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. 3
B. 4
C. 1
D. 2

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 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 Rydberg formula for the wavelength of the emitted light is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

3. **Calculate Wavelength for Lyman Series**:
For the first line of the Lyman series:
- \( n_f = 1 \)
- \( n_i = 2 \)
Thus,
\[ \frac{1}{\lambda_1} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
Therefore,
\[ \lambda_1 = \frac{4}{3R} \]

4. **Calculate Wavelength for Balmer Series**:
For the second line of the Balmer series:
- \( n_f = 2 \)
- \( n_i = 4 \)
Thus,
\[ \frac{1}{\lambda_2} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = RZ^2 \left( \frac{1}{4} - \frac{1}{16} \right) = RZ^2 \left( \frac{4 - 1}{16} \right) = RZ^2 \cdot \frac{3}{16} \]
Therefore,
\[ \lambda_2 = \frac{16}{3RZ^2} \]

5. **Set the Wavelengths Equal**:
Since \( \lambda_1 = \lambda_2 \):
\[ \frac{4}{3R} = \frac{16}{3RZ^2} \]

6. **Cancel Common Terms**:
Cancel \( \frac{3R}{3R} \) from both sides:
\[ 4 = \frac{16}{Z^2} \]

7. **Solve for \( Z^2 \)**:
Rearranging gives:
\[ Z^2 = \frac{16}{4} = 4 \]

8. **Find \( Z \)**:
Taking the square root:
\[ Z = 2 \]

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

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