Photoelectron Spectroscopy - AP Chemistry

SECTION 1

Historical Context & Motivation

Understanding the internal electronic structure of atoms has been one of the grand challenges of modern chemistry and physics. In the late nineteenth and early twentieth centuries, scientists knew atoms contained electrons, but they had no direct experimental method to determine how those electrons were arranged or how tightly each was bound to the nucleus. Emission and absorption spectroscopy provided indirect clues—revealing transitions between energy levels—yet these techniques could not isolate the binding energy of every individual electron. The development of photoelectron spectroscopy (PES) changed that picture entirely, giving chemists a direct window into the quantized energy landscape of atoms and molecules.

1887 • Hertz Observes the Photoelectric Effect

Heinrich Hertz noticed that ultraviolet light striking a metal surface facilitated spark discharge, the first recorded observation of the photoelectric effect. The phenomenon defied classical wave theory because the energy of ejected electrons depended on light frequency, not intensity.

1905 • Einstein's Quantum Explanation

Albert Einstein proposed that light consists of discrete energy packets—photons—each carrying energy E = hν. This quantization explained why only photons above a threshold frequency could eject electrons, earning Einstein the 1921 Nobel Prize.

1957 • Kai Siegbahn Develops ESCA

Swedish physicist Kai Siegbahn pioneered Electron Spectroscopy for Chemical Analysis (ESCA), using X-ray photons to eject core electrons from atoms. This technique could resolve binding energies of inner-shell electrons with high precision.

1962 • Turner & UV Photoelectron Spectroscopy

David W. Turner introduced ultraviolet photoelectron spectroscopy (UPS) to measure the binding energies of valence electrons in molecules. This complemented Siegbahn's X-ray technique by providing finer detail on outer-shell electrons.

1981 • Nobel Prize for PES

Kai Siegbahn received the Nobel Prize in Physics for his development of high-resolution electron spectroscopy. PES became an indispensable analytical tool across chemistry, materials science, and surface physics.

The central question PES addresses is deceptively simple: how much energy is required to remove each electron from an atom? Answering this question reveals the shell and subshell structure predicted by quantum mechanics, provides direct evidence for electron configurations, and confirms periodic trends in ionization energy. On the AP Chemistry exam, PES data appears as spectra that you must interpret to deduce the identity and electronic configuration of an element—making this technique one of the most testable topics in Unit 1.

SECTION 2

Core Principles & Definitions

Photoelectron spectroscopy operates on a beautifully direct principle: bombard a sample with photons of known energy, measure the kinetic energy of the ejected electrons, and calculate the difference to determine how tightly each electron was bound. The technique transforms Einstein's photoelectric equation from a theoretical curiosity into a practical laboratory tool. Before diving into the mathematics, it is essential to build a solid conceptual foundation around several interconnected ideas.

Binding Energy (BE)

The minimum energy needed to remove an electron from its orbital in an atom. Electrons closer to the nucleus have higher binding energies because they experience a greater effective nuclear charge (Z_eff).

Photon Source

A monochromatic X-ray or UV source provides photons of a single, precisely known energy (hν). X-ray sources (e.g., Al Kα at 1486.6 eV) eject core and valence electrons; UV sources probe only valence electrons.

Kinetic Energy (KE)

The leftover energy an ejected electron carries as motion. An electron analyzer measures KE, and the binding energy is computed as BE = hν − KE. Electrons with low KE were tightly bound.

Relative Peak Heights

On a PES spectrum, the height (or area) of each peak is proportional to the number of electrons in that subshell. A peak twice as tall as another indicates twice as many electrons.

Energy-Level Ordering

PES spectra display peaks from highest to lowest binding energy (left to right). This ordering directly maps to the subshell filling order: 1s, 2s, 2p, 3s, 3p, etc.

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KEY TAKEAWAY

Think of PES as an airport security scanner for atoms. Just as an X-ray scanner reveals the layers inside your luggage—outer pockets, inner compartments, and the well-packed center—PES reveals the electron layers of an atom. Photons act like the scanner beam, and the energy it takes to dislodge each item (electron) tells you how deeply it was buried. Items near the surface pop out easily (low binding energy), while items packed in the core require much more energy to extract. The number of items at each depth corresponds to the peak height on the spectrum.

