To identify the orbital angular momentum quantum number, $l$, for the state $S_2$ of a hydrogen-like species such as $Li^{2+}$, we can refer to the given information and the known equations for the energy levels of hydrogen-like atoms. Hydrogen-like atoms or ions have only one electron and their energy in a particular state is given by the equation:

$$E_n = -\frac{Z^2}{n^2} E_0$$

where:

• $E_n$ is the energy of the electron in the nth energy level,
• $Z$ is the atomic number of the species (for $Li$, $Z = 3$),
• $n$ is the principal quantum number,
• $E_0$ is the energy of the ground state of hydrogen ($-13.6$ eV).

The problem states that $S_2$ has energy equal to the ground state energy of the hydrogen atom. The ground state of hydrogen corresponds to $n = 1$ and $E_0 = -13.6$ eV. However, for $Li^{2+}$, with $Z = 3$, the same energy level could be attained at a different value of $n$ since the $Z^2$ factor magnifies the energy levels with increasing atomic number. Let's calculate the principal quantum number, $n$, for the $Li^{2+}$ ion that would give it an energy equal to $E_0$:

For $Li^{2+}$ ion to have the ground state energy of a hydrogen atom, we can set the energies equal and solve for $n$:

$$-\frac{Z^2}{n^2}E_0 = E_0$$

Substituting $Z = 3$ and simplifying:

$$-\frac{9}{n^2} = 1$$

From which we find, $n^2 = 9$ and thus $n = 3$.

Furthermore, the problem states that $S_1$ has one radial node and upon absorbing light, it transitions to $S_2$ which also has one radial node. Radial nodes are related to the principal quantum number $n$ and the angular quantum number $l$ by the formula:

Number of radial nodes = $n - l - 1$

Given that $S_2$ has one radial node, we can plug $n = 3$ into the radial node formula to solve for $l$:

$$1 = 3 - l - 1$$

This simplifies to:

$$l = 3 - 2$$

$$l = 1$$

Therefore, the orbital angular momentum quantum number of the state $S_2$ is 1, which corresponds to option B.