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) based on the wavelengths of specific lines in their emission spectra.

### Step 1: Understanding the Lyman and Balmer Series
The Lyman series corresponds to transitions where the final energy level (n_final) is 1, while the Balmer series corresponds to transitions where n_final is 2. We will consider the first line of the Lyman series (transition from n=2 to n=1) and the second line of the Balmer series (transition from n=4 to n=2).

### Step 2: Energy Levels of Hydrogen-like Ions
The energy of an electron in a hydrogen-like ion is given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 3: Calculate the Wavelength for the Lyman Series
For the first line of the Lyman series (n_initial = 2 to n_final = 1):
- The energy difference \( \Delta E_{21} \) is:
\[ \Delta E_{21} = E_2 - E_1 = -\frac{13.6 Z_1^2}{2^2} - \left(-\frac{13.6 Z_1^2}{1^2}\right) \]
\[ = -\frac{13.6 Z_1^2}{4} + 13.6 Z_1^2 = 13.6 Z_1^2 \left(1 - \frac{1}{4}\right) = 13.6 Z_1^2 \cdot \frac{3}{4} \]

Using the relation \( E = h \nu = \frac{hc}{\lambda} \):
\[ \frac{hc}{\lambda_1} = 13.6 Z_1^2 \cdot \frac{3}{4} \]
Thus, we can express the wavelength as:
\[ \frac{1}{\lambda_1} = \frac{13.6 Z_1^2 \cdot 3}{4hc} \]

### Step 4: Calculate the Wavelength for the Balmer Series
For the second line of the Balmer series (n_initial = 4 to n_final = 2):
- The energy difference \( \Delta E_{42} \) is:
\[ \Delta E_{42} = E_4 - E_2 = -\frac{13.6 Z_2^2}{4^2} - \left(-\frac{13.6 Z_2^2}{2^2}\right) \]
\[ = -\frac{13.6 Z_2^2}{16} + \frac{13.6 Z_2^2}{4} = 13.6 Z_2^2 \left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 Z_2^2 \cdot \frac{3}{16} \]

Using the relation \( E = h \nu = \frac{hc}{\lambda} \):
\[ \frac{hc}{\lambda_2} = 13.6 Z_2^2 \cdot \frac{3}{16} \]
Thus, we can express the wavelength as:
\[ \frac{1}{\lambda_2} = \frac{13.6 Z_2^2 \cdot 3}{16hc} \]

### Step 5: Setting the Wavelengths Equal
Since the wavelengths are identical:
\[ \frac{1}{\lambda_1} = \frac{1}{\lambda_2} \]
This gives us:
\[ \frac{13.6 Z_1^2 \cdot 3}{4hc} = \frac{13.6 Z_2^2 \cdot 3}{16hc} \]

### Step 6: Simplifying the Equation
Canceling out common terms:
\[ \frac{Z_1^2}{4} = \frac{Z_2^2}{16} \]
Cross-multiplying gives:
\[ 16 Z_1^2 = 4 Z_2^2 \]
Dividing both sides by 4:
\[ 4 Z_1^2 = Z_2^2 \]

### Step 7: Finding the Ratio of Atomic Numbers
Taking the square root of both sides:
\[ \frac{Z_1}{Z_2} = \frac{1}{2} \]

### Final Answer
The ratio of the atomic numbers \( Z_1 : Z_2 \) is:
\[ Z_1 : Z_2 = 1 : 2 \]