If the wavelength of series limit of Lyman series for `He^(+)` ions is `x overset(0)(A)`, then what will be the wavelength of series limit of Balmer series for `Li^(+2)` ion?

Options

A. `(9x)/(4) overset(0)(A)`
B. `(16x)/(9) overset(0)(A)`
C. `(5x)/(4)overset(0)(A)`
D. `(4x)/(9)overset(0)(A)`

Explanation

To solve the problem, we need to find the wavelength of the series limit of the Balmer series for the \(Li^{2+}\) ion, given the wavelength of the series limit of the Lyman series for \(He^{+}\) ions.

### Step-by-Step Solution:

1. **Understanding the Series Limits**:
- For the Lyman series, the transition occurs from \(n_2 = \infty\) to \(n_1 = 1\).
- For the Balmer series, the transition occurs from \(n_2 = \infty\) to \(n_1 = 2\).

2. **Using the Rydberg Formula**:
The Rydberg formula for the wavelength (\(\lambda\)) of the emitted light is given by:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(R_H\) is the Rydberg constant,
- \(Z\) is the atomic number,
- \(n_1\) and \(n_2\) are the principal quantum numbers.

3. **Calculating for \(He^{+}\) (Z = 2)**:
- For the Lyman series:
- \(n_1 = 1\)
- \(n_2 = \infty\)
- Substituting into the formula:
\[ \frac{1}{\lambda_{Lyman}} = R_H \cdot (2^2) \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
\[ \frac{1}{\lambda_{Lyman}} = R_H \cdot 4 \cdot 1 = 4R_H \]
Thus,
\[ \lambda_{Lyman} = \frac{1}{4R_H} \]
Given that \(\lambda_{Lyman} = x\) Å, we have:
\[ x = \frac{1}{4R_H} \]

4. **Calculating for \(Li^{2+}\) (Z = 3)**:
- For the Balmer series:
- \(n_1 = 2\)
- \(n_2 = \infty\)
- Substituting into the formula:
\[ \frac{1}{\lambda_{Balmer}} = R_H \cdot (3^2) \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]
\[ \frac{1}{\lambda_{Balmer}} = R_H \cdot 9 \cdot \frac{1}{4} = \frac{9R_H}{4} \]
Thus,
\[ \lambda_{Balmer} = \frac{4}{9R_H} \]

5. **Relating the Two Wavelengths**:
From our earlier results:
\[ \lambda_{Lyman} = x \Rightarrow R_H = \frac{1}{4x} \]
Therefore,
\[ \lambda_{Balmer} = \frac{4}{9R_H} = \frac{4}{9 \cdot \frac{1}{4x}} = \frac{16x}{9} \]

### Final Answer:
The wavelength of the series limit of the Balmer series for \(Li^{2+}\) ion is:
\[ \lambda_{Balmer} = \frac{16}{9} x \text{ Å} \]