Step-by-step Solution:

1. Analyze the initial state S1

• Species: The problem deals with the hydrogen-like species $Li^{2+}$. For Lithium, the atomic number is $Z=3$.
• Symmetry: The state S1 is described as "spherically symmetric". This implies that the azimuthal quantum number, $l$, is 0. Orbitals with $l=0$ are 's' orbitals.
• Radial Nodes: The state S1 has "one radial node". The number of radial nodes in an orbital is given by the formula: Number of radial nodes = n - l - 1, where 'n' is the principal quantum number and 'l' is the azimuthal quantum number.
• Determine Principal Quantum Number (n) for S1: We know the number of radial nodes is 1 and $l=0$. Let the principal quantum number for S1 be $n_1$. $1 = n_1 - 0 - 1$ $n_1 = 2$ Therefore, the state S1 corresponds to the 2s orbital of the $Li^{2+}$ ion.

2. Calculate the energy of state S1

• The energy of an electron in a hydrogen-like species is given by the formula: E_n = -R_H rac{Z^2}{n^2} where $R_H$ is the Rydberg constant, Z is the atomic number, and n is the principal quantum number.
• The ground state energy of the hydrogen atom (Z=1, n=1) is: E_{H,gs} = -R_H rac{1^2}{1^2} = -R_H This is the unit we need to express our answer in.
• Now, let's calculate the energy of state S1 for $Li^{2+}$ ($Z=3$, $n_1=2$): E_{S1} = -R_H rac{3^2}{2^2} = -R_H rac{9}{4} = -2.25 R_H

3. Express the energy of S1 in the required units

• The question asks for the energy of state S1 "in units of the hydrogen atom ground state energy". This means we need to find the ratio $\frac{E_{S1}}{E_{H,gs}}$. $\frac{E_{S1}}{E_{H,gs}} = \frac{-2.25 R_H}{-R_H} = 2.25$
• Alternatively, we can substitute $E_{H,gs}$ into the expression for $E_{S1}$: $E_{S1} = 2.25 \times (-R_H) = 2.25 \times E_{H,gs}$ So, the energy of state S1 is 2.25 times the ground state energy of a hydrogen atom. The value is 2.25.

4. (Optional Confirmation) Analyze the final state S2

• The problem states that the energy of state S2 is equal to the ground state energy of the hydrogen atom: $E_{S2} = E_{H,gs} = -R_H$.
• For the $Li^{2+}$ ion, the energy of state S2 (with principal quantum number $n_2$) is: $E_{S2} = -R_H \frac{Z^2}{n_2^2} = -R_H \frac{3^2}{n_2^2}$
• Equating the two expressions for $E_{S2}$: $-R_H = -R_H \frac{9}{n_2^2}$ $1 = \frac{9}{n_2^2} \implies n_2^2 = 9 \implies n_2 = 3$
• State S2 also has one radial node. Let its azimuthal quantum number be $l_2$. $Number of radial nodes = n_2 - l_2 - 1 = 1$ $3 - l_2 - 1 = 1$ $2 - l_2 = 1 \implies l_2 = 1$
• So, state S2 is the 3p orbital. The transition from S1 (2s) to S2 (3p) is an allowed electronic transition since $\Delta l = +1$. This confirms our interpretation of the problem is consistent.

Conclusion:

The energy of state S1 is 2.25 in units of the hydrogen atom ground state energy. This corresponds to option C.