Of the following transitions in the hydrogen atom, the one which gives on emission line of heighest frequency is

Options

A. `n=1` to `n=2`
B. `n=3` to `n=10`
C. `n=10` to `n=3`
D. `n=2` to `n=1`

Explanation

To solve the question of which transition in the hydrogen atom gives an emission line of the highest frequency, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the transitions**: The transitions given are:
- A: n = 1 to n = 2
- B: n = 3 to n = 10
- C: n = 10 to n = 3
- D: n = 2 to n = 1

2. **Understand emission vs. absorption**:
- Emission occurs when an electron transitions from a higher energy level to a lower energy level, releasing energy in the form of a photon.
- Absorption occurs when an electron moves from a lower energy level to a higher energy level, requiring energy input.

3. **Analyze each transition**:
- **Transition A (n = 1 to n = 2)**: This is an absorption transition (electron moves to a higher energy level).
- **Transition B (n = 3 to n = 10)**: This is also an absorption transition (electron moves to a higher energy level).
- **Transition C (n = 10 to n = 3)**: This is an emission transition (electron moves to a lower energy level).
- **Transition D (n = 2 to n = 1)**: This is also an emission transition (electron moves to a lower energy level).

4. **Use Rydberg's formula**: The frequency of the emitted photon can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

5. **Calculate the frequency for each emission transition**:
- **For Transition C (n = 10 to n = 3)**:
\[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{10^2} \right) = R \left( \frac{1}{9} - \frac{1}{100} \right) = R \left( \frac{100 - 9}{900} \right) = R \left( \frac{91}{900} \right) \]
- **For Transition D (n = 2 to n = 1)**:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

6. **Compare the magnitudes**:
- The transition with the higher frequency corresponds to the larger value of \(\frac{1}{\lambda}\).
- For Transition C, \(\frac{91}{900}\) is smaller than \(\frac{3}{4}\) from Transition D.

7. **Conclusion**: Since Transition D (n = 2 to n = 1) has a higher \(\frac{1}{\lambda}\) value, it corresponds to the highest frequency emission line.

### Final Answer:
The transition that gives an emission line of the highest frequency is **n = 2 to n = 1 (Option D)**.