the wavelength of the first line of lyman series for hydrogen atom is equal to that of the second line of balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is

Options

A. 4
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the atomic number \( Z \) of a hydrogen-like ion, given that the wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for the hydrogen-like ion.

### Step-by-Step Solution:

1. **Identify the Lyman Series Transition**:
The first line of the Lyman series corresponds to the transition from \( n_2 = 2 \) to \( n_1 = 1 \) in a hydrogen atom. The formula for the wavelength is given by:
\[ \frac{1}{\lambda_L} = R \cdot \frac{Z^2}{n_1^2} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For hydrogen (\( Z = 1 \)):
\[ \frac{1}{\lambda_L} = R \cdot \frac{1^2}{1^2} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]

2. **Identify the Balmer Series Transition**:
The second line of the Balmer series corresponds to the transition from \( n_2 = 4 \) to \( n_1 = 2 \). The formula for the wavelength is:
\[ \frac{1}{\lambda_B} = R \cdot \frac{Z^2}{n_1^2} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the Balmer series:
\[ \frac{1}{\lambda_B} = R \cdot \frac{Z^2}{2^2} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \cdot \frac{Z^2}{4} \left( \frac{1}{4} - \frac{1}{16} \right) \]
Simplifying the expression:
\[ \frac{1}{\lambda_B} = R \cdot \frac{Z^2}{4} \left( \frac{4 - 1}{16} \right) = R \cdot \frac{Z^2}{4} \cdot \frac{3}{16} = R \cdot \frac{3Z^2}{64} \]

3. **Set the Wavelengths Equal**:
Since the problem states that \( \lambda_L = \lambda_B \), we can equate the two expressions:
\[ R \cdot \frac{3}{4} = R \cdot \frac{3Z^2}{64} \]
Cancel \( R \) from both sides:
\[ \frac{3}{4} = \frac{3Z^2}{64} \]

4. **Solve for \( Z^2 \)**:
Cross-multiplying gives:
\[ 3 \cdot 64 = 3Z^2 \cdot 4 \]
Simplifying:
\[ 192 = 12Z^2 \]
Dividing both sides by 12:
\[ Z^2 = \frac{192}{12} = 16 \]
Taking the square root:
\[ Z = 4 \]

### Final Answer:
The atomic number \( Z \) of the hydrogen-like ion is \( 4 \).