Which one among the following transitions of electron of hydrogen atom emits radiation of the shortest wavelength ?

Options

A. n=2 to n=1
B. n=3 to n=2
C. n=4 to n=3
D. n=5 to n=4

Explanation

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

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the emitted radiation,
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

The key point is that the shorter the wavelength (\( \lambda \)), the larger the value of \( \frac{1}{\lambda} \) will be. This means we need to find the transition that gives the largest value of \( \frac{1}{n_f^2} - \frac{1}{n_i^2} \).

### Step-by-Step Solution:

1. **Identify the transitions**: The possible transitions are typically given in the options, for example:
- Transition 1: \( n_i = 2 \) to \( n_f = 1 \)
- Transition 2: \( n_i = 3 \) to \( n_f = 2 \)
- Transition 3: \( n_i = 4 \) to \( n_f = 3 \)
- Transition 4: \( n_i = 5 \) to \( n_f = 4 \)

2. **Calculate \( \frac{1}{\lambda} \) for each transition**:
- **Transition 1**: \( n_i = 2 \) to \( n_f = 1 \)
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- **Transition 2**: \( n_i = 3 \) to \( n_f = 2 \)
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
- **Transition 3**: \( n_i = 4 \) to \( n_f = 3 \)
\[ \frac{1}{\lambda_3} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
- **Transition 4**: \( n_i = 5 \) to \( n_f = 4 \)
\[ \frac{1}{\lambda_4} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25 - 16}{400} \right) = R \left( \frac{9}{400} \right) \]

3. **Compare the values**:
- Transition 1: \( \frac{3}{4} \)
- Transition 2: \( \frac{5}{36} \)
- Transition 3: \( \frac{7}{144} \)
- Transition 4: \( \frac{9}{400} \)

4. **Determine the largest value**:
- The largest value corresponds to Transition 1 (\( n_i = 2 \) to \( n_f = 1 \)), which means this transition emits radiation of the shortest wavelength.

### Final Answer:
The transition of the electron in the hydrogen atom that emits radiation of the shortest wavelength is from \( n = 2 \) to \( n = 1 \).