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
Energy of the state `S_(1)` in units of the hydrogen atom ground state energy is

Options

A. 0.75
B. `1.50`
C. 2.25
D. `4.50`

Explanation

To solve the problem, we need to determine the energy of the state \( S_1 \) of the ion \( Li^{2+} \) in units of the ground state energy of the hydrogen atom.

### Step-by-Step Solution:

1. **Identify the given information:**
- The species is \( Li^{2+} \), which has a nuclear charge \( Z = 3 \).
- The state \( S_1 \) has one radial node.
- The state \( S_2 \) has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

2. **Understand the concept of radial nodes:**
- The number of radial nodes in an atomic orbital is given by \( n - l - 1 \), where \( n \) is the principal quantum number and \( l \) is the azimuthal quantum number.
- For \( S \) orbitals, \( l = 0 \), so the number of radial nodes is \( n - 1 \).

3. **Determine the principal quantum number for state \( S_1 \):**
- Since \( S_1 \) has one radial node, we have:
\[ n - 1 = 1 \implies n = 2 \]
- Thus, \( S_1 \) corresponds to \( n = 2 \).

4. **Calculate the energy of state \( S_1 \):**
- Using the formula for the energy of an electron in a hydrogen-like atom:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
- For \( Li^{2+} \) (where \( Z = 3 \)) and \( n = 2 \):
\[ E_{S_1} = -\frac{13.6 \times 3^2}{2^2} = -\frac{13.6 \times 9}{4} = -30.6 \text{ eV} \]

5. **Find the ground state energy of the hydrogen atom:**
- The ground state energy of hydrogen (\( H \)) is:
\[ E_{H} = -13.6 \text{ eV} \]

6. **Express the energy of state \( S_1 \) in units of the ground state energy of hydrogen:**
- To express \( E_{S_1} \) in units of \( E_H \):
\[ \text{Energy of } S_1 \text{ in units of } E_H = \frac{E_{S_1}}{E_H} = \frac{-30.6 \text{ eV}}{-13.6 \text{ eV}} = \frac{30.6}{13.6} \approx 2.25 \]

### Final Answer:
The energy of the state \( S_1 \) in units of the hydrogen atom ground state energy is approximately \( 2.25 \).