The shortest wavelength of H-atom in Lyman series is x, then longest wavelength in Balmer series of `He^(+)` is

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Understand the Lyman Series**:
The Lyman series corresponds to transitions where electrons fall to the n=1 energy 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 Rydberg formula is:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Apply the Rydberg Formula for Hydrogen**:
For hydrogen (Z=1), the shortest wavelength (x) in the Lyman series is given by:
\[ \frac{1}{x} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
Simplifying this gives:
\[ \frac{1}{x} = R \cdot 1 \cdot 1 = R \implies x = \frac{1}{R} \]

3. **Understand the Balmer Series**:
The Balmer series corresponds to transitions where electrons fall to the n=2 energy 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.

4. **Apply the Rydberg Formula for He⁺**:
For the He⁺ ion (Z=2), the longest wavelength in the Balmer series is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \( n_1 = 2 \) and \( n_2 = 3 \):
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = 4R \left( \frac{1}{4} - \frac{1}{9} \right) \]

5. **Calculate the Difference**:
To find \( \frac{1}{4} - \frac{1}{9} \), we need a common denominator:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \implies \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda} = 4R \cdot \frac{5}{36} = \frac{20R}{36} = \frac{5R}{9} \]

6. **Find λ**:
Therefore, the wavelength \( \lambda \) is:
\[ \lambda = \frac{9}{5R} \]

7. **Relate λ to x**:
Since we found that \( x = \frac{1}{R} \), we can express \( R \) in terms of \( x \):
\[ R = \frac{1}{x} \]
Substituting this into the equation for \( \lambda \):
\[ \lambda = \frac{9}{5 \cdot \frac{1}{x}} = \frac{9x}{5} \]

### Final Answer:
The longest wavelength in the Balmer series of He⁺ is:
\[ \lambda = \frac{9x}{5} \]