ASP - Photoelectron Spectroscopy (Lesson)

Photoelectron Spectroscopy

Spectroscopy, in general, is a really powerful tool that chemists use to understand the structure of matter. Since the pieces of matter are too small for us to see directly, we rely heavily on various methods of instrumentation, including spectroscopy, to understand the structure of atoms and molecules, the particles that are responsible for all of chemistry and the chemistry that we study on the macroscopic level.

Concept Overview

• Photoelectron spectrophotometers use high-energy radiation (UV or X-rays) to eject electrons from an atom. 
• The photoelectron spectrophotometer inputs only one type of radiation (with a specific energy).
• Because electrons within an atom are in different energy levels, different electrons can require different amounts of energy in order to be ejected.
• Electrons that are closer to the nucleus are more difficult to remove (greater attraction between protons and electrons).
• Valence (outermost) electrons are the easiest to remove. 
• The photoelectron spectrophotometer removes electrons from multiple atoms, so electrons from all levels will be observed (and graphed). 

Before this technology was commonly available, it was known that certain elements create vividly colored flames. If the light emitted was filtered through a diffraction grating, distinct frequencies of light appeared as bands, rather than creating a full spectrum of light.  Each element that created a colored flame had its own unique set of spectral patterns. 

Look to the upper left and lower right of the orange flame for calcium. You can see only orange and red bands.  In the image for copper, yellow orange, green, and blue bands are apparent on either side of the blue green flame in the center.

How are these specific frequencies emitted? As electrons are excited (gain) energy, they now occupy higher energy levels.  This energy can then be emitted (released) as they move back down to the energy level from where they came.  As this energy is emitted it is detected and directly correlates with the amount of energy between those energy levels.

The addition of energy is shown in the diagram as hf, which comes from the equation for energy found as the product a constant times the frequency of the energy.

The constant is known as Planck's constant and has value 6.63 × 10^-34 m² kg/s (J$\cdot$s) so that when multiplied with frequency in 1/s (Hz), the unit is that of energy: Joules.

Much of what people learned about electrons in the early 1900s came from the study of this phenomenon and the data generated. Our understanding of energy levels for atoms and the orbital diagram we use for electron configurations is one of the outcomes.

Photoelectron Spectra

Consider this PES graph for electrons for an atom of neon. Since all electrons are being removed from the atoms, rather than indicating ionization energy, that energy is instead interpreted as the amount of energy between energy levels.

Please note the energy scale is not how it is typically represented on the x-axis. The higher values are to the left and the lower to the right, as if we were in the fourth quadrant. This arrangement allows for the electrons in the energy levels closest to the nucleus having the greatest binding energy to be ordered as we do when writing electron configurations. Just below 85 Megajoules per mole, we see that there are a few electrons that are removed from the atom when this much energy is applied.  When about 4 MJ/mol of energy is supplied produces the same number. When only about 2 megajoules per mole is provided three times as many electrons are freed from the neon atom. These values correspond to the 1s² 2s² 2p⁶ electron configuration.

For contrast, take a look at the PES graph for fluorine. As you look at fluorine, compare the energies where electrons are removed to neon above. 

Did you notice that the 2s² and 2p⁵ electrons require basically the same energy to remove from both F and Ne, but the 1s electrons require less energy to remove? Why would that be the case? Neon has 10 protons in the nucleus but fluorine only has 9 protons. Think back to Coulomb's Law about the force of electrostatic attraction between particles of opposite charge. When the charge is smaller, less force must be applied to separate opposite charges and with fluorine only having 9 protons and electrons compared to Neon's 10 of each, there is less force needed to remove the two 1s electrons for fluorine.

You Try It! Practice Problem

What would the electron configuration be for the element that has this PES graph? Check your answer by clicking on the plus-mark icon. 

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