One photoelectron spectroscopy chamber Photoelectron spectroscopy removes electrons with photons and measures the outgoing electrons’ kinetic energies.

SECTION 3

Visual Explanation — Reading a PES Spectrum

A PES spectrum is a graph with binding energy on the horizontal axis (decreasing from left to right) and relative number of electrons on the vertical axis. Each peak corresponds to a distinct subshell, and its position (binding energy) and height (electron count) together encode the complete electron configuration of the element. The diagram below shows the PES spectrum for nitrogen (Z = 7), which has the configuration 1s²2s²2p³.

The PES spectrum of nitrogen shows three peaks. The leftmost peak (highest binding energy, ~40.2 MJ/mol) corresponds to the 1s subshell with 2 electrons. The middle peak (~2.45 MJ/mol) corresponds to the 2s subshell with 2 electrons. The rightmost and tallest peak (~1.40 MJ/mol) represents the 2p subshell with 3 electrons. Note that the 1s and 2s peaks are the same height (both have 2 electrons), while the 2p peak is 1.5× taller.

When interpreting any PES spectrum on the AP exam, follow a systematic approach. First, count the total number of peaks—each peak represents one subshell. Second, examine relative peak heights to determine how many electrons occupy each subshell. Third, note that binding energy decreases from left to right; peaks on the far left are core electrons (1s), and peaks on the far right are valence electrons. Fourth, sum all the electrons across all peaks to determine the total electron count, which equals the atomic number Z for a neutral atom. In the nitrogen example above, 2 + 2 + 3 = 7 electrons, confirming that the element is nitrogen.

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AP Exam Tip

The x-axis on AP Chemistry PES spectra always has binding energy decreasing from left to right. This convention is opposite to most graphs you have encountered. Always check the axis labels before interpreting the spectrum.

SECTION 4

Mathematical Framework

The quantitative backbone of PES is Einstein's photoelectric equation, adapted for atomic and molecular systems. A photon of known energy strikes the sample, and its energy is partitioned into two components: the energy consumed to liberate the electron from its orbital (the binding energy) and the residual energy carried away by the electron as kinetic energy. This energy conservation relationship is the single most important equation in PES.

PHOTOELECTRIC EQUATION

hν = BE + KE

where h = Planck's constant (6.626 × 10⁻³⁴ J·s), ν = frequency of incident photon (Hz), BE = binding energy of the electron (J or MJ/mol on PES spectra), KE = kinetic energy of the ejected electron (J).

BINDING ENERGY (REARRANGED)

BE = hν − KE

In practice, the photon energy hν is fixed by the instrument's light source, and the electron analyzer measures KE. The binding energy is computed as the difference. Electrons from deeper subshells have larger BE values and correspondingly smaller KE values.

PHOTON ENERGY FROM WAVELENGTH

E = hc / λ

where c = speed of light (3.00 × 10⁸ m/s) and λ = wavelength of the photon (m). This equation converts between wavelength and energy when the source is characterized by wavelength rather than frequency.

On the AP Chemistry exam, PES binding energies are typically reported in megajoules per mole (MJ/mol) rather than joules per photon or electronvolts. This is simply a unit scaling: one mole of photon-electron interactions is considered instead of a single event. The conceptual physics remains identical. You should be comfortable converting between units using Avogadro's number (6.022 × 10²³ mol⁻¹) and the conversion 1 eV = 1.602 × 10⁻¹⁹ J, although most AP problems supply data already in MJ/mol.

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Key Relationship

Binding energy correlates with effective nuclear charge (Z_eff). Electrons experiencing a higher Z_eff (due to being closer to the nucleus or experiencing less shielding) appear as peaks further to the left on the PES spectrum. This is why 1s electrons always form the leftmost peak, and valence electrons form the rightmost peaks.

SECTION 5

Interpreting PES Spectra for Different Elements

The true power of PES is its ability to identify any element and confirm its electron configuration. As you move across the periodic table, each element adds one proton and one electron, producing a characteristic PES fingerprint. The number of peaks tells you how many occupied subshells exist, the relative heights encode the electron count per subshell, and the binding energy positions shift predictably with atomic number. The diagram below compares the PES spectra of three second-period elements—carbon, nitrogen, and oxygen—to illustrate how spectra evolve across a period.

