Which of the following transitions of an electron in hydrogen atom emits radiation of the lowest wavelength?

Options

A. `n_(2)=infty` to `n_(1)=2`
B. `n_(2)=4` to `n_(1)=3`
C. `n_(2)=2` to `n_(1)=1`
D. `n_(2)=5` to `n_(1)=3`

Answer

To determine which transition of an electron in a hydrogen atom emits radiation of the lowest wavelength, we can use the Rydberg formula for the wavelength of emitted radiation:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \(\lambda\) is the wavelength of emitted radiation,
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number (which is 1 for hydrogen),
- \(n_1\) and \(n_2\) are the principal quantum numbers of the electron's initial and final states, respectively.

### Step-by-Step Solution:

1. **Identify the transitions**: We need to evaluate the transitions provided in the question. Let's assume the transitions are:
- Transition 1: \( n_2 = \infty \) to \( n_1 = 2 \)
- Transition 2: \( n_2 = 4 \) to \( n_1 = 2 \)
- Transition 3: \( n_2 = 3 \) to \( n_1 = 2 \)
- Transition 4: \( n_2 = 5 \) to \( n_1 = 3 \)

2. **Apply the Rydberg formula**: For each transition, we will calculate \(\lambda\) using the rearranged formula:

\[ \lambda = \frac{1}{RZ^2} \cdot \frac{n_1^2 n_2^2}{n_2^2 - n_1^2} \]

Since \(R\) and \(Z\) are constants for hydrogen, we can focus on the term \(\frac{n_1^2 n_2^2}{n_2^2 - n_1^2}\).

3. **Calculate for each transition**:

- **Transition 1**: \( n_2 = \infty \) to \( n_1 = 2 \)
\[ \lambda = \frac{1}{R} \cdot \frac{2^2 \cdot \infty^2}{\infty^2 - 2^2} = \frac{4 \cdot \infty}{\infty} = 0 \quad (\text{wavelength approaches } 0) \]

- **Transition 2**: \( n_2 = 4 \) to \( n_1 = 2 \)
\[ \lambda = \frac{1}{R} \cdot \frac{2^2 \cdot 4^2}{4^2 - 2^2} = \frac{16}{16 - 4} = \frac{16}{12} = \frac{4}{3} \quad (\text{calculate the numerical value}) \]

- **Transition 3**: \( n_2 = 3 \) to \( n_1 = 2 \)
\[ \lambda = \frac{1}{R} \cdot \frac{2^2 \cdot 3^2}{3^2 - 2^2} = \frac{4 \cdot 9}{9 - 4} = \frac{36}{5} = 7.2 \quad (\text{calculate the numerical value}) \]

- **Transition 4**: \( n_2 = 5 \) to \( n_1 = 3 \)
\[ \lambda = \frac{1}{R} \cdot \frac{3^2 \cdot 5^2}{5^2 - 3^2} = \frac{9 \cdot 25}{25 - 9} = \frac{225}{16} = 14.0625 \quad (\text{calculate the numerical value}) \]

4. **Compare the wavelengths**:
- Transition 1: \( \lambda \approx 0 \)
- Transition 2: \( \lambda = \frac{4}{3} \approx 1.33 \)
- Transition 3: \( \lambda = 7.2 \)
- Transition 4: \( \lambda = 14.0625 \)

5. **Conclusion**: The transition that emits radiation of the lowest wavelength is the transition from \( n_2 = \infty \) to \( n_1 = 2 \), which effectively has a wavelength approaching zero.

### Final Answer:
The transition that emits radiation of the lowest wavelength is from \( n = \infty \) to \( n = 2 \).