The hydrogen like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one radial node. Upon absorbing light, the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the hydrogen atom
The orbital angular momentum quantum number of the state `S_(2)` is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem regarding the hydrogen-like species \( \text{Li}^{2+} \) and its transition from state \( S_1 \) to state \( S_2 \), we will follow these steps:

### Step 1: Understand the Energy Levels
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 2: Determine the Ground State Energy of Hydrogen
For hydrogen (\( Z = 1 \)):
\[ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \text{ eV} \]

### Step 3: Set Up the Energy Equation for \( \text{Li}^{2+} \)
For \( \text{Li}^{2+} \) (\( Z = 3 \)):
\[ E_n = -\frac{13.6 \times 3^2}{n^2} = -\frac{13.6 \times 9}{n^2} = -\frac{122.4}{n^2} \text{ eV} \]

### Step 4: Equate the Energies
According to the problem, the energy of state \( S_2 \) is equal to the ground state energy of hydrogen:
\[ -\frac{122.4}{n^2} = -13.6 \]

### Step 5: Solve for \( n^2 \)
Cancelling the negative signs and rearranging gives:
\[ \frac{122.4}{n^2} = 13.6 \]
\[ n^2 = \frac{122.4}{13.6} = 9 \]
Thus, \( n = 3 \).

### Step 6: Determine the Radial Nodes
The number of radial nodes \( R \) is given by the formula:
\[ R = n - l - 1 \]
We know from the problem that \( R = 1 \) for state \( S_2 \) and \( n = 3 \):
\[ 1 = 3 - l - 1 \]
Solving for \( l \):
\[ 1 = 2 - l \implies l = 1 \]

### Step 7: Identify the Orbital Angular Momentum Quantum Number
The orbital angular momentum quantum number \( l \) corresponds to:
- \( l = 0 \) for \( s \) orbitals
- \( l = 1 \) for \( p \) orbitals
- \( l = 2 \) for \( d \) orbitals, etc.

Since we found \( l = 1 \), this indicates that the state \( S_2 \) is a \( p \) state.

### Conclusion
The orbital angular momentum quantum number of the state \( S_2 \) is:
\[ \boxed{1} \]