For the transition from `n=2 to n=1` , which of the following will produce shortest wavelength ?

Options

A. H atom
B. D atom
C. `He^(+) ion`
D. `Li^(2+)`ion

Answer

To solve the question regarding which transition from \( n=2 \) to \( n=1 \) produces the shortest wavelength, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**:
The wavelength (\( \lambda \)) of light emitted during an electronic transition can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively.

2. **Identify the Transition**:
In this case, we have a transition from \( n_2 = 2 \) to \( n_1 = 1 \). Thus, we can substitute these values into the formula:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H Z^2 \left( 1 - \frac{1}{4} \right) = R_H Z^2 \left( \frac{3}{4} \right) \]

3. **Determine the Wavelength Dependency**:
From the formula, we can see that the wavelength (\( \lambda \)) is inversely proportional to \( Z^2 \):
\[ \lambda \propto \frac{1}{Z^2} \]
This means that as the atomic number \( Z \) increases, the wavelength decreases.

4. **Evaluate the Options**:
We have the following options based on different ions:
- Hydrogen (\( Z = 1 \))
- Deuterium (\( Z = 1 \))
- Helium ion (\( He^+ \), \( Z = 2 \))
- Lithium ion (\( Li^{2+} \), \( Z = 3 \))

Now, we calculate \( Z^2 \) for each:
- For Hydrogen: \( Z^2 = 1^2 = 1 \)
- For Deuterium: \( Z^2 = 1^2 = 1 \)
- For Helium: \( Z^2 = 2^2 = 4 \)
- For Lithium: \( Z^2 = 3^2 = 9 \)

5. **Identify the Shortest Wavelength**:
Since \( \lambda \) is inversely proportional to \( Z^2 \), the ion with the highest \( Z^2 \) will have the shortest wavelength.
- Lithium ion (\( Li^{2+} \)) has the highest \( Z^2 = 9 \).

6. **Conclusion**:
Therefore, the transition from \( n=2 \) to \( n=1 \) for the lithium ion (\( Li^{2+} \)) will produce the shortest wavelength.

### Final Answer:
The correct option is **Lithium ion (\( Li^{2+} \))**.

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