In the hydrogen atom spectrum, the least energetic photon takes place during the transition from n = 6 energy level to n = ……… energy level.

Options

A. 1
B. 3
C. 5
D. 4

Answer

To solve the problem of determining the energy level to which an electron in a hydrogen atom transitions from n = 6 to emit the least energetic photon, we can follow these steps:

### Step 1: Understand the Energy Level Transition
In a hydrogen atom, the energy levels are quantized and represented by the principal quantum number \( n \). The energy of a photon emitted during a transition between two energy levels is given by the formula:

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

where \( n_1 \) is the lower energy level and \( n_2 \) is the higher energy level.

### Step 2: Identify the Transition
We are given that the electron is transitioning from \( n = 6 \). To find the least energetic photon, we need to identify the lowest possible transition from \( n = 6 \) to another energy level \( n_1 \).

### Step 3: Calculate Energy Differences
The least energetic photon will correspond to the smallest difference in energy levels. The possible transitions from \( n = 6 \) are:
- \( n = 6 \) to \( n = 5 \)
- \( n = 6 \) to \( n = 4 \)
- \( n = 6 \) to \( n = 3 \)
- \( n = 6 \) to \( n = 2 \)
- \( n = 6 \) to \( n = 1 \)

Among these, the transition from \( n = 6 \) to \( n = 5 \) has the smallest difference in energy levels.

### Step 4: Conclusion
Thus, the least energetic photon is emitted during the transition from \( n = 6 \) to \( n = 5 \).

**Final Answer: \( n = 5 \)**

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