The shortest wavelength of `He^(+)` in Balmer series is `x`. Then longest wavelength in the Paschene series of `Li^(+2)` is :-

Options

A. `(36x)/(5)`
B. `(16x)/(7)`
C. `(9x)/(5)`
D. `(5x)/(9)`

Explanation

To solve the problem, we need to find the longest wavelength in the Paschen series of the lithium ion \(Li^{2+}\) given that the shortest wavelength of \(He^{+}\) in the Balmer series is \(x\).

### Step-by-Step Solution:

1. **Understanding the Balmer Series for \(He^{+}\)**:
The Balmer series corresponds to transitions where the final energy level \(n_1 = 2\). The formula for the wavelength \(\lambda\) in the Balmer series is given by the Rydberg formula:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \(Z\) is the atomic number, \(n_1\) is the lower energy level (2 for Balmer series), and \(n_2\) is the higher energy level.

2. **Calculating the Shortest Wavelength for \(He^{+}\)**:
For \(He^{+}\) (where \(Z = 2\)):
- \(n_1 = 2\)
- For the shortest wavelength, \(n_2\) approaches infinity.
Thus, the equation becomes:
\[ \frac{1}{\lambda} = R_H \cdot 2^2 \left( \frac{1}{2^2} - 0 \right) = R_H \cdot 4 \cdot \frac{1}{4} = R_H \]
Therefore, the shortest wavelength \(\lambda\) for \(He^{+}\) is:
\[ \lambda_{He^{+}} = \frac{1}{R_H} = x \]

3. **Understanding the Paschen Series for \(Li^{2+}\)**:
The Paschen series corresponds to transitions where the final energy level \(n_1 = 3\). The longest wavelength occurs when \(n_2 = 4\).

4. **Calculating the Longest Wavelength for \(Li^{2+}\)**:
For \(Li^{2+}\) (where \(Z = 3\)):
- \(n_1 = 3\)
- \(n_2 = 4\)
The formula for the wavelength becomes:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
Substituting \(Z = 3\):
\[ \frac{1}{\lambda} = R_H \cdot 3^2 \left( \frac{1}{9} - \frac{1}{16} \right) \]
Calculating \(\frac{1}{9} - \frac{1}{16}\):
\[ \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \]
Thus, we have:
\[ \frac{1}{\lambda} = R_H \cdot 9 \cdot \frac{7}{144} = \frac{63 R_H}{144} \]
Therefore:
\[ \lambda_{Li^{2+}} = \frac{144}{63 R_H} \]

5. **Relating \(R_H\) to \(x\)**:
Since we established that \(R_H = \frac{1}{x}\):
\[ \lambda_{Li^{2+}} = \frac{144}{63 \cdot \frac{1}{x}} = \frac{144x}{63} = \frac{16x}{7} \]

### Final Answer:
The longest wavelength in the Paschen series of \(Li^{2+}\) is:
\[ \lambda_{Li^{2+}} = \frac{16x}{7} \]