# Counting gamma rays with a scintillation detector

Let's take a look at how a scintillation detector works. How does it convert a gamma ray into an electrical pulse that can be counted?

When a gamma ray approaches an atom, there are several things that can happen.

1. The gamma ray may not interact at all -- which is rather uninteresting, so let's skip that.
2. The gamma ray may be absorbed by the atom, giving its energy completely to an electron which is consequently knocked out of the atom. This is a photoelectric absorption.

3. The gamma ray may transfer some, but not all, of its energy to an electon, again knocking it free from the atom. This is Compton scattering.

One can use conservation of energy and momentum to derive formulae for the energy of the gamma-ray and the electron after an episode of Compton scattering:

Note the general result: the gamma ray always has less energy than it started, but there is an upper limit to the amount of energy it can lose. That means that the energy of the ejected electron may range from zero to a maximum value, which is again less than the energy of the incoming gamma ray.

4. The gamma ray may interact with the nucleus. We won't consider that case here.

Let us concentrate on possibilities 2 and 3. We'll begin with photoelectric absorption. The energy of the freed electron must be

In our lab, some typical values are

• gamma ray energies of 500 to 1200 keV
• ionization energies of 1-10 eV (sodium 5.1 eV, iodine 10.4 eV)

so it should be clear that the energy given to the liberated electron is almost exactly the same as the energy of the incident gamma ray, to a tiny fraction of a percent.

So, at this point, we have one electron with a boatload of kinetic energy. How can we turn this into a signal we can detect? The answer is to convert the electron's kinetic energy into visible light. The mechanism involves many collisions with atoms in a special sort of crystal: sodium iodide, doped with a small amount of thallium to replace a small fraction (0.1% to 0.4%) of the sodium atoms. The function of the thallium is to modify the electronic properties of the crystal. In an pure sodium iodide crystal, one finds the usual situation: a valence band and a conduction band, separated by a large gap, in this case about 5.9 eV.

If an electron were excited into the conduction band, then managed to drop back down into the valence band by emitting energy as a photon, the wavelength of that photon would be around 210 nm, in the near-UV. Such a photon would immediately be absorbed by other atoms in the crystal, which would prevent it from leaving the crystal and being measured.

When a small amount of thallium is added to the crystal, new bands are created at a somewhat lower energies above the valence band. The term "exciton" is sometimes used to describe electrons in this state.

An electron excited into the conduction band may now drop down to the lowest energy in the conduction band (as usual), but now it may drop further by small steps into the exciton states. As the electron makes its way back to the valence band by these small steps, it may emit a photon with a lower energy and longer wavelength.

The lower energy and longer wavelength of this photon are important for the detection process in two ways:

• This photon can travel much farther through the crystal before it interacts with other atoms, since it does not have enough energy to bump electrons out of the valence band into the conduction band. In the case of sodium iodide, the free path length of this photon may be several centimeters, versus micrometers or less for a 5.9-eV photon.
• This visible-light photon may be detected more efficiently by common photosensitive devices such as photomultiplier tubes. In the case of sodium iodide, these exciton-based photons have wavelengths of roughly 350-500 nm, or energies of roughly 2.5-3.5 eV.

How many photons will be emitted for each gamma-ray?
Most of the kinetic energy of the initial electron is NOT converted into visible light. Instead, most of that kinetic energy is turned into heat as the electron bumps into atoms in the crystal. Only about 10 percent of the electron's kinetic energy eventually emerges as photons. Since each photon carries about 3 eV, we can figure out very roughly how many photons will be created from a single gamma ray.
```
initial gamma ray  E   =   662,000 eV      for Cs-137

x 10 percent      =    66,000 eV

/ 3 ev/photon     =    22,000 photons

```
So, each gamma ray may cause tens of thousands of visible-light photons to be emitted from the crystal. Remember that only a fraction of these photons will go in the right direction -- towards the photomultiplier tube.

How long does it take for the electron to lose its energy?
We'll do a very rough estimate here, just to get the order of magnitude. Begin with the initial speed of the electron. We can use the formula for relativistic kinetic energy to figure out the gamma factor for the electron:

We find that gamma = 2.3 and so the initial velocity of the electron must be around v = 0.9 c, very close to the speed of light. At relativistic speeds, it takes only around 10-18 seconds to travel from one atom in a crystal to another. Even if the electron must travel many atoms between interactions, it can make tens of thousands of collisions in a very short time, perhaps 10-13 seconds.

