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

Options

A. `(9x)/(4)Å`
B. `(16x)/(9)Å`
C. `(5x)/(4)Å`
D. `(4x)/(7)Å`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Lyman Series for \( \text{He}^+ \)**:
- The Lyman series corresponds to transitions from higher energy levels (n = 2, 3, ...) to the ground state (n = 1).
- The formula for the wavelength (\( \lambda \)) is given by:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For the Lyman series, \( n_1 = 1 \) and \( n_2 = \infty \).

2. **Applying the Formula for \( \text{He}^+ \)**:
- For \( \text{He}^+ \), the atomic number \( Z = 2 \).
- Thus, we have:
\[ \frac{1}{\lambda} = R_H \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \cdot 4 \cdot (1 - 0) = 4 R_H \]
- Therefore, the wavelength is:
\[ \lambda = \frac{1}{4 R_H} \]
- Given that this wavelength is \( x \) Å, we can write:
\[ x = \frac{1}{4 R_H} \]

3. **Finding the Wavelength for the Balmer Series of \( \text{Li}^{2+} \)**:
- The Balmer series corresponds to transitions from higher energy levels (n = 3, 4, ...) to the first excited state (n = 2).
- For the Balmer series, we set \( n_1 = 2 \) and \( n_2 = \infty \).
- The atomic number for \( \text{Li}^{2+} \) is \( Z = 3 \).

4. **Applying the Formula for \( \text{Li}^{2+} \)**:
- Using the same formula:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For \( \text{Li}^{2+} \):
\[ \frac{1}{\lambda} = R_H \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R_H \cdot 9 \cdot \left( \frac{1}{4} - 0 \right) = \frac{9 R_H}{4} \]
- Thus, the wavelength is:
\[ \lambda = \frac{4}{9 R_H} \]

5. **Relating to the Given Wavelength \( x \)**:
- We know that \( x = \frac{1}{4 R_H} \).
- Therefore, we can express \( R_H \) in terms of \( x \):
\[ R_H = \frac{1}{4x} \]
- Substituting this into the wavelength for \( \text{Li}^{2+} \):
\[ \lambda = \frac{4}{9} \cdot \frac{1}{R_H} = \frac{4}{9} \cdot 4x = \frac{16}{9} x \]

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