If the shortest wavelength of H atom in Lyman series is x, then longest wavelength in Balmer series of H–atom is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understand the Lyman Series**:
- The Lyman series corresponds to transitions where an electron falls to the first energy level (\( n_1 = 1 \)).
- The shortest wavelength in the Lyman series occurs when the electron transitions from an infinitely high energy level (\( n_2 = \infty \)) to \( n_1 = 1 \).

2. **Use the Rydberg Formula for Lyman Series**:
- The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For hydrogen (\( Z = 1 \)), substituting \( n_1 = 1 \) and \( n_2 = \infty \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \]
- Therefore, the shortest wavelength \( \lambda \) in the Lyman series is:
\[ \lambda = \frac{1}{R_H} \]
- Given that this shortest wavelength is \( x \), we have:
\[ x = \frac{1}{R_H} \implies R_H = \frac{1}{x} \]

3. **Understand the Balmer Series**:
- The Balmer series corresponds to transitions where an electron falls to the second energy level (\( n_1 = 2 \)).
- The longest wavelength in the Balmer series occurs when the electron transitions from the third energy level (\( n_2 = 3 \)) to \( n_1 = 2 \).

4. **Use the Rydberg Formula for Balmer Series**:
- Again using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For the Balmer series, substituting \( n_1 = 2 \) and \( n_2 = 3 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- This simplifies to:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = R_H \cdot \frac{5}{36} \]

5. **Substituting \( R_H \)**:
- Now substituting \( R_H = \frac{1}{x} \):
\[ \frac{1}{\lambda} = \frac{1}{x} \cdot \frac{5}{36} \]
- Therefore:
\[ \lambda = \frac{36x}{5} \]

### Final Answer:
The longest wavelength in the Balmer series of the hydrogen atom is:
\[ \lambda = \frac{36x}{5} \]