The hydrogen -like species ` Li^(2+)` is in a spherically symmetric state ` S_1` whth one radisal node . Upon absorbing light the ion undergoes transitoj ot a state ` S_2` has one radial node and its enrgy is equal to the groun sate energy of hhe hydrogen atom.
The orbital angular momentum quantum number of the state ` s_2` is :

Explanation

To solve the problem, we need to determine the orbital angular momentum quantum number (L) of the state \( S_2 \) of the hydrogen-like ion \( Li^{2+} \) after it transitions from state \( S_1 \).

### Step-by-Step Solution:

1. **Understand the given information**:
- The ion \( Li^{2+} \) is in a state \( S_1 \) with one radial node.
- After absorbing light, it transitions to state \( S_2 \) which also has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

2. **Use the formula for radial nodes**:
- The formula for the number of radial nodes \( n - L - 1 \) is given, where \( n \) is the principal quantum number and \( L \) is the orbital angular momentum quantum number.

3. **Set up the equation for the radial node**:
- Since \( S_2 \) has one radial node, we can set up the equation:
\[ n - L - 1 = 1 \]
- Rearranging gives:
\[ n - L = 2 \quad \text{(Equation 1)} \]

4. **Determine the principal quantum number \( n \)**:
- The energy of state \( S_2 \) is equal to the ground state energy of hydrogen, which is given by:
\[ E = -\frac{13.6 Z^2}{n^2} \]
- For hydrogen, \( Z = 1 \) and its ground state energy is \( -13.6 \) eV, so we can equate the energies for \( Li^{2+} \) (where \( Z = 3 \)):
\[ -\frac{13.6 \cdot 3^2}{n^2} = -13.6 \]
- Simplifying gives:
\[ 9 = n^2 \implies n = 3 \]

5. **Substitute \( n \) back into Equation 1**:
- Now substitute \( n = 3 \) into Equation 1:
\[ 3 - L = 2 \]
- Solving for \( L \):
\[ L = 1 \]

6. **Identify the orbital type**:
- Since \( L = 1 \), this corresponds to a \( p \) orbital.

### Conclusion:
The orbital angular momentum quantum number \( L \) of the state \( S_2 \) is \( 1 \).