Which of the following species will produce the shortest wavelength for the transition `n = 2 "to" n= 1` ?

Options

A. Hydrogen atom
B. Singly ionised helium
C. Deuterium atom
D. Dioubly ionised lithium

Explanation

To determine which species will produce the shortest wavelength for the transition from \( n = 2 \) to \( n = 1 \), we can use the Rydberg formula 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 (which is the same for all hydrogen-like species),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states.

### Step 1: Identify the values of \( n_1 \) and \( n_2 \)
For the transition from \( n = 2 \) to \( n = 1 \):
- \( n_1 = 1 \)
- \( n_2 = 2 \)

### Step 2: Write the formula for \( \frac{1}{\lambda} \)
Substituting the values of \( n_1 \) and \( n_2 \) into the Rydberg formula gives:

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \cdot Z^2 \left( 1 - \frac{1}{4} \right) = R_H \cdot Z^2 \cdot \frac{3}{4} \]

### Step 3: Determine the atomic numbers \( Z \) for each species
1. **Hydrogen atom (H)**: \( Z = 1 \)
2. **Singly ionized helium (He\(^+\))**: \( Z = 2 \)
3. **Deuterium (D)**: \( Z = 1 \) (Deuterium is an isotope of hydrogen)
4. **Doubly ionized lithium (Li\(^{2+}\))**: \( Z = 3 \)

### Step 4: Calculate \( Z^2 \) for each species
- For Hydrogen: \( Z^2 = 1^2 = 1 \)
- For Singly ionized Helium: \( Z^2 = 2^2 = 4 \)
- For Deuterium: \( Z^2 = 1^2 = 1 \)
- For Doubly ionized Lithium: \( Z^2 = 3^2 = 9 \)

### Step 5: Compare the values of \( Z^2 \)
The value of \( \frac{1}{\lambda} \) is proportional to \( Z^2 \). The larger the value of \( Z^2 \), the shorter the wavelength.

- Hydrogen: \( 1 \)
- Singly ionized Helium: \( 4 \)
- Deuterium: \( 1 \)
- Doubly ionized Lithium: \( 9 \)

### Step 6: Identify the species with the highest \( Z^2 \)
The species with the highest \( Z^2 \) is the doubly ionized lithium (Li\(^{2+}\)) with \( Z^2 = 9 \).

### Conclusion
Thus, the species that will produce the shortest wavelength for the transition from \( n = 2 \) to \( n = 1 \) is:

**Doubly ionized lithium (Li\(^{2+}\))**

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