All three elements have three peaks (1s, 2s, 2p), but the 2p peak grows taller from C to N to O as electrons fill the 2p subshell. The 1s and 2s peaks remain the same height (2 electrons each). As Z increases, all peaks shift slightly to the left because higher nuclear charge increases all binding energies.

Patterns to Recognize on the AP Exam

Observation	What It Tells You	Example
Number of peaks	Number of occupied subshells	Na has 4 peaks → 1s, 2s, 2p, 3s
Relative peak heights	Number of electrons in each subshell	A peak 3× as tall as the 1s peak → 6 electrons (e.g., 2p⁶)
Sum of all electron counts	Atomic number Z (for neutral atoms)	2 + 2 + 6 + 2 + 1 = 13 → Al
Large gap between two consecutive peaks	Transition from one principal energy level to the next	Gap between 1s and 2s ≫ gap between 2s and 2p
Rightmost peak with 1 electron	Likely an alkali metal (Group 1)	Na → last peak at very low BE, height = 1

Quick-reference patterns for PES spectrum interpretation

SECTION 6

Worked Example — Identifying an Unknown Element

A PES spectrum of an unknown neutral element shows five peaks with the following relative heights (from highest to lowest binding energy): 2, 2, 6, 2, 4. Identify the element and write its full electron configuration.

Identifying an Element from PES Data

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Step 1 — Count the Peaks and Assign Subshells

Five peaks indicate five occupied subshells. In order of decreasing binding energy (left to right on the spectrum), these correspond to the standard filling order: 1s, 2s, 2p, 3s, 3p. Each peak's height gives the electron count for that subshell.
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Step 2 — Map Heights to Electron Counts

Reading the relative heights in order: 1s has 2 electrons, 2s has 2 electrons, 2p has 6 electrons, 3s has 2 electrons, and 3p has 4 electrons.

Electron configuration: 1s²2s²2p⁶3s²3p⁴
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Step 3 — Sum the Electrons to Find Z

Total electrons = 2 + 2 + 6 + 2 + 4 = 16. Since the atom is neutral, the atomic number Z = 16.

Z = 16 → Sulfur (S)
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Step 4 — Verify with Periodic Trends

Sulfur is in Period 3, Group 16. It should have electrons filling through the 3p subshell with 4 electrons in 3p, which matches the spectrum. The 3p peak should appear at a lower binding energy than the 3s peak because 3p electrons experience more shielding from the 3s electrons and are on average farther from the nucleus.

Confirmed: The element is sulfur (S) with configuration [Ne]3s²3p⁴.

SECTION 7

Strengths, Limitations & Comparison with Other Techniques

Like any analytical technique, PES has characteristic strengths and limitations. Understanding these helps you appreciate why PES data appears on the AP exam and where its explanatory power is complemented by other spectroscopic methods.

Strengths	Limitations
Provides direct measurement of binding energies for all electrons, not just valence	Requires high vacuum; samples must be in the gas phase or on clean surfaces
Peak heights give quantitative electron counts per subshell	Cannot distinguish between elements with identical electron configurations (isoelectronic species) without additional data
Confirms electron configuration predictions from quantum mechanics	Resolution of closely spaced subshells (e.g., 3d and 4s) can be challenging
Applicable to all elements across the periodic table	Expensive instrumentation; not a routine benchtop technique
Reveals trends in Z_eff and shielding across and down the periodic table	Does not directly reveal molecular geometry or bonding (complementary techniques like IR or NMR are needed)

Strengths and limitations of photoelectron spectroscopy

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KEY TAKEAWAY

PES is to atomic structure what an MRI is to medicine: it provides a detailed internal map without destroying the overall pattern you are investigating. Just as an MRI reveals different tissue layers at different depths, PES reveals electron subshells at different energy levels. However, just as an MRI cannot tell you how two bones articulate at a joint (you need an X-ray or CT for that), PES alone cannot tell you about molecular shape or bond type—you need complementary spectroscopic techniques for those questions.

SECTION 8

Connections to Ionization Energy & Periodic Trends

PES data is intimately connected to the concept of ionization energy (IE), one of the most important periodic trends you will study. The first ionization energy of an element—the energy required to remove the outermost (most loosely bound) electron—corresponds to the binding energy of the rightmost peak on a PES spectrum. Higher successive ionization energies correspond to peaks progressively further to the left. Thus, PES provides a complete, experimentally measured ionization energy profile for every electron, not just the first.

