According to Bohr theory, which of the following transition in hydrogen atom will give rise to the least energetic proton?

Options

A. n = 6 to n = 1
B. n = 5 to n = 4
C. n = 6 to n = 5
D. n = 5 to n = 3

Explanation

To determine which transition in a hydrogen atom will give rise to the least energetic photon according to Bohr's theory, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Transition Formula**:
According to Bohr's theory, the energy difference (ΔE) between two energy levels in a hydrogen atom can be calculated using the formula:
\[ \Delta E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final energy levels, respectively.

2. **Identify the Transition**:
We need to evaluate the given transitions and identify the values of \( n_1 \) and \( n_2 \) for each transition. The transitions will be of the form \( n_2 \to n_1 \).

3. **Determine the Energy of Each Transition**:
For each transition, calculate the energy difference using the formula. The transition that results in the smallest value of ΔE will correspond to the least energetic photon.

4. **Evaluate the Inverse Relationship**:
The energy difference is inversely related to the squares of the principal quantum numbers. This means that a transition with larger values of \( n_1 \) and \( n_2 \) will yield a smaller energy difference.

5. **Select the Transition with the Largest Quantum Numbers**:
Among the given transitions, identify the one with the largest values of \( n_1 \) and \( n_2 \). The transition with the largest quantum numbers will result in the least energetic photon.

6. **Conclusion**:
After evaluating the options, the transition that has the largest values of \( n_1 \) and \( n_2 \) will be the correct answer.

### Final Answer:
The transition that gives rise to the least energetic photon is option **C**, where \( n_1 = 5 \) and \( n_2 = 6 \).