The wavelength of the first line of the Lyman series of a hydrogen like ion X is identical to that of the second line of the baimer series of another hydrogen like ion Y . Find the ratio of their atomic number.

Answer

To solve the problem, we need to find the ratio of the atomic numbers of two hydrogen-like ions, X and Y, given that the wavelength of the first line of the Lyman series of ion X is identical to that of the second line of the Balmer series of ion Y.

### Step-by-Step Solution:

1. **Identify the transitions for Lyman and Balmer series:**
- 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 for the Lyman series:**
- The wavelength \( \lambda_X \) for the Lyman series (ion X) is given by:
\[ \frac{1}{\lambda_X} = RZ_X^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
- Simplifying this gives:
\[ \frac{1}{\lambda_X} = RZ_X^2 \left( 1 - \frac{1}{4} \right) = RZ_X^2 \left( \frac{3}{4} \right) \]
- Therefore:
\[ \frac{1}{\lambda_X} = \frac{3RZ_X^2}{4} \]

3. **Use the Rydberg formula for the Balmer series:**
- The wavelength \( \lambda_Y \) for the Balmer series (ion Y) is given by:
\[ \frac{1}{\lambda_Y} = RZ_Y^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
- Simplifying this gives:
\[ \frac{1}{\lambda_Y} = RZ_Y^2 \left( \frac{1}{4} - \frac{1}{16} \right) = RZ_Y^2 \left( \frac{4 - 1}{16} \right) = RZ_Y^2 \left( \frac{3}{16} \right) \]
- Therefore:
\[ \frac{1}{\lambda_Y} = \frac{3RZ_Y^2}{16} \]

4. **Set the wavelengths equal:**
- According to the problem, the wavelengths are identical:
\[ \frac{3RZ_X^2}{4} = \frac{3RZ_Y^2}{16} \]
- We can cancel \( 3R \) from both sides:
\[ \frac{Z_X^2}{4} = \frac{Z_Y^2}{16} \]

5. **Cross-multiply to find the ratio of atomic numbers:**
- Cross-multiplying gives:
\[ 16Z_X^2 = 4Z_Y^2 \]
- Dividing both sides by 4:
\[ 4Z_X^2 = Z_Y^2 \]

6. **Take the square root to find the ratio:**
- Taking the square root of both sides gives:
\[ 2Z_X = Z_Y \quad \Rightarrow \quad \frac{Z_X}{Z_Y} = \frac{1}{2} \]

### Final Answer:
The ratio of the atomic numbers \( Z_X : Z_Y \) is \( 1 : 2 \).