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title: "Simulated photoelectron spectra (PES) for isolated, ground-state atoms of \\(\\text{Ne}\\) (atomic number \\(10\\)) and \\(\\text{Na}\\) (atomic number \\(11\\)) are shown below.  Which of the following best explains why the valence electron peak for \\(\\text{Na}\\) appears at a substantially lower binding energy (\\(\\approx 0.50 \\text{ MJ/mol}\\)) than the valence electron peak for \\(\\text{Ne}\\) (\\(\\approx 2.1 \\text{ MJ/mol}\\)), despite \\(\\text{Na}\\) having a greater nuclear charge (\\(Z = 11\\)) than \\(\\text{Ne}\\) (\\(Z = 10\\))?"
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url: "https://nerd-notes.com/ubq/123584/"
date_modified: "2026-09-28T12:00:17+00:00"
---

# Simulated photoelectron spectra (PES) for isolated, ground-state atoms of \(\text{Ne}\) (atomic number \(10\)) and \(\text{Na}\) (atomic number \(11\)) are shown below.

Which of the following best explains why the valence electron peak for \(\text{Na}\) appears at a substantially lower binding energy (\(\approx 0.50 \text{ MJ/mol}\)) than the valence electron peak for \(\text{Ne}\) (\(\approx 2.1 \text{ MJ/mol}\)), despite \(\text{Na}\) having a greater nuclear charge (\(Z = 11\)) than \(\text{Ne}\) (\(Z = 10\))?

Simulated photoelectron spectra (PES) for isolated, ground-state atoms of \(\text{Ne}\) (atomic number \(10\)) and \(\text{Na}\) (atomic number \(11\)) are shown below.

Which of the following best explains why the valence electron peak for \(\text{Na}\) appears at a substantially lower binding energy (\(\approx 0.50 \text{ MJ/mol}\)) than the valence electron peak for \(\text{Ne}\) (\(\approx 2.1 \text{ MJ/mol}\)), despite \(\text{Na}\) having a greater nuclear charge (\(Z = 11\)) than \(\text{Ne}\) (\(Z = 10\))?

![A grayscale photoelectron spectroscopy diagram consisting of two vertically stacked rectangular panels sharing a common horizontal axis. The horizontal axis is oriented at the bottom with values decreasing logarithmically from left to right, with major tick marks labeled 1000, 100, 10, 1, and 0.1, and an axis title labeled Binding Energy (MJ/mol). The vertical axis on each panel is labeled Relative Number of Electrons. The top panel is labeled Ne and displays three solid black vertical line peaks: peak 1 at 84 MJ/mol with a relative height of 2, peak 2 at 4.7 MJ/mol with a relative height of 2, and peak 3 at 2.1 MJ/mol with a relative height of 6. The bottom panel is labeled Na and displays four solid black vertical line peaks: peak 1 at 104 MJ/mol with a relative height of 2, peak 2 at 6.8 MJ/mol with a relative height of 2, peak 3 at 3.7 MJ/mol with a relative height of 6, and peak 4 at 0.50 MJ/mol with a relative height of 1. No other peaks, curves, gridlines, labels, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790596816-khrLVh.jpg)

- **A.** The \(3s\) electron in \(\text{Na}\) experiences greater electron-electron repulsion within its subshell than the \(2p\) electrons in \(\text{Ne}\), which destabilizes the orbital and decreases the energy required to remove the electron.
- **B.** The \(3s\) subshell in \(\text{Na}\) has a lower angular momentum quantum number than the \(2p\) subshell in \(\text{Ne}\), which allows the \(3s\) electron to penetrate closer to the nucleus and be more easily removed.
- **C.** The valence electron in \(\text{Na}\) occupies the \(n = 3\) shell, where it is at a greater average distance from the nucleus and is shielded by \(10\) core electrons, resulting in a weaker Coulombic attraction to the nucleus than that experienced by the \(n = 2\) valence electrons in \(\text{Ne}\).
- **D.** The \(2p\) electrons in \(\text{Ne}\) experience complete shielding from the \(1s\) and \(2s\) electrons, whereas the single \(3s\) electron in \(\text{Na}\) is unshielded by core electrons and held only by weak dispersion forces.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123584/*