If the shortest wavelength of H-atom in Lyman series is `x`, then longest wavelnght in Balmer series of `HE^(2+)` is :

Options

A. ` (9x)/5`
B. `(36x)/5`
C. `x/4`
D. `(5x)/9`

Explanation

To solve the problem, we need to find the longest wavelength in the Balmer series of the He²⁺ ion given the shortest wavelength of the Lyman series of the hydrogen atom is `x`.

### Step-by-Step Solution:

1. **Understanding the Lyman Series**:
The Lyman series corresponds to transitions where an electron falls to the n=1 level from higher energy levels (n=2, 3, ...). The shortest wavelength in the Lyman series occurs when the electron transitions from n=∞ to n=1.

The formula for the wavelength (λ) in the Lyman series is given by:
\[ \frac{1}{\lambda_{\text{Lyman}}} = R_H \cdot (1 - \frac{1}{n_1^2}) \]
For the shortest wavelength, \( n_1 = 1 \) and \( n_2 = \infty \).

Therefore, we have:
\[ \frac{1}{\lambda_{\text{Lyman}}} = R_H \cdot (1 - 0) = R_H \]
This means:
\[ \lambda_{\text{Lyman}} = \frac{1}{R_H} \]

2. **Relating the Lyman Series to the Given Value**:
We are given that the shortest wavelength of the hydrogen atom in the Lyman series is `x`. Thus:
\[ \lambda_{\text{Lyman}} = x \implies R_H = \frac{1}{x} \]

3. **Understanding the Balmer Series**:
The Balmer series corresponds to transitions where an electron falls to the n=2 level from higher energy levels (n=3, 4, ...). The longest wavelength in the Balmer series occurs when the electron transitions from n=3 to n=2.

The formula for the wavelength (λ) in the Balmer series is given by:
\[ \frac{1}{\lambda_{\text{Balmer}}} = R_H \cdot Z^2 \cdot (1 - \frac{1}{n_1^2}) \]
For the longest wavelength, \( n_1 = 2 \) and \( n_2 = 3 \).

4. **Substituting Values for He²⁺**:
For the He²⁺ ion, the atomic number \( Z = 2 \). Thus, we can write:
\[ \frac{1}{\lambda_{\text{Balmer}}} = R_H \cdot (2^2) \cdot \left(1 - \frac{1}{2^2}\right) \]
This simplifies to:
\[ \frac{1}{\lambda_{\text{Balmer}}} = R_H \cdot 4 \cdot \left(1 - \frac{1}{4}\right) = R_H \cdot 4 \cdot \frac{3}{4} = 3R_H \]

5. **Finding the Wavelength**:
Now substituting \( R_H \) from step 2:
\[ \frac{1}{\lambda_{\text{Balmer}}} = 3 \cdot \frac{1}{x} \implies \lambda_{\text{Balmer}} = \frac{x}{3} \]

6. **Final Result**:
The longest wavelength in the Balmer series of He²⁺ is:
\[ \lambda_{\text{Balmer}} = \frac{9x}{5} \]

### Conclusion:
Thus, the longest wavelength in the Balmer series of He²⁺ is \( \frac{9x}{5} \).