The photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen like atom X in the second excited state. Then, the hydrogen-like atom X makes a transition of nth orbit.

Options

A. `X = He^(+) , n = 4`
B. `X = Li^(+ +) , n = 6`
C. `X = He^(+) , n = 6`
D. `X = Li^(+ +) , n = 9`

Explanation

To solve the problem step by step, we need to analyze the information given and apply the appropriate formulas related to atomic physics, particularly the energy levels of hydrogen-like atoms.

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions where an electron falls to the n=1 energy level in hydrogen. The second line of the Lyman series corresponds to a transition from n=3 to n=1.

### Step 2: Calculate the Energy of the Photon
The energy of the photon emitted during the transition from n=3 to n=1 in hydrogen can be calculated using the formula:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the second line of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 3 \)

Substituting these values:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \, \text{eV} \left( 1 - \frac{1}{9} \right) = 13.6 \, \text{eV} \left( \frac{8}{9} \right) \]
Calculating this gives:
\[ E = \frac{13.6 \times 8}{9} \approx 12.09 \, \text{eV} \]

### Step 3: Energy Absorption by Hydrogen-like Atom X
The energy absorbed by the hydrogen-like atom X in the second excited state (which corresponds to n=3) can be expressed as:
\[ E = 13.6 \, z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \( n_1 = 3 \) (second excited state) and \( n_2 \) is the final state we need to find.

### Step 4: Set Up the Equation
Equating the energy absorbed to the energy of the photon:
\[ 12.09 = 13.6 \, z^2 \left( \frac{1}{3^2} - \frac{1}{n^2} \right) \]
This simplifies to:
\[ 12.09 = 13.6 \, z^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]

### Step 5: Solve for z and n
Assuming \( z = 2 \) for Helium (He) and \( z = 3 \) for Lithium (Li), we will check both cases.

#### Case 1: For Helium (z = 2)
\[ 12.09 = 13.6 \cdot 2^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
\[ 12.09 = 54.4 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
\[ \frac{12.09}{54.4} = \frac{1}{9} - \frac{1}{n^2} \]
Calculating \( \frac{12.09}{54.4} \approx 0.222 \):
\[ 0.222 = \frac{1}{9} - \frac{1}{n^2} \]
This leads to a contradiction as it results in a negative value for \( \frac{1}{n^2} \).

#### Case 2: For Lithium (z = 3)
\[ 12.09 = 13.6 \cdot 3^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
\[ 12.09 = 122.4 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
\[ \frac{12.09}{122.4} = \frac{1}{9} - \frac{1}{n^2} \]
Calculating \( \frac{12.09}{122.4} \approx 0.0987 \):
\[ 0.0987 = \frac{1}{9} - \frac{1}{n^2} \]
\[ \frac{1}{9} \approx 0.1111 \]
Thus:
\[ \frac{1}{n^2} = 0.1111 - 0.0987 \approx 0.0124 \]
Calculating \( n^2 \):
\[ n^2 \approx \frac{1}{0.0124} \approx 81 \quad \Rightarrow \quad n \approx 9 \]

### Conclusion
The hydrogen-like atom X is Lithium (Li) with \( z = 3 \), and the transition occurs to the \( n = 9 \) orbit.