Teacher Notes

Photoelectron Spectroscopy

Guided-Inquiry Learning Activity for AP® Chemistry

Additional Materials Required

Periodic table (one per student group)

Teacher Tips

  • “Photoelectron Spectroscopy” is one of three learning activities in Spectroscopy Guided-Inquiry Learning Activities for AP® Chemistry, Catalog No. AP7710, from Flinn Scientific. The other two guided-inquiry activities in the AP7710 learning activity package are “Mass Spectrometry”and “Electron and Vibrational Spectroscopy.”
  • We recommend a collaborative, peer-learning environment for students to complete this guided-inquiry learning activity. Effective group sizes may include 3–4 students. To ensure active participation by all members of the group it is helpful to assign rotating roles to the students and call on different students in each group to be the “reporter” for class discussion and explanation of key parts of the activity.
  • The organization of this activity into three parts provides convenient stopping points to discuss key questions and relevant learning goals. See, for example, Questions 12–14 in Part A, Questions 4, 8 and 12 in Part B and Questions 5, 7, 9, 10, 13, 14 and 17 in Part C.
  • Ionization energies may also be called binding energies.
  • The PES data described in this activity comes from low-resolution photoelectron spectroscopy of elements in the gas phase. Modern photoelectron spectroscopy offers higher resolution and is more commonly used to analyze the surface composition and ionization or oxidation state of solids under high-vacuum conditions. Spectrophotometers for this purpose typically have more advanced features and are differentiated by the energy of the electromagnetic radiation source and the sensititivity of the electron analyzer. UPS refers to ultraviolet photoelectron spectroscopy while the abbreviation XPS is used for X-ray photoelectron spectroscopy. An ultraviolet source is generally capable of ionizing and analyzing only valence electrons, while XPS has enough energy to ionize both core and valence electrons. High-resolution photoelectron spectroscopy will show splitting or coupling of peaks from one subshell.
  • The evidence and reasoning in this learning activity should be used to recall, review and reinforce periodic trends in ionization energy, atomic radius, and the properties of matter. There are two main periodic trends in the value of the first or lowest ionization energy. (a) Ionization energy increases from left to right across a period or row in the periodic table due to the increase in the nuclear charge within the same principal energy level. (b) Ionization energy decreases from top to bottom in a group or column of the periodic table because of the increasing size of the atoms with the increase in the principal energy level.
  • Common misconceptions and conceptual hurdles or stumbling blocks regarding on photoelectron spectroscopy are summarized below and may need to be reinforced for students.
    • Ionization energy decreases from left to right along the x-axis in a photoelectron spectrum, with zero energy at the far right.
    • Electrons with lower ionization energies will have greater kinetic energies due to the photoelectric effect.
    • Assuming the energy of the radiation source is high enough, each electron in an atom has an equal probability of being ionized or ejected from the sample, but each atom can eject only one electron.
    • The potential energy of electrons in an atom is always negative, and the ionization energy is the “positive” energy that must be supplied to raise the energy of an electron to a “zero” or unbound state far away from the nucleus (and with no additional kinetic energy). Electrons that are closer to the nucleus are in lower potential energy states and thus have greater ionization energies.

Further Extensions

Alignment with AP® Chemistry Curriculum Framework

Essential Knowledge
1B1: The atom is composed of negatively charged electrons, which can leave the atom, and a positively charged nucleus that is made of protons and neutrons. The attraction of the electrons to the nucleus is the basis of the structure of the atom. Coulomb’s law is qualitatively useful for understanding the structure of the atom.
1B2: The electronic structure of the atom can be described using electron configuration that reflects the concept of electrons in quantized energy levels or shells. The energetics of the electrons in the atom can be understood by consideration of Coulomb’s law.
1C1: Many properties of atoms exhibit periodic trends that are reflective of the periodicity of electronic structure.

