The user wants me to identify the initial electronic state S1 of a $Li^{2+}$ ion based on the given properties.

Step-by-step Derivation:

1. Analyze the properties of the initial state S1:

• The species is the hydrogen-like ion $Li^{2+}$. For Lithium, the atomic number is Z = 3.
• The state S1 is described as "spherically symmetric". In quantum mechanics, an orbital is spherically symmetric if its wavefunction depends only on the distance from the nucleus (r) and not on the angles (θ, φ). This is characteristic of s-orbitals, for which the azimuthal quantum number $l = 0$.
• The state S1 has "one radial node". The number of radial nodes in an atomic orbital is given by the formula: $\text{Number of radial nodes} = n - l - 1$ where 'n' is the principal quantum number and 'l' is the azimuthal quantum number.

2. Determine the quantum numbers for state S1:

• Since S1 is spherically symmetric, we have $l=0$.
• We are given that S1 has one radial node.
• Substituting these values into the formula for radial nodes: $1 = n - 0 - 1$
• Solving for n, we get: $n = 1 + 1 = 2$
• So, for state S1, the principal quantum number is $n=2$ and the azimuthal quantum number is $l=0$.

3. Identify the orbital corresponding to state S1:

• An electronic state with $n=2$ and $l=0$ corresponds to the 2s orbital.
• Therefore, the state S1 is 2s.

4. (Optional) Analyze the final state S2 for consistency:

• The ion absorbs light and transitions to state S2. This means energy is absorbed, and S2 must be a higher energy state than S1.
• State S2 also has one radial node. Let its quantum numbers be n' and l'. So, $n' - l' - 1 = 1$, which means $n' - l' = 2$.
• The energy of S2 is equal to the ground state energy of the hydrogen atom.
• The energy of an electron in a hydrogen-like species is given by: $E_n = -13.6 \frac{Z^2}{n^2}$ eV.
• For S2 in $Li^{2+}$ (Z=3), the energy is $E_{S2} = -13.6 \frac{3^2}{(n')^2} = -13.6 \frac{9}{(n')^2}$ eV.
• The ground state energy of a hydrogen atom (Z=1, n=1) is $E_{H, gs} = -13.6 \frac{1^2}{1^2} = -13.6$ eV.
• Equating the two energies: $-13.6 \frac{9}{(n')^2} = -13.6$.
• This simplifies to $(n')^2 = 9$, which gives $n' = 3$.
• Now, using the radial node condition for S2: $n' - l' = 2 \implies 3 - l' = 2 \implies l' = 1$.
• So, state S2 is the 3p orbital ($n'=3, l'=1$).
• The transition is from 2s (S1) to 3p (S2). This is a valid spectroscopic transition as it follows the selection rule $\Delta l = \pm 1$. This confirms our analysis is consistent.

5. Conclusion:

• The question asks for the state S1. From our analysis in steps 1-3, state S1 is the 2s orbital.
• Comparing this with the options:
  • A: 1s (n=1, l=0, 0 radial nodes)
  • B: 2s (n=2, l=0, 1 radial node, spherically symmetric)
  • C: 2p (n=2, l=1, 0 radial nodes, not spherically symmetric)
  • D: 3s (n=3, l=0, 2 radial nodes, spherically symmetric)
• Option B matches our findings.