the formula:

\[ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( n_1 \) is the final energy level and \( n_2 \) is the initial energy level.

### Step-by-Step Solution:

**Option A: Transition from n=∞ to n=1**
1. Here, \( n_1 = 1 \) and \( n_2 = \infty \).
2. Substitute into the formula:
\[ E = 13.6 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = 13.6 \left( 1 - 0 \right) = 13.6 \text{ eV} \]

**Option B: Transition from n=2 to n=1**
1. Here, \( n_1 = 1 \) and \( n_2 = 2 \).
2. Substitute into the formula:
\[ E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \left( \frac{3}{4} \right) = 10.2 \text{ eV} \]

**Option C: Transition from n=3 to n=2**
1. Here, \( n_1 = 2 \) and \( n_2 = 3 \).
2. Substitute into the formula:
\[ E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ = 13.6 \left( \frac{9 - 4}{36} \right) = 13.6 \left( \frac{5}{36} \right) \approx 1.88 \text{ eV} \]

**Option D: Transition from n=4 to n=3**
1. Here, \( n_1 = 3 \) and \( n_2 = 4 \).
2. Substitute into the formula:
\[ E = 13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = 13.6 \left( \frac{16 - 9}{144} \right) = 13.6 \left( \frac{7}{144} \right) \approx 0.66 \text{ eV} \]

### Conclusion:
After calculating the energy emitted for each transition:
- Option A: 13.6 eV
- Option B: 10.2 eV
- Option C: 1.88 eV
- Option D: 0.66 eV

The transition that emits the minimum energy is **Option D** with an energy of **0.66 eV**.