The wavelength will be minimum for which of the following electronic transition in an unielectron species?

Options

A. `n = 6 " to " n = 4 `
B. `n = 4 " to " n = 2 `
C. `n = 3 " to " n = 1`
D. `n = 2 " to " n = 1`

Answer

To determine the electronic transition that results in the minimum wavelength for an unielectron species, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**:
The wavelength (λ) of the emitted or absorbed light during an electronic transition can be calculated using the formula:
\[ \frac{1}{\lambda} = R_z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is a constant (Rydberg constant), \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

2. **Identify the Relationship**:
From the formula, we can see that the wavelength (λ) is inversely related to the term \( \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \). Therefore, to minimize λ, we need to maximize this term.

3. **Calculate for Each Transition**:
We will calculate \( \frac{1}{\lambda} \) for each of the given transitions.

- **Transition 1**: \( n_i = 6, n_f = 4 \)
\[ \frac{1}{\lambda_1} = R_z^2 \left( \frac{1}{4^2} - \frac{1}{6^2} \right) = R_z^2 \left( \frac{1}{16} - \frac{1}{36} \right) = R_z^2 \left( \frac{9 - 4}{144} \right) = R_z^2 \left( \frac{5}{144} \right) \]

- **Transition 2**: \( n_i = 4, n_f = 2 \)
\[ \frac{1}{\lambda_2} = R_z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_z^2 \left( \frac{1}{4} - \frac{1}{16} \right) = R_z^2 \left( \frac{4 - 1}{16} \right) = R_z^2 \left( \frac{3}{16} \right) \]

- **Transition 3**: \( n_i = 3, n_f = 1 \)
\[ \frac{1}{\lambda_3} = R_z^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_z^2 \left( 1 - \frac{1}{9} \right) = R_z^2 \left( \frac{9 - 1}{9} \right) = R_z^2 \left( \frac{8}{9} \right) \]

- **Transition 4**: \( n_i = 2, n_f = 1 \)
\[ \frac{1}{\lambda_4} = R_z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_z^2 \left( 1 - \frac{1}{4} \right) = R_z^2 \left( \frac{4 - 1}{4} \right) = R_z^2 \left( \frac{3}{4} \right) \]

4. **Compare Values**:
Now we can compare the values of \( \frac{1}{\lambda} \) calculated for each transition:
- Transition 1: \( \frac{5}{144} \)
- Transition 2: \( \frac{3}{16} \)
- Transition 3: \( \frac{8}{9} \)
- Transition 4: \( \frac{3}{4} \)

The largest value of \( \frac{1}{\lambda} \) corresponds to the smallest wavelength.

5. **Conclusion**:
The transition with the largest value of \( \frac{1}{\lambda} \) is Transition 3 (from \( n=3 \) to \( n=1 \)), which means it has the minimum wavelength.

### Final Answer:
The wavelength will be minimum for the electronic transition from \( n=3 \) to \( n=1 \).