The second line of lyman series of hydrogen coincides with the `6^(th)` line of paschen series of an ionic species X find X assuming R to be same for both ghydrogen and X ?

Options

A. `He^(+)`
B. `Li^(+2)`
C. `Li^(+)`
D. `Be^(+3)`

Answer

To solve the problem, we need to find the ionic species \( X \) given that the second line of the Lyman series of hydrogen coincides with the sixth line of the Paschen series of \( X \). We will use the Rydberg formula for both hydrogen and the ionic species \( X \).

### Step-by-Step Solution:

1. **Identify the transitions**:
- The second line of the Lyman series corresponds to a transition from \( n_2 = 3 \) to \( n_1 = 1 \).
- The sixth line of the Paschen series corresponds to a transition from \( n_2 = 6 \) to \( n_1 = 3 \).

2. **Rydberg formula**:
The Rydberg formula is given by:
\[ R = Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers.

3. **Apply the formula for hydrogen**:
For hydrogen (\( Z_H = 1 \)):
\[ R_H = 1^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ R_H = 1 \left( 1 - \frac{1}{9} \right) = 1 \left( \frac{8}{9} \right) = \frac{8}{9} \]

4. **Apply the formula for ionic species \( X \)**:
For the ionic species \( X \) (\( Z_X \)):
\[ R_X = Z_X^2 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) \]
\[ R_X = Z_X^2 \left( \frac{1}{9} - \frac{1}{36} \right) = Z_X^2 \left( \frac{4}{36} - \frac{1}{36} \right) = Z_X^2 \left( \frac{3}{36} \right) = Z_X^2 \left( \frac{1}{12} \right) \]

5. **Set the equations equal**:
Since \( R_H = R_X \):
\[ \frac{8}{9} = Z_X^2 \left( \frac{1}{12} \right) \]

6. **Solve for \( Z_X^2 \)**:
\[ Z_X^2 = \frac{8}{9} \times 12 = \frac{96}{9} = \frac{32}{3} \]

7. **Calculate \( Z_X \)**:
\[ Z_X = \sqrt{\frac{32}{3}} \approx 3.27 \]
Since \( Z \) must be a whole number, we round to the nearest whole number, which is \( 3 \).

8. **Identify the ionic species**:
The atomic number \( Z = 3 \) corresponds to lithium (\( Li \)). Since we are looking for an ionic species, we consider \( Li^{2+} \).

### Conclusion:
The ionic species \( X \) is \( Li^{2+} \).