Orbital Angular Momentum Quantum Number of State S2

To find the orbital angular momentum quantum number ($l$) of the state $\text{S}_2$, we analyze the information given in the passage:

1. Energy of State $\text{S}_2$:
   The energy of a hydrogen-like species with atomic number $Z$ in a state with principal quantum number $n$ is given by:
   $E_n = -E_H \frac{Z^2}{n^2}$
   where $E_H = 13.6 \text{ eV}$ is the ground state ionization energy of the hydrogen atom ($Z = 1, n = 1$).

   For $\text{Li}^{2+}$, the atomic number is $Z = 3$.

   It is given that the energy of state $\text{S}_2$ is equal to the ground state energy of the hydrogen atom ($-E_H$):
   $-E_H \frac{3^2}{n_2^2} = -E_H$

   Solving for $n_2$:
   $\frac{9}{n_2^2} = 1 \implies n_2^2 = 9 \implies n_2 = 3$

2. Radial Nodes of State $\text{S}_2$:
   The number of radial nodes for an orbital is given by the formula:
   $\text{Radial nodes} = n - l - 1$

   We are given that state $\text{S}_2$ has $1$ radial node. Substituting $n_2 = 3$:
   $1 = 3 - l - 1$
   $1 = 2 - l \implies l = 1$

Thus, the orbital angular momentum quantum number of state $\text{S}_2$ is $l = 1$.

Correct Option: B