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
The sate `S_(1)` is

Options

A. 1s
B. 2s
C. 2p
D. 3s

Explanation

To solve the problem step-by-step, we need to analyze the information given about the hydrogen-like species \( \text{Li}^{2+} \) and its states \( S_1 \) and \( S_2 \).

### Step 1: Understanding the Given Information
We know that:
- The ion \( \text{Li}^{2+} \) is in a spherically symmetric state \( S_1 \) with 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.

### Step 2: Identifying the Orbital Type
Since the state \( S_1 \) is described as spherically symmetric, we can conclude that it corresponds to an s orbital. In quantum mechanics, the angular momentum quantum number \( l \) for s orbitals is 0.

### Step 3: Relating Radial Nodes to Quantum Numbers
The number of radial nodes \( n_r \) can be calculated using the formula:
\[ n_r = n - l - 1 \]
where \( n \) is the principal quantum number and \( l \) is the angular momentum quantum number.

### Step 4: Applying the Information
From the problem, we know that \( S_1 \) has one radial node:
\[ n_r = 1 \]
Substituting the values into the formula:
\[ 1 = n - 0 - 1 \]
This simplifies to:
\[ n = 2 \]

### Step 5: Determining the State \( S_1 \)
Since \( n = 2 \) and the orbital is an s orbital (as established earlier), we can denote the state \( S_1 \) as:
\[ S_1 = 2s \]

### Step 6: Conclusion
Thus, the state \( S_1 \) is \( 2s \).

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

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