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
The state `S_(1)` is

Options

A. 1s
B. 2s
C. 2p
D. 3s, 4d

Explanation

To solve the problem, we need to determine the state \( S_1 \) of the hydrogen-like ion \( \text{Li}^{2+} \) based on the information given about its radial nodes and the transition to state \( S_2 \).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The ion \( \text{Li}^{2+} \) is in a spherically symmetric state \( S_1 \) with one radial node.
- It transitions to state \( S_2 \), which has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

2. **Using the Formula for Radial Nodes:**
- The number of radial nodes \( R_n \) is given by the formula:
\[ R_n = n - l - 1 \]
where \( n \) is the principal quantum number and \( l \) is the azimuthal quantum number.

3. **Analyzing State \( S_2 \):**
- We know that state \( S_2 \) has one radial node. Therefore, we can set up the equation:
\[ 1 = n_2 - l_2 - 1 \]
- Rearranging gives:
\[ n_2 - l_2 = 2 \]
- Since \( S_2 \) has energy equal to the ground state of hydrogen, we know that for hydrogen, the ground state corresponds to \( n = 1 \). However, for \( \text{Li}^{2+} \), we need to find the corresponding quantum numbers.

4. **Determining the Quantum Numbers for \( S_2 \):**
- The ground state energy of hydrogen is \( -13.6 \, \text{eV} \), and for hydrogen-like ions, the energy is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \, \text{eV} \]
- For \( \text{Li}^{2+} \) (where \( Z = 3 \)), the ground state energy corresponds to \( n = 1 \):
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -122.4 \, \text{eV} \]

5. **Finding \( n_2 \) and \( l_2 \):**
- Since \( S_2 \) has one radial node, we can assume \( n_2 = 3 \) (as it must be higher than the ground state) and \( l_2 = 1 \) (which corresponds to the \( p \) subshell):
\[ 3 - 1 - 1 = 1 \quad \text{(This satisfies the radial node condition)} \]

6. **Determining State \( S_1 \):**
- Now, we need to find \( S_1 \) which has one radial node:
\[ 1 = n_1 - l_1 - 1 \]
- Rearranging gives:
\[ n_1 - l_1 = 2 \]
- If we assume \( n_1 = 2 \), then \( l_1 = 0 \) (which corresponds to the \( s \) subshell):
\[ 2 - 0 - 1 = 1 \quad \text{(This satisfies the radial node condition)} \]

7. **Conclusion:**
- Therefore, the state \( S_1 \) is \( 2s \).

### Final Answer:
The state \( S_1 \) is \( 2s \).

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