Photoelectron Spectroscopy
Help Questions
AP Chemistry › Photoelectron Spectroscopy
A simplified PES for an atom shows three peaks with binding energy increasing to the left. The leftmost peak has height 2, the middle peak has height 2, and the rightmost peak has height 3. Which subshell is represented by the rightmost peak?
3s
1s
2s
3p
2p
Explanation
This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy shows energy to remove electrons, higher for core levels. The plot increases energy left, with right peaks for valence. Peak heights are proportional to electrons in subshells. A tempting distractor is E 3p, assuming period 3, but three peaks with rightmost 3 match 2p³ for nitrogen-like atoms. A transferable strategy is that inner electrons have higher binding energy, and p subshell heights range from 1 to 6.
A PES for an atom shows six peaks with binding energy increasing to the left. From left to right, the relative peak heights are 2, 2, 6, 2, 6, and 1. Which element is most consistent with this PES?
Cl
Ar
Na
K
Ca
Explanation
This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy quantifies attraction, greater for inner shells. Spectra show left-higher energy, from 1s outward. Heights indicate subshell electrons, for configuration matching. A tempting distractor is D Na, with four peaks, but six peaks of 2,2,6,2,6,1 match potassium's 1s²2s²2p⁶3s²3p⁶4s¹. A transferable strategy is that inner electrons have higher binding energy, and the rightmost peak identifies valence electrons.
A simplified PES for an atom shows four peaks (binding energy increases to the left). The peak heights from left to right are 2, 2, 6, and 6. A student concludes that the atom must be in the third period because the rightmost peak is the largest. Which statement best evaluates the student’s conclusion?
The conclusion is correct because the largest peak must correspond to the 3p subshell.
The conclusion is incorrect because binding energy increases to the right in PES.
The conclusion is incorrect because peak height indicates number of electrons, not the period by itself.
The conclusion is correct because a larger rightmost peak always indicates a higher principal energy level.
The conclusion is correct because the rightmost peak always corresponds to the 1s subshell.
Explanation
This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy measures nuclear pull on electrons, higher for inner shells. Spectra plot increasing energy left, from core to valence. Peak heights correspond to electron counts per subshell. A tempting distractor is C, thinking largest peak is always 3p, but height indicates electrons, not period, so the conclusion errs without considering peak count. A transferable strategy is that inner electrons have higher binding energy, and analyze both height and position for accurate interpretation.
A simplified PES shows peaks labeled 1s, 2s, 2p, 3s, and 3p. The 2p peak is at a higher binding energy than the 3s peak. Which statement best explains this ordering?
The 2p peak is taller than the 3s peak, so it must have higher binding energy.
Electrons in the 2p subshell have lower binding energy because p electrons are always less tightly bound than s electrons.
Electrons in the 3s subshell experience less shielding than electrons in the 2p subshell.
Electrons in the 3s subshell have the same binding energy as electrons in the 2p subshell because both are s and p orbitals.
Electrons in the 2p subshell are, on average, closer to the nucleus than electrons in the 3s subshell and are held more tightly.
Explanation
This question tests photoelectron spectroscopy. Electrons in different principal energy levels (shells) have different binding energies, with inner shells having higher binding energy. The 2p electrons are in the n=2 shell while 3s electrons are in the n=3 shell, making 2p electrons closer to the nucleus on average and more tightly bound. This results in higher binding energy for 2p compared to 3s, regardless of the s vs p orbital type. Students might incorrectly think peak height determines binding energy (choice C) or that orbital type matters more than shell number (choice E). Remember that binding energy primarily depends on distance from nucleus: inner electrons have higher binding energy than outer electrons.
A simplified PES of an atom shows peaks labeled 1s, 2s, 2p, and 3s only (no 3p peak is present). The relative heights are 1s: 2, 2s: 2, 2p: 6, 3s: 2. Which element is most consistent with this spectrum?
Na
Mg
Ne
Al
Si
Explanation
This question tests photoelectron spectroscopy. The spectrum shows only four peaks (1s, 2s, 2p, 3s) with no 3p peak, meaning there are no electrons in the 3p subshell. The peak heights sum to 12 electrons total (2+2+6+2), corresponding to magnesium (Mg) with configuration 1s²2s²2p⁶3s². The absence of a 3p peak is crucial - it means all valence electrons are in the 3s subshell. Students might incorrectly choose aluminum (Al) thinking it could have an empty 3p, but Al would show a 3p peak with one electron. Remember that PES only shows peaks for occupied subshells; empty subshells produce no signal.
A simplified PES for an atom shows five peaks labeled 1s, 2s, 2p, 3s, and 3p. The 3p peak has relative height 1 (much smaller than the 3s peak height of 2). Which element is most consistent with this spectrum?