Concept	PES Interpretation	Periodic Trend
First Ionization Energy (IE₁)	Binding energy of the rightmost peak (lowest BE)	Increases across a period; decreases down a group
Effective Nuclear Charge (Z_eff)	Higher Z_eff shifts all peaks to higher binding energies (left)	Increases across a period (more protons, similar shielding)
Electron Shielding	Inner subshell electrons reduce Z_eff for outer electrons → large BE gaps between shells	New shells add significant shielding (explains Group 1 low IE)
Subshell Energy Splitting	Within a shell, s electrons have higher BE than p electrons (less shielding, more penetration)	Explains exceptions at Groups 13 and 16 (IE dips)

Connecting PES to ionization energy and periodic trends

Two well-known exceptions to the general left-to-right increase in first ionization energy across a period are explained beautifully by PES data. The dip from Be (Group 2) to B (Group 13) occurs because boron's outermost electron is in the 2p subshell, which has a lower binding energy than the 2s subshell of beryllium. The dip from N (Group 15) to O (Group 16) arises because oxygen's fourth 2p electron must pair with an existing electron in a 2p orbital, creating electron-electron repulsion that lowers the binding energy. Both of these anomalies are directly visible in PES spectra: the rightmost peak of boron sits at a lower binding energy than that of beryllium, and the rightmost peak of oxygen sits at a lower binding energy than that of nitrogen.

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Looking Ahead

In more advanced courses (physical chemistry and quantum mechanics), PES data is connected to Koopmans' theorem, which states that the ionization energy of an electron approximates the negative of its orbital energy calculated by Hartree-Fock theory. This bridge between experiment and theory is one of the most powerful validations of quantum mechanical models of atomic structure.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A PES spectrum of a neutral atom shows four peaks with relative heights of 2, 2, 6, and 1 (listed from highest to lowest binding energy). Which element does this spectrum represent?

A. Fluorine (Z = 9)
B. Neon (Z = 10)
C. Sodium (Z = 11)
D. Magnesium (Z = 12)

Show answer (C)

The correct answer is C (Sodium). The four peaks correspond to the subshells 1s, 2s, 2p, and 3s with electron counts of 2, 2, 6, and 1, respectively. The total electron count is 2 + 2 + 6 + 1 = 11, which corresponds to sodium (Z = 11). Choice A (Fluorine, Z = 9) is incorrect because fluorine has 9 electrons with configuration 1s²2s²2p⁵, giving three peaks with heights 2, 2, 5. Choice B (Neon, Z = 10) is incorrect because neon has 10 electrons (1s²2s²2p⁶) and only three peaks. Choice D (Magnesium, Z = 12) is incorrect because magnesium has configuration 1s²2s²2p⁶3s² with a 3s peak height of 2, not 1.

Question 2: BASIC CALCULATION

An X-ray photoelectron spectrometer uses photons with energy 1486.6 eV. An electron is ejected from a sample with a measured kinetic energy of 1379.1 eV. What is the binding energy of that electron?

A. 2865.7 eV
B. 107.5 eV
C. 1379.1 eV
D. 1486.6 eV

Show answer (B)

The correct answer is B. Using the photoelectric equation BE = hν − KE = 1486.6 eV − 1379.1 eV = 107.5 eV. Choice A (2865.7 eV) incorrectly adds hν + KE instead of subtracting. Choice C (1379.1 eV) confuses the kinetic energy of the ejected electron with the binding energy. Choice D (1486.6 eV) is simply the photon energy, not the binding energy. Remember: BE and KE always sum to the total photon energy, so BE must always be less than hν.

Question 3: INTERMEDIATE

The PES spectra of two consecutive elements in Period 3 are compared. Element X has a rightmost peak at a binding energy of 1.01 MJ/mol, and Element Y has a rightmost peak at 0.74 MJ/mol, even though Y has a higher atomic number. Which pair of elements best explains this observation?

