What is the shortest wavelength line in the Paschen series of `Li^(2+)` ion?

Options

A. `R/9`
B. `9/R`
C. `1/R`
D. `(9R)/4`

Answer

To find the shortest wavelength line in the Paschen series of the `Li^(2+)` ion, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Paschen Series**: The Paschen series corresponds to electron transitions from higher energy levels (n ≥ 4) to the n = 3 energy level in a hydrogen-like atom. For `Li^(2+)`, we are looking for transitions to n = 3.

2. **Use the Rydberg Formula**: The wavelength of the emitted light can be calculated using the Rydberg formula for hydrogen-like ions:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

3. **Identify the Values**: For `Li^(2+)`, the atomic number \( Z = 3 \). For the shortest wavelength in the Paschen series, we consider the transition from \( n_2 = \infty \) to \( n_1 = 3 \).

4. **Substitute the Values into the Formula**:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \frac{1}{\lambda} = R \cdot Z^2 \cdot \frac{1}{3^2} \]
\[ \frac{1}{\lambda} = R \cdot 3^2 \cdot \frac{1}{9} \]
\[ \frac{1}{\lambda} = R \]

5. **Calculate the Wavelength**:
\[ \lambda = \frac{1}{R} \]
This gives us the shortest wavelength in the Paschen series for the `Li^(2+)` ion.

6. **Conclusion**: The shortest wavelength line in the Paschen series of the `Li^(2+)` ion is given by:
\[ \lambda = \frac{1}{R} \]