Mg
Al
P
Na
Si
Explanation
This question tests photoelectron spectroscopy. The spectrum shows five peaks with the 3p peak having height 1, indicating just one electron in the 3p subshell. Combined with a 3s peak of height 2, this represents the electron configuration ending in 3s²3p¹, which is aluminum (Al) with 13 total electrons. The complete configuration is 1s²2s²2p⁶3s²3p¹. Students might incorrectly choose sodium (Na) or magnesium (Mg), but these wouldn't have any 3p electrons and would show only four peaks. Remember that the presence and height of peaks directly indicate which subshells contain electrons and how many.
A simplified PES for an atom shows three peaks. The highest binding energy peak has height 2. The other two peaks are at lower binding energies and have heights 2 and 3. The peaks correspond to 1s, 2s, and 2p electrons. Which element is most consistent with the spectrum?
Be
B
Ne
N
Li
Explanation
This problem involves photoelectron spectroscopy analysis. In PES, the highest binding energy peak corresponds to 1s electrons (closest to nucleus), and peak height indicates electron count in each subshell. The spectrum shows peaks with heights 2, 2, and 3, representing 1s (2e⁻), 2s (2e⁻), and 2p (3e⁻) respectively. Adding these gives 7 total electrons, identifying the element as nitrogen (N) with electron configuration 1s² 2s² 2p³. Boron might seem plausible since it also has electrons in all three subshells, but it would show a 2p peak height of only 1, not 3. The strategy is to use peak heights to count total electrons and match to the atomic number of elements in the answer choices.
An atom’s PES shows four peaks. From highest to lowest binding energy, the peak heights are 2, 2, 6, and 1. The peaks correspond to the subshells 1s, 2s, 2p, and 3s (not necessarily in that order). Which subshell corresponds to the peak with height 1?
3p
2p
3s
1s
2s
Explanation
This problem tests photoelectron spectroscopy concepts. In PES, binding energy decreases as you move from inner to outer electron shells (1s > 2s > 2p > 3s), and peak height corresponds to the number of electrons in each subshell. The peak heights 2, 2, 6, and 1 (from highest to lowest binding energy) represent the electron counts in each subshell. Since s subshells hold maximum 2 electrons and p subshells hold maximum 6 electrons, the heights correspond to: 1s (2), 2s (2), 2p (6), and 3s (1). The peak with height 1 represents a partially filled subshell with only one electron, which must be the 3s subshell. Students might incorrectly think the smallest peak corresponds to 1s because it's the "first" subshell, but 1s is always fully filled with 2 electrons in neutral atoms beyond hydrogen. Remember: the outermost electrons have the lowest binding energy and can be partially filled.
A simplified PES for an atom shows three peaks labeled 1s, 2s, and 2p. The 2s peak occurs at a higher binding energy than the 2p peak. Which best explains why the 2s electrons have higher binding energy than the 2p electrons in the same principal energy level?
2p electrons have a greater nuclear charge than 2s electrons.
2s electrons experience less shielding and are held more tightly by the nucleus than 2p electrons.
2s electrons have a higher principal energy level than 2p electrons.
2p electrons are more penetrating and therefore are held more tightly than 2s electrons.
2s electrons have lower binding energy because they are in an s subshell.
Explanation
This question tests photoelectron spectroscopy. Within the same principal energy level, s electrons have higher binding energy than p electrons because s orbitals penetrate closer to the nucleus and experience less shielding from inner electrons. The 2s electrons spend more time near the nucleus than 2p electrons, making them more tightly bound despite being in the same shell. This greater penetration means 2s electrons feel a stronger effective nuclear charge. Students might incorrectly think 2p electrons are held more tightly (choice A) or confuse principal energy levels (choice C). Remember that within the same shell, penetration follows the pattern s > p > d > f, leading to binding energies in the same order.
Two atoms, X and Y, are in the same period. Their simplified PES data show that the 1s peak of Y occurs at a higher binding energy than the 1s peak of X. Which conclusion is best supported by this observation?
Atom Y must have fewer valence electrons than atom X.
Atom Y must have a larger atomic radius than atom X.
Atom Y has more core electrons than atom X.
Atom Y has a greater nuclear charge than atom X.
Atom Y has fewer protons than atom X.
Explanation
This question tests photoelectron spectroscopy. When comparing atoms in the same period, higher binding energy for the same electron shell indicates stronger nuclear attraction. Since atom Y's 1s electrons have higher binding energy than atom X's 1s electrons, atom Y must have more protons (greater nuclear charge) pulling on those electrons. Being in the same period means they have the same number of electron shells, so the difference must be in nuclear charge. Students might incorrectly think higher binding energy means fewer protons (choice C) or relates to atomic radius (choice D). Remember that across a period, increasing nuclear charge leads to higher binding energies for all electron shells.