A. Mg (Z = 12) and Al (Z = 13)
B. Na (Z = 11) and Mg (Z = 12)
C. Si (Z = 14) and P (Z = 15)
D. Ar (Z = 18) and K (Z = 19)

Show answer (A)

The correct answer is A. The rightmost peak represents the outermost electron. Magnesium (1s²2s²2p⁶3s²) has its outermost electrons in the 3s subshell, while aluminum (1s²2s²2p⁶3s²3p¹) has its outermost electron in the 3p subshell. Despite Al having a higher Z, the 3p electron is higher in energy (less tightly bound) than a 3s electron because 3p orbitals experience less nuclear penetration and more shielding. This explains the drop in first IE from Mg to Al, paralleling the Be-to-B drop in Period 2. Choice B is wrong because Na to Mg both involve 3s electrons and IE increases normally. Choice C is wrong because Si and P both involve 3p electrons, and IE increases from Si to P without a subshell change. Choice D is wrong because Ar and K are in different periods, not consecutive elements within Period 3.

Question 4: APPLIED

7 pts

The table below shows the binding energies and relative number of electrons for the PES spectrum of an unknown element. Peak 1: BE = 151.0 MJ/mol, relative electrons = 2 Peak 2: BE = 17.4 MJ/mol, relative electrons = 2 Peak 3: BE = 10.9 MJ/mol, relative electrons = 6 Peak 4: BE = 1.95 MJ/mol, relative electrons = 2 Peak 5: BE = 1.01 MJ/mol, relative electrons = 2 (a) Identify the element. Justify your answer. (b) Write the full electron configuration and the noble gas shorthand configuration for this element. (c) Explain why there is a much larger gap in binding energy between Peak 3 and Peak 4 than between Peak 4 and Peak 5. (d) Predict how this element's first ionization energy compares to that of sodium (IE₁ = 0.50 MJ/mol). Justify your prediction using the PES data.

Show answer

Rubric (7 points): (a) Identification (2 points) • 1 point: Total electrons = 2 + 2 + 6 + 2 + 2 = 14. • 1 point: Z = 14 → the element is silicon (Si). (b) Electron configuration (2 points) • 1 point: Full configuration: 1s²2s²2p⁶3s²3p². — Note: The PES data shows Peaks 4 and 5 both with 2 electrons. Peak 4 (1.95 MJ/mol) = 3s² and Peak 5 (1.01 MJ/mol) = 3p². The student must recognize this matches 5 peaks = 5 subshells, though 3p has only 2 of its maximum 6 electrons. • 1 point: Noble gas shorthand: [Ne]3s²3p². (c) Binding energy gap explanation (2 points) • 1 point: The gap between Peak 3 (2p, n = 2) and Peak 4 (3s, n = 3) represents a transition from one principal energy level to the next. Electrons in the n = 2 shell are much closer to the nucleus and experience far greater effective nuclear charge than those in n = 3. • 1 point: Within the same principal level (3s vs. 3p), the energy difference is small because both subshells experience similar Z_eff, with only modest differences due to penetration effects. (d) Comparison to sodium's IE₁ (1 point) • 1 point: Silicon's first ionization energy (≈ 1.01 MJ/mol, the BE of the rightmost peak) is significantly higher than sodium's (0.50 MJ/mol). This is because silicon has Z = 14 vs. sodium's Z = 11, giving silicon a much higher effective nuclear charge on its valence electrons. Both are in Period 3, but silicon is further to the right, and the additional protons are not fully offset by the shielding of the additional electrons.

Question 5: CRITICAL THINKING

7 pts

A student collects PES data for four neutral atoms: W, X, Y, and Z. The binding energy (in MJ/mol) of the 1s electrons and the first ionization energy (IE₁, corresponding to the rightmost/lowest-energy peak in the PES spectrum) for each element are given below. | Element | 1s BE (MJ/mol) | IE₁ (MJ/mol) | |---------|----------------|---------------| | W | 1.36 | 0.801 | | X | 1.68 | 1.09 | | Y | 2.03 | 1.40 | | Z | 2.41 | 1.31 | All four elements are consecutive elements in the same period of the periodic table, listed in order of increasing atomic number. (a) Explain the general trend in 1s binding energy from W to Z. (b) The first ionization energies generally increase from W to Z but show a slight dip from Y to Z. Identify which period these elements belong to and assign a specific element to each of W, X, Y, and Z. Justify your answer using both the 1s binding energies and the IE₁ trend. (c) Using your assignments from part (b), explain why the IE₁ of Z is slightly lower than that of Y despite Z having a higher atomic number. (d) A student claims: "Because element Z has the highest atomic number of the four, it must also have the highest 1s binding energy." Is this claim valid? Explain why the trend in 1s binding energy does not show the same anomaly as the IE₁ trend.

