---
title: "The complete photoelectron spectra of two isoelectronic chemical species, gas-phase \\(\\text{Ne}\\) and gas-phase \\(\\text{Na}^+\\), are shown in the figure. Which of the following correctly identifies the spectrum that corresponds to \\(\\text{Na}^+\\) and provides the best explanation for the difference in peak positions between the two spectra?"
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url: "https://nerd-notes.com/ubq/119548/"
date_modified: "2026-08-21T03:27:00+00:00"
---

# The complete photoelectron spectra of two isoelectronic chemical species, gas-phase \(\text{Ne}\) and gas-phase \(\text{Na}^+\), are shown in the figure. Which of the following correctly identifies the spectrum that corresponds to \(\text{Na}^+\) and provides the best explanation for the difference in peak positions between the two spectra?

The complete photoelectron spectra of two isoelectronic chemical species, gas-phase \(\text{Ne}\) and gas-phase \(\text{Na}^+\), are shown in the figure. Which of the following correctly identifies the spectrum that corresponds to \(\text{Na}^+\) and provides the best explanation for the difference in peak positions between the two spectra?

![A grayscale figure containing two vertically stacked photoelectron spectra sharing identical axes. The horizontal axis is labeled 'Binding Energy (MJ/mol)' and decreases from left to right with scale markings at 100, 10, and 1. The vertical axis is labeled 'Relative Number of Electrons' with unit tick marks at 2, 4, and 6. The top panel is labeled 'Spectrum X' and displays three distinct solid vertical peak lines: the first peak is at 84.0 MJ/mol with a height of 2 units; the second peak is at 4.68 MJ/mol with a height of 2 units; the third peak is at 2.08 MJ/mol with a height of 6 units. The bottom panel is labeled 'Spectrum Y' and displays three distinct solid vertical peak lines: the first peak is at 104 MJ/mol with a height of 2 units; the second peak is at 6.84 MJ/mol with a height of 2 units; the third peak is at 3.67 MJ/mol with a height of 6 units. No other lines, labels, gridlines, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787282819-F3yJRe.jpg)

- **A.** Spectrum Y, because \(\text{Na}^+\) has 11 protons compared to 10 protons in \(\text{Ne}\), resulting in a greater effective nuclear charge that exerts a stronger Coulombic attraction on electrons in all occupied subshells.
- **B.** Spectrum Y, because \(\text{Na}^+\) has a smaller ionic radius than \(\text{Ne}\), which decreases electron shielding in the \(1s\) subshell and increases valence electron-electron repulsions.
- **C.** Spectrum X, because the positive charge on \(\text{Na}^+\) increases core electron shielding, which lowers the net attractive force between the nucleus and the valence electrons.
- **D.** Spectrum X, because the loss of an electron to form \(\text{Na}^+\) decreases the nuclear charge, allowing all remaining electrons to be removed with less energy.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119548/*