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 transitions of electrons in hydrogen and hydrogen-like atoms, specifically focusing on the Lyman series and energy levels.

### Step 1: Identify the Transition in the Lyman Series
The second line of the Lyman series corresponds to the transition from n = 3 to n = 1 in a hydrogen atom.

### Step 2: Calculate the Energy Released in the Transition
The energy of an electron in a hydrogen atom at a specific energy level n is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the transition from n = 3 to n = 1:
- Energy at n = 3:
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, \text{eV} \]
- Energy at n = 1:
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
- Energy released (ΔE) during the transition:
\[ \Delta E = E_1 - E_3 = -13.6 - (-1.51) = -13.6 + 1.51 \approx -12.09 \, \text{eV} \]

### Step 3: Energy Absorbed by the Hydrogen-like Atom X
The hydrogen-like atom X is initially in the second excited state, which corresponds to n = 3. When it absorbs the photon, it transitions to a higher energy level n = n₀. The energy absorbed by atom X must equal the energy released by the hydrogen atom:
\[ E_{\text{absorbed}} = -\Delta E = 12.09 \, \text{eV} \]

### Step 4: Relate the Energy Levels of Hydrogen-like Atom X
For a hydrogen-like atom, the energy levels are given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \]
Where Z is the atomic number of the hydrogen-like atom. The energy absorbed can be expressed as:
\[ E_{\text{absorbed}} = E_{n_0} - E_{3} \]
Substituting the energy expressions:
\[ 12.09 = \left(-\frac{13.6 Z^2}{n_0^2}\right) - \left(-\frac{13.6 Z^2}{3^2}\right) \]
This simplifies to:
\[ 12.09 = 13.6 Z^2 \left(\frac{1}{9} - \frac{1}{n_0^2}\right) \]

### Step 5: Solve for Z and n₀
Rearranging gives:
\[ \frac{12.09}{13.6 Z^2} = \frac{1}{9} - \frac{1}{n_0^2} \]
To find Z, we can assume that the energy absorbed is equal to the energy difference between the two states. By comparing the ratios of the energy levels, we can set:
\[ \frac{Z}{n_0} = \frac{1}{3} \]
From this, we can derive:
\[ Z = \frac{n_0}{3} \]

### Step 6: Substitute and Solve
If we assume n₀ = 9 (as derived from the energy calculations), then:
\[ Z = \frac{9}{3} = 3 \]
This indicates that the hydrogen-like atom X is lithium (Z = 3).

### Final Answer
The hydrogen-like atom X is lithium (Z = 3) and it transitions to n₀ = 9.