what is the shortest wavelength line in Paschen series of `Li^(2+)ion` (R is Rydberg constant)

Options

A. `1/R`
B. `3/R`
C. `16`/`7R`
D. 4R

Answer

To find the shortest wavelength line in the Paschen series of the \( Li^{2+} \) ion, we can use the Rydberg formula for hydrogen-like ions. The formula is given by:

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

where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number of the ion,
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step-by-Step Solution:

1. **Identify the Atomic Number (Z)**:
For \( Li^{2+} \), the atomic number \( Z \) is 3.

2. **Determine the Values of \( n_1 \) and \( n_2 \)**:
In the Paschen series, the transition occurs to \( n_1 = 3 \). The value of \( n_2 \) can be any integer greater than \( n_1 \). To find the shortest wavelength, we need the smallest possible value of \( n_2 \), which is 4.

3. **Substitute Values into the Rydberg Formula**:
Using \( n_1 = 3 \) and \( n_2 = 4 \):
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
\[ \frac{1}{\lambda} = R \cdot 3^2 \left( \frac{1}{9} - \frac{1}{16} \right) \]

4. **Calculate the Difference**:
First, calculate \( \frac{1}{9} - \frac{1}{16} \):
\[ \frac{1}{9} = \frac{16}{144}, \quad \frac{1}{16} = \frac{9}{144} \]
\[ \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \]

5. **Plug This Back into the Formula**:
Now substitute back into the equation:
\[ \frac{1}{\lambda} = R \cdot 9 \cdot \frac{7}{144} \]
\[ \frac{1}{\lambda} = \frac{63R}{144} \]

6. **Calculate \( \lambda \)**:
Taking the reciprocal gives:
\[ \lambda = \frac{144}{63R} \]
Simplifying this:
\[ \lambda = \frac{16}{7R} \]

### Final Answer:
The shortest wavelength line in the Paschen series of the \( Li^{2+} \) ion is:
\[ \lambda = \frac{16}{7R} \]