Answer to Question #10838 Submitted to "Ask the Experts"

Category: Instrumentation and Measurements

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Q

Using a PMT (photomultiplier tube) for scintillation, when a gamma ray interacts with an NaI crystal, is light emitted in the ionizing effect when an electron is removed from the shell of an atom of the crystal? Can light itself cause ionization? The word photon is normally applied to electromagnetic radiation, correct? In the spectrum, light is not considered as ionizing radiation. In the manuals I have read they just start with a photon entering the PMT and the energy is amplified. This is where I get confused. This I know, Eienstein came up with this photoelectric effect.

A

When gamma radiation interacts in a scintillation crystal an electron may be raised from the valence band in the crystal (the valance band represents the usual bound state of valence electrons in the crystalline material) to the conduction band (normally empty in a perfect insulating crystal). Electrons in the conduction band are free to migrate around, and some of these may get trapped by impurity sites (referred to as activator sites), such as are represented by the dopant (activator) materials purposely added during manufacture of the crystal. In the case of the common scintillator, NaI(Tl), thallium (Tl) represents the activator material of common choice.

If the electron migrating in the conduction band of the crystal gets trapped in an excited state of the activator, as is often the case, it will typically drop quickly into a lower energy ground state associated with the activator trapping site. This electron transition is accompanied by an emission of electromagnetic energy in the form of light. The light-transparent window of the photomultiplier tube will transmit part of this light that may be incident on it. The window is backed with a thin metal or metallic mixture that has what is usually referred to as a low work function; this metallic material represents the photocathode of the PMT. A low work function simply means that it doesn’t take much energy to remove an electron from the metal. In common photocathodes (such as commonly used bialkali cathodes; e.g., potassium/cesium/antimony), the work function is considerably lower than the typical photon energy emitted during the deexcitation process at the activator site, thus allowing for reasonably efficient production of photoelectrons from the photocathode. It might require 1 to 2 electron volts of energy to remove the electron from the photocathode, and the light photon from the scintillator will typically have an energy of about 3 electron volts. The photoelectrons are, as you cite, a demonstration of Einstein’s discovery.

You are correct in associating a photon with electromagnetic radiation; thus, both gamma radiation and visible light may be referred to as photons. Visible light is not normally referred to as ionizing radiation because in most situations it is not capable of ionizing materials with which it interacts, but there are exceptions as in this case with the photocathode material.

After the electrons are produced from the photocathode, they go through multiple stages of acceleration and multiplication by being attracted to sequential electrodes (called dynodes) of increasing voltage, with more electrons being set free at each dynode and collection of a pulse of many electrons being accomplished at the final electrode, the tube anode.

I hope this helps to clarify your thoughts about how scintillators/PMTs work.

George Chabot, PhD

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