Learning Objectives
1.5 The student is able to explain the distribution of electrons in an atom or ion based upon data.
1.6 The student is able to analyze data relating to electron energies for patterns and relationships.
1.7 The student is able to describe the electronic structure of the atom using PES data, ionization energy data, and/or Coulomb’s law to construct explanations of how the energies of electrons within shells in atoms vary.
1.8 The student is able to explain the distribution of electrons using Coulomb’s law to analyze measured energies.
1.9 The student is able to predict and/or justify trends in atomic properties based on location on the periodic table and/or the shell model.

Correlation to Next Generation Science Standards (NGSS)

Science & Engineering Practices

Developing and using models
Analyzing and interpreting data
Constructing explanations and designing solutions
Obtaining, evaluation, and communicating information

Disciplinary Core Ideas

HS-PS1.A: Structure and Properties of Matter
HS-PS1.B: Chemical Reactions

Crosscutting Concepts

Cause and effect
Scale, proportion, and quantity
Structure and function

Performance Expectations

HS-PS1-1. Use the periodic table as a model to predict the relative properties of elements based on the patterns of electrons in the outermost energy level of atoms.
HS-PS1-2. Construct and revise an explanation for the outcome of a simple chemical reaction based on the outermost electron states of atoms, trends in the periodic table, and knowledge of the patterns of chemical properties.
HS-PS1-3. Plan and conduct an investigation to gather evidence to compare the structure of substances at the bulk scale to infer the strength of electrical forces between particles.
HS-PS2-6. Communicate scientific and technical information about why the molecular-level structure is important in the functioning of designed materials.

Answers to Questions

Part A. Coulomb’s Law and the Ionization Energy of Electrons

Analysis of atomic emission spectra leads to the conclusion that electrons in an atom reside in quantized energy levels or shells. The shells have discrete energies based on the distance separating the electrons from the nucleus and the charge of the nucleus. These energies may be qualitatively compared using Coulomb’s law, which states that the force (F) experienced by two charged objects, q1 and q2, is inversely related to the square of the distance (r) separating them (Equation 2). If the two objects have opposite charges, then the force between them is attractive; if the charges are the same then the force is repulsive.

  1. Rank diagrams A, B and C representing the distances between a proton (+) and an electron (−) in order of smallest force of attraction to largest. Explain how you made your determination

    Electrostatic attractive force: C < B < A. The magnitude of the force increases as the distance between the particles decreases.

  2. Which electron in diagrams A–C would require the most energy to be removed from the vicinity of the proton?

    The electron in diagram A would require the most energy to be separated from the nucleus because it is attracted to or held closer to the nucleus by the greatest force.

  3. If the distance between the proton and electron in diagram C were increased to 0.60 nm, would you predict the force of attraction to be stronger or weaker than the original?

    The force of attraction would be about 4X weaker if the distance were doubled.

  4. Consider the following diagrams representing different numbers of protons in an atomic nucleus. Which electron would require the most energy to be removed?

    The electrostatic attractive force increases (becomes stronger) when the charge on the particles increases. The electron depicted in F would require the most energy to be removed since it experiences the greatest attractive force to the nucleus.

  5. Describe the types of forces experienced by electrons X and Y in the following diagram.


    X and Y are both attracted to the nucleus. The electrons also experience a mutual repulsive force between them.

  6. Which electron would require more energy to be removed from the system of charges?

    Electron X is closer and more attracted to the positively charged “nucleus” and would therefore require more energy to be separated from it.
    Consider the following “shell diagram” for a hypothetical atom (see Figure 2).

    {13773_Answers_Figure_2_Shell diagram of electron structure}
  7. Which electrons in Figure 2 are in the same energy level?

    A and B are shown in the same energy level.

  8. What type of force is experienced between the electrons and the nucleus?

    All of the electrons are attracted to the nucleus.

  9. What type of force is experienced between different electrons in Figure 2.

    The electrons mutually repel each other.

