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 determine which transition from \( n=2 \) to \( n=1 \) produces the shortest wavelength, we can use the formula derived from Bohr's theory for hydrogen-like atoms:

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

Where:
- \( \lambda \) is the wavelength,
- \( R_H \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) is the lower energy level (1 in this case),
- \( n_2 \) is the higher energy level (2 in this case).

### Step-by-Step Solution:

1. **Identify the Transition**:
We are looking at the transition from \( n=2 \) to \( n=1 \).

2. **Substitute Values**:
For this transition, \( n_1 = 1 \) and \( n_2 = 2 \). The formula becomes:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \cdot Z^2 \left( 1 - \frac{1}{4} \right) = R_H \cdot Z^2 \cdot \frac{3}{4} \]

3. **Determine Atomic Numbers**:
- For Hydrogen (H): \( Z = 1 \)
- For Deuterium (D): \( Z = 1 \) (it is an isotope of hydrogen)
- For Helium ion (He\(^+\)): \( Z = 2 \)
- For Lithium ion (Li\(^{2+}\)): \( Z = 3 \)

4. **Calculate \( \frac{1}{\lambda} \)** for each atom:
- For H:
\[ \frac{1}{\lambda} = R_H \cdot (1^2) \cdot \frac{3}{4} = \frac{3}{4} R_H \]
- For D:
\[ \frac{1}{\lambda} = R_H \cdot (1^2) \cdot \frac{3}{4} = \frac{3}{4} R_H \]
- For He\(^+\):
\[ \frac{1}{\lambda} = R_H \cdot (2^2) \cdot \frac{3}{4} = 3 R_H \]
- For Li\(^{2+}\):
\[ \frac{1}{\lambda} = R_H \cdot (3^2) \cdot \frac{3}{4} = \frac{27}{4} R_H \]

5. **Compare Values**:
The values of \( \frac{1}{\lambda} \) are:
- H: \( \frac{3}{4} R_H \)
- D: \( \frac{3}{4} R_H \)
- He\(^+\): \( 3 R_H \)
- Li\(^{2+}\): \( \frac{27}{4} R_H \)

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

6. **Conclusion**:
The highest value is for Li\(^{2+}\) with \( \frac{27}{4} R_H \), indicating that the transition from \( n=2 \) to \( n=1 \) in Li\(^{2+}\) produces the shortest wavelength.

Thus, the answer is **Li\(^{2+}\)** (option no. 4).