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 step by step, 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 1: Understand the Lyman Series
The Lyman series corresponds to transitions where an electron falls to the \( n=1 \) energy level from higher levels (\( n=2, 3, 4, \ldots \)). The shortest wavelength in the Lyman series occurs when the transition is from \( n=\infty \) to \( n=1 \).

### Step 2: Use the Rydberg Formula for Lyman Series
The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the shortest wavelength in the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = \infty \)

Substituting these values into the formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \left( 1 - 0 \right) = R_H \]
Thus, the shortest wavelength \( \lambda \) in the Lyman series is:
\[ \lambda = \frac{1}{R_H} \]
Given that this is equal to \( x \), we have:
\[ x = \frac{1}{R_H} \]

### Step 3: Understand the Balmer Series
The Balmer series corresponds to transitions where an electron falls to the \( n=2 \) energy level from higher levels (\( n=3, 4, 5, \ldots \)). The longest wavelength in the Balmer series occurs when the transition is from \( n=3 \) to \( n=2 \).

### Step 4: Use the Rydberg Formula for Balmer Series
For the longest wavelength in the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the right-hand side:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus, we have:
\[ \frac{1}{\lambda} = R_H \cdot \frac{5}{36} \]
This implies:
\[ \lambda = \frac{36}{5 R_H} \]

### Step 5: Substitute \( R_H \) with \( x \)
From Step 2, we know that \( R_H = \frac{1}{x} \). Substituting this into the equation for \( \lambda \):
\[ \lambda = \frac{36}{5 \cdot \frac{1}{x}} = \frac{36x}{5} \]

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