In simple terms, the photoelectric effect is the complete absorption of an X-ray photon by an inner-shell (typically K- or L-shell) electron of an atom. The photon transfers all its energy to the electron, which is then ejected from the atom.

Pass your radiology physics exam first time

Step-by-step mechanism

The photoelectric effect follows a simple 3 step process.

Incident photon interaction
- An X-ray photon with energy E_photon strikes an inner-shell electron (usually the K-shell).
- The photon must have at least as much energy as the electron’s binding energy (E_b) to eject it.
Ejection of the photoelectron
- The photon’s energy is absorbed and converted into the kinetic energy of the ejected electron: E_{photoelectron} = E_photon − E_b
- The photon ceases to exist as its energy is fully absorbed.
Atomic vacancy and relaxation
- The atom is left ionised with a vacancy in an inner shell.
- An electron from a higher shell drops down to fill the vacancy.
- The energy released is emitted as characteristic radiation (X-ray photon) or transferred to another electron (emission of an Auger electron). **This characteristic photon has very low energy because the difference in binding energy between inner shells in small atoms (such as those in tissue) is very small. Don’t confuse this with the high energy characteristic X-ray produced in tungsten (much higher atomic number) in the anode.

Energy conservation: Total energy of the incoming photon = binding energy + kinetic energy of the ejected electron + any emitted characteristic photon.

Outcome of photoelectric effect steps	Description
Photon absorbed	No photon transmitted → contributes to image contrast (white areas).
Photoelectron released	Causes local ionisation and tissue dose.
Characteristic radiation / Auger electrons	Emitted locally; low energy; further contributes to absorbed dose.

Energy and Atomic Number Dependence

Perhaps one of the most asked question in exams. This formula is the basis of many many questions. Know it well!

The probability of photoelectric absorption depends strongly on both photon energy (E) and atomic number (Z).

Probability of photoelectric effect ∝ Z³ / E³

NB:

Higher Z → much greater probability (bone, iodine, barium).
Lower photon energy → much greater probability.
As energy increases, the photoelectric effect falls rapidly — this is why it dominates at low photon energies and high-Z materials, but not in soft tissues at higher energies.

There is one exception to the above equation. At certain photon energies the likelihood of the photoelectric effect occurring rises drastically before dropping exponentially again. These energies are known as K-edges.

Absorption Edges (K-edge)

At the photon energy equal to the binding energy of a particular shell, there is a sudden jump in absorption probability: the absorption or K-edge.

For tungsten, K-edge ≈ 69.5 keV.
For iodine, ≈ 33 keV.
For barium, ≈ 37 keV.

This happens because at these energies a whole new shell of electrons become available for the photoelectric effect to take place. Remember, the incident photon energy must exceed the inner shell binding energy. Selecting beam energies just above the K-edge of contrast agents maximises absorption and hence image visibility (e.g. 70–80 kVp in contrast studies).

In the next section we’ll look at Compton scatter. I want to include this comparison table here though, in case you’ve come to this article after having written a mock exam and you only plan on reading this page.

Photoelectric effect vs Compton scatter

Feature	Photoelectric	Compton
Photon energy	Low	Moderate–high
Electron shell	Inner (bound)	Outer (loosely bound)
Photon outcome	Completely absorbed	Scattered, lower energy
Image effect	Increases contrast	Reduces contrast, increases noise
Dose contribution	High (complete absorption)	Moderate
Dependence	∝ Z³/E³	∝ 1/E and electron density, ~independent of Z
Dominance	Low-energy photons, high-Z materials	High-energy photons, low-Z materials

Key Takeaways and Exam Tips:

Photoelectric effect: photon fully absorbed by inner-shell electron.
Ejected photoelectron energy: E_photon − E_binding
Probability ∝ Z³/E³: strongly favours high-Z materials and low energies.
Dominant <50 keV, major source of contrast and patient dose.
K-edges explain why iodine and barium are effective contrast agents.
Characteristic X-rays and Auger electrons are secondary effects.
Common FRCR question: “Describe the photoelectric effect and explain its relevance to image contrast and patient dose.” “Why does Compton scatter predominate at higher photon energies?” “Why does increasing kVp reduce image contrast?”

Up Next

Next, we’ll move on to Compton Scatter. How do photon–electron collisions produce scatter radiation, how does this affects image contrast, and how it can be controlled in practice?