  10. Compare the force of the nucleus on the electrons in both shells in Figure 2. Based on Coulomb’s law, which electron(s) experience a greater force? Explain.

    Electrons A and B are closer to the nucleus than electron C and thus experience a greater attractive force to the nucleus.

  11. Predict which electron(s) would require the least amount of energy to be removed (ionized) from the atom. Explain your reasoning.

    Electron C would require less energy than A or B to be removed from the atom.

  12. Suppose photoelectron spectroscopy yielded ionization energy values of 0.76 MJ/mol and 5.25 MJ/mol for the electrons in this hypothetical atom. How much energy, in MJ, is required to remove electron B from one mole of atoms? Note: 1 MJ = 1 Megajoule = 1 x 106 J.

    The energy required to remove electron B from one mole of atoms is 5.25 MJ.

Part B. Principles of Photoelectron Spectroscopy
{13773_Answers_Figure_3_The photoelectric process}
  1. Figure 3 represents the basic physical process that takes place in photoelectron spectroscopy. What is the energy per mole of the photon in this example?

    The energy of the photon is 25 MJ per mole.

  2. What is the kinetic energy per mole of the photoelectron that is released from A?

    The kinetic energy of the ejected electron is 13 MJ per mole.

  3. Determine the ionization energy per mole of the ejected electron.

    The ionization energy is equal to 12 MJ/mole, corresponding to the initial energy of the photon (25 MJ/mole) minus the kinetic energy (13 MJ/mole).

  4. The photoelectron spectrum is usually plotted as shown below in Figure 4 for A. Ionization energy is on the x-axis versus signal intensity on the y-axis. Note the position of zero on the x-axis. (a) Write a sentence describing how the value of the ionization energy increases along the x-axis in a photoelectron spectrum. (b) Which direction on the x-axis corresponds to greater kinetic energy of an emitted photoelectron?

    a) The ionization energy increases from right to left along the x-axis in a photoelectron spectrum.
    b) The kinetic energy will be greater for electrons that have lower ionization energies (see the arrow).

    {13773_Answers_Figure_4_A simulated photoelectron spectrum}
  5. What determines the position of the peak in the spectrum? What is the numerical value of the peak for atom (element) A?

    The position of the peak is based on the ionization energy of the emitted electron, 12 MJ/mole for A.

  6. How many energy levels are represented by the photoelectron spectrum in Figure 4? Explain your reasoning.

    The photoelectron spectrum has only one peak, corresponding to one energy level.

  7. Is it possible to determine how many electrons are present in A based on this spectrum? Explain.

    The signal intensity corresponds to the probability that an electron has been released from A. It cannot be used to say how many electrons are present in A or in a specific energy level.

  8. The simulated photoelectron spectrum for a second hypothetical element is shown below. What does the number of peaks indicate about the energy levels for the electrons in this element?
    {13773_Answers_Figure_5_Simulated photoelectron spectrum}

    The photoelectron spectrum shows two peaks, corresponding to electrons in two different energy levels.

  9. What are the ionization energies for the electrons in Figure 5?

    The ionization energies for the two peaks (two sets of electrons) are 12.1 and 33.5 MJ/mole.

  10. Which ionization energy value corresponds to an electron that is closer to the nucleus?

    An electron that is closer to the nucleus has a higher ionization energy, 33.5 MJ/mole in this case.

  11. Compare the peak heights for the signals in this spectrum. Predict the relative number of electrons per atom originating from each energy level.

    The relative peak heights are 2:1. Thus there are twice as many electrons with higher ionization energy. These electrons would originate from the “lower” energy level that is closer to the nucleus.

  12. Complete the following shell diagram for this hypothetical element based on the photoelectron spectrum in Figure 5.
Part C. Photoelectron Spectra (PES) and Electron Configurations of the Elements

Table 1 summarizes PES data for elements 1−4 (hydrogen through beryllium).
  1. Compare the ionization energy values for hydrogen and helium. What is the main factor accounting for the difference in ionization energy of these two elements?