How long does it take for excited atoms to generate a photon?
The decay time for an exciton (excited location in the crystal lattice) is around 250 nano-seconds for sodium iodide with thallium doping. This is so much longer than the time required for the initial electron to dump all its energy into the crystal lattice that, for practical purposes, all the photons due to a single gamma-ray will be emitted simultaneously.

The incoming visible photon knocks just one electron from the photocathode. That single electron is accelerated through a high voltage, causing it to strike a metal surface (a "dynode") at high speed. It knocks several electrons free, and they are in turn accelerated towards a second metal surface. This process is repeated a number of times, increasing the number of electrons at each stage. By the end of the photomultiplier tube, the single electron has turned into a cloud of many thousands of electrons, a signal which can easily be measured.

The size of each packet of charge striking the final dynode (the anode) in a photomultiplier tube indicates the number of photons which struck the photocathode; that, in turn, indicates the energy of the electron which was bumping into atoms in the crystal; that, in turn, indicates the energy of the original gamma ray. There are, after all, quite a few steps in the process:

1. a gamma ray strikes at atom in the crystal, giving one electron much of its energy (hundreds of keV)
2. this high-energy electron strikes many other atoms in the crystal, creating many electron/hole pairs
3. some of the electron/hole pairs recombine via excited states at special sites in the crystal; each such recombination yields one visible-light photon
4. some of the photons leave the crystal and strike the photocathode of the detector, knocking one electron free
5. the liberated electrons are accelerated inside the photomultiplier tube before striking a dynode and knocking free multiple electrons
6. repeat last step 5-10 times
7. a large cloud of electrons strikes the anode of the photomultiplier tube, giving it a negative charge which can be measured by a capacitor

#### Features in a gamma-ray spectrum

So, when we look at the spectrum of a typical gamma-ray source, such as Cs-137, what should we expect to see?

If we had a PERFECT instrument, we might expect to see a spectrum like this:

since Cs-137 produces gamma rays with energies of 32.06, 36.40 and 661.7 keV.

But what we ACTUALLY see looks quite different:

What has happened?

• Each photopeak has been broadened, due to several factors in the chain of events leading to detection. The Full-Width at Half-Maximum of this particular spectrum is -- using the 662 keV line -- about 50 keV. Another way to describe the broadening is to compare the width of the line to the central energy of the line: 50 kev / 662 keV = 0.08 , or about 8 percent.

This broadening places a limit on the energy resolution of the instrument.

• Compton scattering by gamma rays responsible for each photopeak produce electrons which have a wide range of energies, with a maximum possible value which is less than the original gamma's energy. These electrons are responsible for the Compton edge and the Compton plateau.
• Gamma rays which do not enter the crystal, directly, but instead fly into the lead shielding or other materials around the detector, may experience Compton scattering in that material. Those which are scattered by small angles continue to penetrate the shielding and never reach the detector. Those which are scattered by large angles -- backscattered -- will lose a large fraction of their energy, but may then enter the detector. Most of these "second-hand" gamma rays will have close to the minimum possible energy permitted by Compton scattering, and give rise to the backscatter peak.
• Some of the gamma rays may strike lead atoms in the shielding, ejecting an electron from the innermost K shell of electrons. To replace that electron, an electron from the adjacent L shell may drop down to the K shell. As it does so, it emits a X-ray with the energy difference between L-shell (16 keV) and K-shell (88 keV) electrons: about 72 keV. Some of these X-rays from the lead shielding may then enter the photomultiplier tube. (See comparison further below of spectra of sources with and without lead shielding).

We usually place the radioactive source in the middle of a little structure made of lead bricks, to prevent radiation from escaping into the room at large. Some of the gamma rays from the source interact with the lead atoms in the bricks, creating lower-energy scattered gamma rays and also X-rays with the characteristic energy of the L-shell-to-K-shell transition in lead. The interactions with the lead shielding lead to several noticeable features in the spectrum. I removed the lead bricks and placed the source on an empty cardboard box so that it was at the same distance from the source as usual. Look at the difference in the spectrum: