What is the wavelength in nm of the spectral line associated with a transition from n=3 to n= 2 for the `Li^(2+)` ion?

Options

A. 219
B. 656
C. `73.0`
D. 486

Answer

To find the wavelength of the spectral line associated with the transition from n=3 to n=2 for the \( \text{Li}^{2+} \) ion, we can use the Rydberg formula for hydrogen-like ions:

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

Where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant (\( R = 1.09 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for \( \text{Li}^{2+} \), \( Z = 3 \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states (here, \( n_1 = 2 \) and \( n_2 = 3 \)).

### Step-by-Step Solution:

1. **Identify the values**:
- \( R = 1.09 \times 10^7 \, \text{m}^{-1} \)
- \( Z = 3 \) (for lithium ion)
- \( n_1 = 2 \)
- \( n_2 = 3 \)

2. **Substitute the values into the Rydberg formula**:
\[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
\[ \frac{1}{\lambda} = 1.09 \times 10^7 \times 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

3. **Calculate \( Z^2 \)**:
\[ Z^2 = 3^2 = 9 \]

4. **Calculate \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \)**:
\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{3^2} = \frac{1}{9} \]
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

5. **Substitute back into the equation**:
\[ \frac{1}{\lambda} = 1.09 \times 10^7 \times 9 \times \frac{5}{36} \]

6. **Calculate the right-hand side**:
\[ \frac{1}{\lambda} = 1.09 \times 10^7 \times 9 \times \frac{5}{36} = 1.09 \times 10^7 \times \frac{45}{36} \]
\[ = 1.09 \times 10^7 \times 1.25 = 1.3625 \times 10^7 \, \text{m}^{-1} \]

7. **Calculate \( \lambda \)**:
\[ \lambda = \frac{1}{1.3625 \times 10^7} \approx 7.34 \times 10^{-8} \, \text{m} \]

8. **Convert to nanometers**:
\[ \lambda = 7.34 \times 10^{-8} \, \text{m} \times 10^9 \, \text{nm/m} = 73.4 \, \text{nm} \]

### Final Answer:
The wavelength of the spectral line associated with the transition from \( n=3 \) to \( n=2 \) for the \( \text{Li}^{2+} \) ion is approximately **73 nm**.