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 \( \text{Li}^{2+} \) given the shortest wavelength in the Balmer series of \( \text{He}^{+} \) is \( x \).

### Step-by-step Solution:

1. **Understand the Rydberg Formula**:
The Rydberg formula for the wavelength of emitted light during electronic transitions is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the energy levels involved in the transition.

2. **Determine the Shortest Wavelength for \( \text{He}^{+} \)**:
For the Balmer series, the transition occurs from \( n_2 = \infty \) to \( n_1 = 2 \).
Thus, substituting these values into the Rydberg formula for \( \text{He}^{+} \) (where \( Z = 2 \)):
\[ \frac{1}{x} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]
Simplifying:
\[ \frac{1}{x} = R \cdot 4 \left( \frac{1}{4} - 0 \right) = R \]
Therefore, we have:
\[ x = \frac{1}{R} \]

3. **Calculate the Longest Wavelength in the Paschen Series for \( \text{Li}^{2+} \)**:
For the Paschen series, the transition occurs from \( n_2 = 4 \) to \( n_1 = 3 \).
For \( \text{Li}^{2+} \) (where \( Z = 3 \)):
\[ \frac{1}{\lambda} = R \cdot 3^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
Simplifying:
\[ \frac{1}{\lambda} = R \cdot 9 \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda} = R \cdot 9 \left( \frac{16 - 9}{144} \right) = R \cdot 9 \cdot \frac{7}{144} = R \cdot \frac{63}{144} \]
Therefore:
\[ \lambda = \frac{144}{63R} \]

4. **Relate the Wavelengths**:
Now, we can relate \( \lambda \) to \( x \):
\[ \lambda = \frac{144}{63R} = \frac{144}{63} \cdot \frac{1}{x} \]
Simplifying:
\[ \lambda = \frac{144}{63} \cdot x \]

5. **Final Calculation**:
To express \( \lambda \) in terms of \( x \):
\[ \lambda = \frac{144}{63} x = \frac{16}{7} x \]

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