    Hydrogen and helium are similar in size, but the charge on the nucleus is +2 for He and +1 for H. Almost twice as much energy is required to remove an electron from He because it is attracted more strongly to the nucleus with two protons.

  2. Helium has two electrons but only one peak in its photoelectron spectrum. Explain.

    Both electrons in He occupy the same energy level and have the same ionization energy.

  3. For each element in Table 1, the electrons giving rise to Peak A are all in the same energy level (n = 1). Explain how the charge on the nucleus and the distance of these electrons from the nucleus might (or might not) account for the trend in these ionization energy values for elements 1−4.

    The electrons responsible for Peak A occupy the first or lowest energy level (n = 1) that is closest to the nucleus, leading to the assumption that distance is not the primary factor for the increase in this ionization energy for H through Be. The charge on the nucleus increases from +1 to +4 in this series. Since the attractive force is proportional to the charge, the ionization energy is greater for atoms that have more protons.

  4. For each element, identify the peak corresponding to the first (lower) ionization energy. Account for the difference in energy for the first ionization of elements 1−4.

    The photoelectron spectra of H and He have only one peak—Peak A represents the first ionization energy. For Li and Be, the lower ionization energy is Peak B, which represents the second energy level. Electrons in the second energy level are further away from the nucleus and thus require less energy to be removed from the atom. The first ionization energies for Li and Be are lower than those for H and He. This is opposite the trend observed for the increase in ionization energy for He compared to H.

  5. Compare the ionization energy for Peak A in the photoelectron spectrum of hydrogen versus Peak B for lithium. Draw simple shell diagrams to illustrate the difference in ionization energy for these peaks.
  6. The photoelectron spectrum of lithium is shown in Figure 6. Identify the peak representing the first ionization energy of lithium.
    {13773_Answers_Figure_6_Photoelectron spectrum of lithium}
  7. Compare the energy and relative height of the peaks in Figure 6. Write out the accepted electron configuration for lithium and assign the peaks in the PES of lithium to these atomic orbitals.

    The electron configuration of lithium is 1s22s1. The 6.26 MJ/mole peak in the PES spectrum of lithium is assigned to the 1s electrons, and the 0.52 MJ/mole peak is assigned to the 2s electron.

  8. Consider the element boron with five electrons in a neutral atom. Predict the number of peaks, and their relative height, in the photoelectron spectrum of boron.

    Up until now we have only discussed main shells corresponding to principal energy levels. Students might think boron will have two peaks in its PES spectrum, in a 2:3 ratio (from highest to lowest IE). If students add the concept of subshells, they should predict three peaks. See Question 9.

  9. The actual photoelectron spectrum of boron is shown below. Describe any similarities and differences in the information obtained from this spectrum with the predictions you made in Question 8.
    {13773_Answers_Figure_7_Photoelectron spectrum of boron}

    If students understand from previous knowledge and experience the accepted electron configuration of boron (1s22s22p1), they should expect three peaks with the 2s and 2p peaks similar in energy, as shown.

  10. The n = 2 energy level of an atom consists of two subshells (2s and 2p orbitals). What does the presence of a third peak in the PES of boron indicate about its electronic structure? Use the terms shells and subshells in your answer, and be as specific as possible concerning the difference in energy of the three peaks, as well as their relative height.

    The third electron in the second main energy level (shell) has a slightly smaller IE and is responsible for the lowest IE peak in the spectrum. This is in contrast to a simple shell model which predicts two peaks with a ratio of 2:3 electrons. The shell model must be refined to take into account a new discrete energy level or subshell within the second principal or main energy level. The second electron shell has two subshells which are similar in overall energy but with discrete values.

