If the shortest wave length of Lyman series of `H` atom is `x`, then the wave length of the first line of Balmer series of `H` atom will be-

Options

A. `9x//5`
B. `36x//5`
C. `5x//9`
D. `5x//36`

Answer

To solve the problem, we need to find the wavelength of the first line of the Balmer series of the hydrogen atom given that the shortest wavelength of 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 \( n=1 \) energy level from higher energy levels. The shortest wavelength (or the first line) occurs when the electron transitions from \( n=\infty \) to \( n=1 \).

The formula for the wavelength (\( \lambda_L \)) in the Lyman series is given by:
\[ \frac{1}{\lambda_L} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \):
\[ \frac{1}{\lambda_L} = R_H \left( 1 - 0 \right) = R_H \]
Therefore, the shortest wavelength of the Lyman series is:
\[ \lambda_L = \frac{1}{R_H} \]

2. **Understand the Balmer Series**: The Balmer series corresponds to transitions where an electron falls to the \( n=2 \) energy level from higher energy levels. The first line of the Balmer series occurs when the electron transitions from \( n=3 \) to \( n=2 \).

The formula for the wavelength (\( \lambda_B \)) in the Balmer series is given by:
\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating this gives:
\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \]
Therefore:
\[ \lambda_B = \frac{36}{5 R_H} \]

3. **Relate the Two Wavelengths**: We know from the Lyman series that:
\[ \lambda_L = \frac{1}{R_H} \]
Therefore, we can express \( R_H \) as:
\[ R_H = \frac{1}{\lambda_L} \]
Substituting this into the expression for \( \lambda_B \):
\[ \lambda_B = \frac{36}{5} \cdot \lambda_L \]

4. **Substitute \( \lambda_L \) with \( x \)**: Given that the shortest wavelength of the Lyman series is \( x \):
\[ \lambda_B = \frac{36}{5} \cdot x \]

5. **Final Answer**: Thus, the wavelength of the first line of the Balmer series of the hydrogen atom is:
\[ \lambda_B = \frac{36}{5} x \]

### Conclusion:
The wavelength of the first line of the Balmer series of the hydrogen atom is \( \frac{36}{5} x \).