If the shortest wavelength of Lyman series of hydrogen atom is x, then the wavelength of first member of Balmer series of hydrogen 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 member of the Balmer series of the hydrogen atom, given the shortest wavelength of the Lyman series is \( x \).

### Step-by-Step Solution:

1. **Understanding the Lyman Series**: The Lyman series corresponds to transitions where the electron falls to the first energy level (\( n = 1 \)). The shortest wavelength in the Lyman series occurs when the electron transitions from \( n = \infty \) to \( n = 1 \).

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = \infty \)

Thus, the formula simplifies to:

\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{1^2} - 0 \right) = R \]

Therefore, the shortest wavelength \( \lambda \) (denoted as \( x \)) is:

\[ \lambda = \frac{1}{R} = x \]

2. **Understanding the Balmer Series**: The Balmer series corresponds to transitions where the electron falls to the second energy level (\( n = 2 \)). The first member of the Balmer series occurs when the electron transitions from \( n = 3 \) to \( n = 2 \).

For the first member of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Using the formula again:

\[ \frac{1}{\lambda_2} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting the values:

\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

3. **Finding a Common Denominator**: To simplify \( \frac{1}{4} - \frac{1}{9} \):

The common denominator of 4 and 9 is 36.

\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]

Therefore,

\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

Substituting back into the equation for \( \lambda_2 \):

\[ \frac{1}{\lambda_2} = R \cdot \frac{5}{36} \]

4. **Finding \( \lambda_2 \)**: Now, we can find \( \lambda_2 \):

\[ \lambda_2 = \frac{36}{5R} \]

5. **Relating \( R \) to \( x \)**: Since we know that \( R = \frac{1}{x} \), we can substitute this into the equation:

\[ \lambda_2 = \frac{36}{5 \cdot \frac{1}{x}} = \frac{36x}{5} \]

### Final Answer:
The wavelength of the first member of the Balmer series of the hydrogen atom is:

\[ \lambda_2 = \frac{36x}{5} \]