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)`

Explanation

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

### Step 1: Identify the atomic number and the series
The atomic number (Z) of lithium (Li) is 3. The Paschen series corresponds to transitions where the electron falls to the n=3 energy level.

### Step 2: Write the formula for the wavelength
The formula for the wavelength (λ) in the hydrogen-like atom is given by:

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

where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Set the values for the Paschen series
In the Paschen series, the transition starts from \( n_1 = 3 \). To find the shortest wavelength, we consider the transition from \( n_2 \) to infinity (i.e., \( n_2 \to \infty \)).

### Step 4: Substitute the values into the formula
Substituting \( n_1 = 3 \) and \( n_2 = \infty \) 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:

\[ \frac{1}{\lambda} = R \cdot Z^2 \cdot \frac{1}{3^2} \]

### Step 5: Calculate \( Z^2 \)
For lithium, \( Z = 3 \), so:

\[ Z^2 = 3^2 = 9 \]

### Step 6: Substitute \( Z^2 \) back into the equation
Now substituting \( Z^2 \) into the equation:

\[ \frac{1}{\lambda} = R \cdot 9 \cdot \frac{1}{9} = R \]

### Step 7: Solve for \( \lambda \)
Thus, we find:

\[ \lambda = \frac{1}{R} \]

### Conclusion
The shortest wavelength line in the Paschen series of the `Li^(2+)` ion is given by:

\[ \lambda = \frac{1}{R} \]