According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the less energetic photon?

Options

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

Explanation

To determine which transition in the hydrogen atom gives rise to the least energetic photon according to Bohr's theory, we need to analyze the transitions based on their principal quantum numbers (n). The energy of a photon emitted during a transition is inversely proportional to the square of the principal quantum number (n).

### Step-by-Step Solution:

1. **Understand the Energy Relation**: According to Bohr's theory, the energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number. The energy difference (\( \Delta E \)) between two levels \( n_1 \) and \( n_2 \) is given by:
\[ \Delta E = E_{n_2} - E_{n_1} = -\frac{13.6 \, \text{eV}}{n_2^2} + \frac{13.6 \, \text{eV}}{n_1^2} \]

2. **Identify the Transitions**: Let's denote the transitions given in the question as follows:
- Transition 1: \( n_1 = 6 \) to \( n_2 = 5 \)
- Transition 2: \( n_1 = 5 \) to \( n_2 = 3 \)
- Transition 3: \( n_1 = 6 \) to \( n_2 = 1 \)
- Transition 4: \( n_1 = 5 \) to \( n_2 = 4 \)

3. **Calculate Energy Differences**:
- For Transition 1 (\( n_1 = 6 \) to \( n_2 = 5 \)):
\[ \Delta E_1 = -\frac{13.6}{5^2} + \frac{13.6}{6^2} = -\frac{13.6}{25} + \frac{13.6}{36} \]
\[ \Delta E_1 = 0.012 \, \text{eV} \]

- For Transition 2 (\( n_1 = 5 \) to \( n_2 = 3 \)):
\[ \Delta E_2 = -\frac{13.6}{3^2} + \frac{13.6}{5^2} = -\frac{13.6}{9} + \frac{13.6}{25} \]
\[ \Delta E_2 \approx 0.07 \, \text{eV} \]

- For Transition 3 (\( n_1 = 6 \) to \( n_2 = 1 \)):
\[ \Delta E_3 = -\frac{13.6}{1^2} + \frac{13.6}{6^2} = -13.6 + \frac{13.6}{36} \]
\[ \Delta E_3 \approx 0.09 \, \text{eV} \]

- For Transition 4 (\( n_1 = 5 \) to \( n_2 = 4 \)):
\[ \Delta E_4 = -\frac{13.6}{4^2} + \frac{13.6}{5^2} = -\frac{13.6}{16} + \frac{13.6}{25} \]
\[ \Delta E_4 \approx 0.02 \, \text{eV} \]

4. **Compare Energy Values**: After calculating the energy differences, we can summarize:
- \( \Delta E_1 \approx 0.012 \, \text{eV} \)
- \( \Delta E_2 \approx 0.07 \, \text{eV} \)
- \( \Delta E_3 \approx 0.09 \, \text{eV} \)
- \( \Delta E_4 \approx 0.02 \, \text{eV} \)

5. **Conclusion**: The transition that gives rise to the least energetic photon is Transition 1 (from \( n = 6 \) to \( n = 5 \)) with an energy of approximately \( 0.012 \, \text{eV} \).

### Final Answer:
The transition that gives rise to the less energetic photon is from \( n = 6 \) to \( n = 5 \).