---
title: "A student examines the photoelectron spectra of gaseous oxygen, \\(\\text{O}\\), and gaseous neon, \\(\\text{Ne}\\). Which of the following correctly compares the relative position of the \\(1s\\) peak in the spectrum of \\(\\text{Ne}\\) to that in the spectrum of \\(\\text{O}\\) and provides the correct physical justification?"
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url: "https://nerd-notes.com/ubq/120164/"
date_modified: "2026-08-21T16:02:41+00:00"
---

# A student examines the photoelectron spectra of gaseous oxygen, \(\text{O}\), and gaseous neon, \(\text{Ne}\). Which of the following correctly compares the relative position of the \(1s\) peak in the spectrum of \(\text{Ne}\) to that in the spectrum of \(\text{O}\) and provides the correct physical justification?

A student examines the photoelectron spectra of gaseous oxygen, \(\text{O}\), and gaseous neon, \(\text{Ne}\). Which of the following correctly compares the relative position of the \(1s\) peak in the spectrum of \(\text{Ne}\) to that in the spectrum of \(\text{O}\) and provides the correct physical justification?

- **A.** The \(1s\) peak for \(\text{Ne}\) is at a higher binding energy than that for \(\text{O}\) because \(\text{Ne}\) has a completely filled valence shell, which imparts electronic stability to all occupied orbitals.
- **B.** The \(1s\) peak for \(\text{Ne}\) is at a higher binding energy than that for \(\text{O}\) because \(\text{Ne}\) has a greater nuclear charge, resulting in a stronger Coulombic attraction between the nucleus and the \(1s\) electrons.
- **C.** The \(1s\) peak for \(\text{Ne}\) is at a lower binding energy than that for \(\text{O}\) because the greater total number of electrons in \(\text{Ne}\) creates increased shielding that weakens the attraction to the \(1s\) electrons.
- **D.** The \(1s\) peak for \(\text{Ne}\) is at a lower binding energy than that for \(\text{O}\) because the electron pairing in the \(2p\) subshell of \(\text{Ne}\) causes interelectronic repulsions that destabilize the core subshells.

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