Show answer

Rubric (7 points): (a) 1s binding energy trend (1 point) • 1 point: As atomic number increases from W to Z, the 1s binding energy increases because the nuclear charge increases. The 1s electrons experience very little shielding from other electrons (they are the innermost electrons), so the effective nuclear charge felt by 1s electrons increases nearly one-for-one with each additional proton, resulting in progressively higher binding energies. (b) Identification of elements (3 points) • 1 point: The four elements are in Period 2 of the periodic table. The 1s binding energies (1.36, 1.68, 2.03, 2.41 MJ/mol) correspond to the known approximate 1s binding energies for B (1.36), C (1.68), N (2.03), and O (2.41 MJ/mol), which are Period 2 elements with Z = 5 through 8. • 1 point: The IE₁ values generally increase across the period (0.801 → 1.09 → 1.40) but show a characteristic dip from X to Y (1.40 → 1.31 MJ/mol). The known IE₁ values are: B = 0.801, C = 1.09, N = 1.40, O = 1.31 MJ/mol. The dip from N (1.40) to O (1.31) matches the dip from X to Y. • 1 point: Therefore: W = B (Z = 5), X = C (Z = 6), Y = N (Z = 7), Z = O (Z = 8). Wait — the dip occurs from Y to Z in the data, not from X to Y. Re-examining: IE₁ values are W = 0.801, X = 1.09, Y = 1.40, Z = 1.31. The dip is from Y to Z. Known values: N (1.40) to O (1.31) is the dip. So W = B (Z = 5), X = C (Z = 6), Y = N (Z = 7), Z = O (Z = 8). (c) IE₁ of Z (oxygen) < IE₁ of Y (nitrogen) (2 points) • 1 point: Nitrogen (Y) has the electron configuration 1s² 2s² 2p³, with a half-filled 2p subshell in which each 2p orbital is singly occupied. There is no electron-electron pairing repulsion in any 2p orbital, so the outermost electrons are relatively tightly held. • 1 point: Oxygen (Z) has the configuration 1s² 2s² 2p⁴, meaning one of the 2p orbitals must contain a pair of electrons. The electron-electron repulsion between the two electrons sharing that orbital destabilizes the fourth 2p electron, making it easier to remove. This pairing energy penalty outweighs the small increase in nuclear charge from Z = 7 to Z = 8, so oxygen has a lower IE₁ than nitrogen. (d) Validity of the claim about 1s binding energy (1 point) • 1 point: The claim is valid. Because all four elements are in Period 2, the 1s subshell is a core subshell for all of them (except that for these light elements, 1s electrons experience essentially no shielding). The 1s binding energy increases steadily with atomic number as nuclear charge increases: B (1.36) < C (1.68) < N (2.03) < O (2.41 MJ/mol). Element Z (oxygen) has the highest atomic number and therefore the highest 1s binding energy. Unlike IE₁ (which involves the outermost 2p electrons and is affected by pairing repulsion), the 1s binding energy follows the nuclear charge trend monotonically because the 1s electrons have no other electrons shielding them from the nucleus.

SUMMARY

Lesson Summary

Summary

Photoelectron spectroscopy (PES) is founded on the photoelectric effect: when a photon of known energy (hν) strikes an atom, an electron is ejected with kinetic energy KE, and the binding energy is calculated as BE = hν − KE. A PES spectrum plots binding energy (decreasing left to right) versus relative number of electrons (peak height), with each peak corresponding to one subshell in the atom's electron configuration.

To interpret a spectrum, count peaks (= number of occupied subshells), read relative heights (= electron count per subshell), and sum all electrons to determine atomic number Z. The rightmost peak gives the first ionization energy, and trends in binding energy across elements confirm periodic patterns in effective nuclear charge and electron shielding. Anomalies in ionization energy (e.g., the dips at Groups 13 and 16) are directly visible in the spectra as shifts in the outermost peak position, providing powerful experimental evidence for the quantum mechanical model of the atom.