  11. Assign each peak in Figure 7 to electrons and atomic orbitals in the accepted electron configuration of boron.

    The first (highest ionization energy) peak corresponds to electrons in the 1s orbital, the second peak is the 2s orbital, and the third peak is the 2p orbital. The five electrons are distributed as follows: two electrons in 1s, two in 2s and one in 2p. The small difference in energy for the 2s and 2p electrons indicates the electrons in these orbitals are approximately the same distance from the nucleus, and much farther away than the 1s orbital.

  12. The 1s and 2s orbitals can each accommodate two electrons. How many electrons can occupy the 2p subshell in an atom?

    The 2p subshell consists of three equivalent 2p orbitals and can therefore accommodate a total of six electrons, two per atomic orbital.

  13. Complete Table 2 to predict the number of peaks, and the relative number of electrons, for the remaining elements in row 2 of the periodic table. Arrange the peaks from lowest to highest ionization energy to show the relative number of electrons.
  14. Compared to the elements in Table 2, the PES of sodium (element 11) contains a new, fourth peak. Would you expect this peak to have a lower or higher ionization energy value than the lowest IE peak for neon? Explain based on the arrangement of elements in the periodic table and the shell structure of atoms.

    The first and second energy levels are “complete” for neon (atomic number 10), with two electrons in the 1s orbital and the maximum of eight electrons that can be accommodated in the 2s and 2p orbitals. The “next” electron in sodium enters a higher energy level, 3s, that is further away from the nucleus. This electron is in the n = 3 principal energy level and has a lower ionization energy than electrons in the n = 2 shell.

    Table 3 summarizes atomic orbital assignments for the ionization energy values obtained from the photoelectron spectra of elements 11−20.
  15. Based on the ionization energy data in Table 3, which orbitals appear to be closer in absolute energy—a 1s and 2s orbital, or a 2s and 3s orbital? Explain based on energy as a function of distance from the nucleus.

    The 1s and 2s orbitals differ in absolute energy by approximately 100−350 MJ/mol for the elements included in this table. The absolute energy difference for the 2s and 3s orbitals is much smaller, ranging from 6.3 MJ/mol for Na to 38 MJ/mol for Ca. We conclude that orbitals get closer in energy as they get farther away from the nucleus. Coulomb’s law predicts that the electrostatic attractive force between an electron and the positively charged nucleus decreases by the square of the distance from the nucleus. The force drops off very sharply, and the ionization energy changes by a smaller amount as electrons get farther away from the nucleus.

  16. The potential energies of electrons in an atom are always negative, meaning they are more stable than isolated or “free” electrons. This leads to an apparently contradictory statement that “electrons that are higher in energy have lower ionization energies.” Write a short 1−3 sentence explanation for ionization energy that eliminates potential confusion about these terms.

    Energy must always be supplied to remove an electron from an atom—this is the ionization energy. Electrons have negative potential energy due to the electrostatic attractive force with the nucleus. “Adding” ionization energy raises the energy of an electron from its negative potential energy “well” to a state of “zero” potential energy when it is far removed from the nucleus. Electrons that are closer to the nucleus have a greater attractive force and lower potential energies, so more energy is required to ionize them.

  17. The relative energies for the atomic orbitals of Mg are shown. Draw lines to show the approximate relative energies of the 1s, 2s, 2p, 3s and 3p orbitals of Al.


AP® Chemistry Curriculum Framework: 2013−2014; The College Board, 2011.

Student Pages

Photoelectron Spectroscopy

Guided-Inquiry Learning Activity for AP® Chemistry


The chemical properties of elements are based on the number of electrons in the neutral atom and the arrangement of electrons into shells and subshells reflecting specific or quantized energy levels. Photoelectron spectroscopy and ionization energy measurements provide direct evidence for the electronic structure of atoms.


