The hydrogen -like species `Li^(2+)` is in a spherically sysmmetric state `S_(1)` with one node ,Upon ansorbing light , the ion undergoes transition to a state `S_(2)` The state `s_(2)` has one radial node and its energy is equal is to the ground state energy of the hydrogen atom
Energy of the state `S_(1)` in units of the hydrogen atom ground state enegy is

Options

A. 0.75
B. 1.5
C. 2.25
D. 4.5

Explanation

To find the energy of the state \( S_1 \) of the hydrogen-like species \( \text{Li}^{2+} \) in units of the ground state energy of the hydrogen atom, we can follow these steps:

### Step 1: Identify the Quantum Numbers
The state \( S_1 \) is described as having one radial node. For a spherically symmetric state (s orbital), the angular momentum quantum number \( l = 0 \). The number of radial nodes \( n_r \) is given by the formula:
\[ n_r = n - l - 1 \]
Given that \( n_r = 1 \) (one radial node) and \( l = 0 \), we can substitute these values into the equation:
\[ 1 = n - 0 - 1 \implies n = 2 \]
Thus, the principal quantum number \( n \) for state \( S_1 \) is 2.

### Step 2: Calculate the Energy of \( \text{Li}^{2+} \)
The energy of a hydrogen-like atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where \( Z \) is the atomic number. For \( \text{Li}^{2+} \), \( Z = 3 \) and we have found \( n = 2 \):
\[ E_2 = -\frac{13.6 \times 3^2}{2^2} = -\frac{13.6 \times 9}{4} = -\frac{122.4}{4} = -30.6 \, \text{eV} \]

### Step 3: Calculate the Energy of the Ground State of Hydrogen
The ground state energy of hydrogen (\( n = 1 \)) is:
\[ E_1 = -13.6 \, \text{eV} \]

### Step 4: Calculate the Ratio of Energies
To express the energy of state \( S_1 \) in units of the ground state energy of hydrogen, we calculate the ratio:
\[ \text{Ratio} = \frac{E_{S_1}}{E_{\text{H}}} = \frac{-30.6}{-13.6} = \frac{30.6}{13.6} \]
Calculating this gives:
\[ \text{Ratio} = \frac{30.6}{13.6} = 2.25 \]

### Conclusion
The energy of the state \( S_1 \) in units of the ground state energy of the hydrogen atom is \( \boxed{2.25} \).

---