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 hydrogen is equal to that of the second line of the Balmer series for the hydrogen-like ion.

### Step-by-Step Solution:

1. **Identify the Wavelengths**:
- The first line of the Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \).
- The second line of the Balmer series corresponds to the transition from \( n = 4 \) to \( n = 2 \).

2. **Use the Rydberg Formula**:
The wavelength \( \lambda \) for a transition can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, and \( Z \) is the atomic number.

3. **Calculate for the Lyman Series**:
For the first line of the Lyman series:
- \( n_1 = 1 \), \( n_2 = 2 \)
\[ \frac{1}{\lambda_L} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]

4. **Calculate for the Balmer Series**:
For the second line of the Balmer series:
- \( n_1 = 2 \), \( n_2 = 4 \)
\[ \frac{1}{\lambda_B} = R \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \cdot Z^2 \left( \frac{1}{4} - \frac{1}{16} \right) = R \cdot Z^2 \cdot \frac{3}{16} \]

5. **Set the Wavelengths Equal**:
Since the wavelengths are equal:
\[ \frac{3}{4} = Z^2 \cdot \frac{3}{16} \]

6. **Solve for \( Z^2 \)**:
Cancel \( 3 \) from both sides:
\[ \frac{1}{4} = Z^2 \cdot \frac{1}{16} \]
Multiply both sides by \( 16 \):
\[ 4 = Z^2 \]
Thus, \( Z = 2 \).

### Conclusion:
The atomic number \( Z \) of the hydrogen-like ion is \( 2 \).