Atomic spectroscopy nomenclature: In atomic physics, is a state with orbital quantum number l = 0 called a p state?
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ATrue
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BFalse
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CTrue only for hydrogen-like atoms
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DTrue at high principal quantum number n
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ETrue when spin–orbit coupling is neglected
Answer
Correct Answer: False
Explanation
Introduction / Context:Atomic states are labeled by spectroscopic notation tied to the orbital angular momentum quantum number l. This notation is universal across atomic physics, spectroscopy, and quantum chemistry, and is essential for interpreting line spectra and selection rules.
Given Data / Assumptions:
- Orbital quantum number l takes values 0, 1, 2, 3, …
- Standard letter mapping is used.
- The claim is that l = 0 corresponds to a p state.
Concept / Approach:
The accepted mapping is: l = 0 → s, l = 1 → p, l = 2 → d, l = 3 → f, and so on (g, h, … for higher l). Therefore, l = 0 is an s state, not a p state. This holds regardless of the atom (hydrogenic or many-electron) and does not depend on n or spin–orbit effects; those affect term symbols but not the s, p, d, f letter code for l.
Step-by-Step Solution:
Recall mapping: 0→s, 1→p, 2→d, 3→f.Compare statement: “l = 0 is p” → contradicts mapping.Conclude the statement is false.Verification / Alternative check:
Any elementary spectroscopy table or periodic table electron configurations confirm s orbitals correspond to l = 0 (e.g., 1s, 2s, 3s).
Why Other Options Are Wrong:
- Conditions about hydrogen-like atoms, high n, or neglecting spin–orbit do not change the l-to-letter mapping.
Common Pitfalls:
- Confusing principal quantum number n with orbital quantum number l.
Final Answer:
False