Which of the following transitions in a hydrogen atom emits photon of the highest frequency ?

Options

A. n = 1 to n = 2
B. n = 2 to n = 1
C. n = 2 to n = 6
D. n = 6 to n = 2

Answer

To determine which transition in a hydrogen atom emits a photon of the highest frequency, we can follow these steps:

### Step 1: Understand the Concept of Energy Levels
In a hydrogen atom, electrons occupy discrete energy levels, denoted by quantum numbers (n). The energy of these levels increases with n. A transition from a higher energy level (higher n) to a lower energy level (lower n) results in the emission of a photon.

### Step 2: Identify the Given Transitions
We need to analyze the transitions provided in the options. Let's denote them as follows:
1. Transition from n = 2 to n = 1
2. Transition from n = 3 to n = 1
3. Transition from n = 4 to n = 2
4. Transition from n = 6 to n = 2

### Step 3: Use the Formula for Frequency of Emitted Photon
The frequency (f) of the emitted photon during a transition can be calculated using the formula:
\[ f = R \cdot c \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \]
where R is the Rydberg constant and c is the speed of light.

### Step 4: Calculate Frequency for Each Transition
1. **For n = 2 to n = 1**:
\[ f_1 = R \cdot c \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot c \left( 1 - \frac{1}{4} \right) = R \cdot c \cdot \frac{3}{4} \]

2. **For n = 3 to n = 1**:
\[ f_2 = R \cdot c \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \cdot c \left( 1 - \frac{1}{9} \right) = R \cdot c \cdot \frac{8}{9} \]

3. **For n = 4 to n = 2**:
\[ f_3 = R \cdot c \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \cdot c \left( \frac{1}{4} - \frac{1}{16} \right) = R \cdot c \cdot \frac{3}{16} \]

4. **For n = 6 to n = 2**:
\[ f_4 = R \cdot c \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = R \cdot c \left( \frac{1}{4} - \frac{1}{36} \right) = R \cdot c \left( \frac{9 - 1}{36} \right) = R \cdot c \cdot \frac{8}{36} = R \cdot c \cdot \frac{2}{9} \]

### Step 5: Compare the Frequencies
Now we compare the calculated frequencies:
- \( f_1 = R \cdot c \cdot \frac{3}{4} \)
- \( f_2 = R \cdot c \cdot \frac{8}{9} \)
- \( f_3 = R \cdot c \cdot \frac{3}{16} \)
- \( f_4 = R \cdot c \cdot \frac{2}{9} \)

### Step 6: Determine the Highest Frequency
To find the highest frequency, we can compare the fractions:
- \( \frac{3}{4} \) is approximately 0.75
- \( \frac{8}{9} \) is approximately 0.89
- \( \frac{3}{16} \) is approximately 0.1875
- \( \frac{2}{9} \) is approximately 0.2222

The highest frequency corresponds to the transition from n = 3 to n = 1, which is \( f_2 \).

### Conclusion
The transition in a hydrogen atom that emits a photon of the highest frequency is from n = 3 to n = 1.

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