  • Photoelectric effect
  • Planck’s law
  • Quantum theory
  • Coulomb’s law
  • Ionization energy
  • Photoelectron spectroscopy
  • Electronic structure
  • Atomic orbitals


Photoelectron spectroscopy is defined as the measurement of the relative number of electrons of different energies that are ejected from atoms when they are bombarded with high-energy electromagnetic radiation. Photoelectron spectra are generated based on the photoelectric effect, which was first observed for metals in the 19th century. When light of the appropriate wavelength and energy is shined on a metal, electrons may be given off or ejected from the metal surface. If the light is below a certain minimum or threshold frequency, no electrons are produced, regardless of the intensity of the light source. This apparent paradox could not be explained by the laws of classical physics. In 1905, Albert Einstein applied the new quantum theory of light to explain the photoelectric effect. Einstein’s explanation of the nature of the photoelectric effect represented a watershed event in the history of science. The development of quantum physics and quantum mechanics in the years that followed ultimately transformed and led to our current understanding of atomic and electron structure.

Experiments with the photoelectric effect led to two major findings: 1) light must be above a certain frequency for electron emission to occur, and 2) as the frequency of light increases above this minimum, the average kinetic energy of the dislodged particles increases. These findings were interpreted by Albert Einstein using Planck’s law. This law describes light in terms of photons possessing a quantum or bundle of energy, where the energy (E) of a photon is proportional to its frequency (v).

E = hv                         Planck’s law

The photoelectric effect is only observed when the frequency of light is greater than a specific threshold value, which depends on the nature of the metal. If the energy of a photon is greater than the work function, defined as the minimum energy needed to displace an electron from a metal, the metal will give up an electron. If more photons having this energy strike the metal, additional electrons will be lost, but their kinetic energy will not change. The kinetic energy of photoelectrons ejected from a metal depends on the frequency (energy) of the light source and the work function of the metal. Albert Einstein was awarded the Nobel Prize for Physics in 1921 for his work on the photoelectric effect, which bridged the gap between the laws of classical and quantum physics.

Photoelectron spectroscopy (PES) utilizes the principles of the photoelectric effect to generate spectra revealing the electronic structure (energy levels) of atoms. A photoelectron spectrophotometer consists of three basic parts: a vacuum chamber for the sample, a radiation or light source, and an electron analyzer to separate electrons based on their kinetic energies (see Figure 1).

{13773_Background_Figure_1_Basic features of a photoelectron spectrophotometer}

Samples of gas-phase atoms are exposed to electromagnetic radiation of sufficient energy, typically shortwave ultraviolet light or X-rays, to eject electrons from the atoms. Although each atom can emit only a single photoelectron using this technique, a sample always contains a multitude of atoms. Different atoms within the sample will emit electrons from different energy levels. The kinetic energy (KE) of the resulting photoelectrons is measured with an electron analyzer. PES uses monochromatic (single frequency) electromagnetic radiation. Based on the law of conservation of energy, the kinetic energy (KE) of a photoelectron is equal to the energy of the incident light (hv) minus the ionization energy (IE), which is the energy required to remove an electron from a particular energy level within an atom.


From Equation 1, the ionization energies of the electrons within an atom can be calculated. Because the incident light is of sufficient energy to ionize both core and valence electrons, there is an equal probability that each electron in an atom will be ejected. The number of electrons ejected is proportional to the number of electrons present at each energy level. A photoelectron spectrum displays the signal intensity, corresponding to the relative number of electrons ejected at a particular energy, versus the ionization energy. For a spectrum displaying two or more peaks, a comparison of the peak heights allows for the relative number of electrons at each energy level to be determined.

Experiment Overview

The purpose of this guided-inquiry learning activity is to examine experimental evidence for the electron configuration of neutral atoms. The activity is divided into three parts

  1. Review Coulomb’s law to predict the relative ionization energies of electrons in an atom.
  2. Investigate the basic principles of photoelectron spectroscopy and identify the way information is presented in a photoelectron spectrum.
  3. Analyze the photoelectron spectra of elements 1−20 to predict and explain their electron configurations.

Student Worksheet PDF


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