In this experiment we will study the following:

• how gammas interact with matter;
• how gammas produce light in a NaI(Tl) crystal;
• how a photomultiplier tube (PMT) uses light to produce electrons and amplifies the number of electrons;
• what gives rise to the polarity and size of PMT pulses seen on a scope;
• how a pulse height analyzer (PHA) produces a histogram of pulse heights, (i.e. a spectrum); and
• the energy spectrum of some common isotopes.

In order for gammas to be detected, they must lose energy in the detector. Since gammas are electromagnetic radiation, we must understand how such radiation interacts with matter. There are three distinct modes of interaction: photoelectric effect, Compton scattering, and pair-production.

The photoelectric effect is when a gamma (i.e, a photon) is absorbed by an atom and an electron is released. The electron, initially in a bound state, uses some of the energy of the photon to escape the orbit of the atom (this is the binding energy, or the work function for that atom) and the rest of the energy goes into excess kinetic energy.

Compton scattering is when a gamma elastically scatters from a free electron. The process produces a photon of reduced energy and an electron with kinetic energy. Note that Rayleigh scattering, or elastic scattering, may be considered a special case of Compton scattering with the electron replaced by a more massive object, e.g., a nucleus. A schematic of Compton scattering is shown in Fig. 1.

{ ${/download/attachments/164078984/fig_1.png?version=1&modificationDate=1499973046000&api=v2}$ Figure 1: A gamma of initial energy E scatters from an electron at angle _θ _(relative to the incident direction), carrying away energy E'

The relation between the incoming and outgoing gamma energies is given by the Compton scattering formula,

E′=E1+Emc2(1−cos⁡θ)E′=E1+Emc2(1−cosθ) { $E' = \dfrac{E}{1+\frac{E}{mc^2}\left(1-\cos\theta\right)}$ ,

(1)

where mc2=0.511MeVmc2=0.511MeV { $mc^2 = 0.511 \;\mathrm{MeV}$ is the rest-energy of the electron, EE { $E$  is the energy of the incident gamma, E′E′ { $E'$  is the energy of the scattered gamma, and θθ { $\theta$  is the angle of scatter (measured relative to the incident direction.) Finally, when a gamma passes near a massive nucleus, pair production can occur whereby the energy of the gamma is converted to mass (an electron-positron pair) with kinetic energy. For this process to occur, the gamma must have at least twice the rest-energy of the electron, or 2 x 0.511 MeV.

When a gamma passes through a material, there is a probability for interaction. One way to quantify this probability for interaction is through a quantity called the linear attenuation coefficient, λλ { $\lambda$ . Mathematically, if N0N0 { $N_0$  gammas are incident on a material, then it is expected that only N=N0e−λxN=N0e−λx { $N = N_0 e^{-\lambda x}$  gammas will remain (i.e. will not interact) after passing through a thickness xx { $x$ . The total linear attenuation coefficient (and the contribution from each of the above interactions) in sodium iodide (NaI) as a function of gamma energy is shown in Fig. 2.

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Figure 2: Linear attenuation coefficients for gammas in sodium iodide (NaI).

It is important to note that in each of the modes of interaction, the gamma’s energy is converted to an electron having kinetic energy. The electron – and charged particles in general – traveling through matter lose energy in a manner different from gammas.

A vast majority of the volume of bulk matter contains electrons. Those electrons feel a Coulomb force as another electron moves through the bulk matter. The cumulative forces acting on the moving electron bring it to rest in a short distance.

Question 1: If a 1 MeV gamma enters a NaI crystal, what is the most likely mode of interaction? (Hint: see Fig. 2.)

If the gamma undergoes a “head-on” collision with an electron in the crystal, the gamma will be scattered by 180°. Applying Eq. (1), show that this electron will have kinetic energy of about 0.8 MeV.

Question 2: Is this sufficient energy to kick the electron loose from the bound state in the crystal? Hint: Compare this energy to the ionization potential of hydrogen.

Question 3: Is this electron relativistic? Hint: Use the Lorentz factor,

γ=11−v2/c2=1+kineticenergyrestenergyγ=11−v2/c2−−−−−−−−√=1+kineticenergyrestenergy { $\gamma = \dfrac{1}{\sqrt{1-{v^2}/{c^2}}} = 1 + \dfrac{\mathrm{kinetic\; energy}}{\mathrm{rest\; energy}}$ .

(2)

Such an electron will move only about 1 mm in NaI before losing all of its kinetic energy. Approximate experimental values of the range of electrons in NaI as a function of electron energy are shown in Fig. 3.

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Figure 3: Average range of electrons in sodium iodide as a function of electron energy.

In this experiment, the gammas are detected by a sodium iodide crystal doped with a small amount of thallium, NaI(Tl). The gammas entering the NaI crystal may undergo the photoelectric effect, but more probably will Compton scatter from the electrons in the crystal. (See Fig. 2 for a sense of the “probability of interaction”, which is proportional to the linear attenuation coefficient.)

There are two products of this scattering: a scattered electron and a gamma of reduced energy. The electron loses energy very quickly in the crystal. The rate of energy loss, -d_E_/d_x_, is governed by the Bethe formula (see e.g. Radiation Detection and Measurement by G. Knoll for details). In the process of moving through the crystal, the scattered electron excites electrons in the crystal's valence band into the conduction band, leaving a trail of electron-hole pairs. Holes migrate to the thallium impurity sites and ionize them. The ionized thallium atoms capture electrons, ultimately return to the ground state, and give off visible light. Meanwhile, the remaining gamma moves through the crystal losing energy only when it scatters from another electron.

With each successive scattering, the remaining gamma has less energy. At some point the gamma energy is small enough that the photoelectric cross section is dominant and the remaining gamma gives up its energy entirely to an ejected electron, which loses energy in the same manner as the electrons mentioned above.

The light produced in the scintillator passes through a glass window and into the photo-multiplier tube where it liberates an electron by the photoelectric effect in the photo-cathode of the tube. The photo-electrons are accelerated through a series of metal plates called dynodes. When an energetic electron strikes a dynode, several more electrons are ejected and are, in turn, accelerated to the next dynode. After typically 10 to 14 dynodes, the electron gain may be about 106. The electrons are collected at the anode of the photomultiplier tube. (See Fig. 4.)

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Figure 4: Schematic of photomultiplier (PM) tube.

Question 4: What polarity (positive or negative) would you expect at the dynode output? At the anode output?

When looking at a decay scheme, each column of horizontal lines represents the nuclear energy levels of a specific isotope of some element. Diagonal arrows indicate the process involved in changing from one element to another. Vertical arrows indicate changes in nuclear energy-level within one element. Gammas are emitted when a nucleus changes from a higher to a lower energy nuclear state within the same isotope.

In order to observe pulses from the photomultiplier tube, perform the following steps:

1. Place a Cs-137 source near the detector. - Connect the high voltage cable from the HV input on the PMT base to the HV output on the back of the UCS-30 module. - Turn on the UCS-30 module and make sure that it is connected to the computer through the USB cable. - Turn on the USX software. (The module must be turned on first. If it is not, the software will not detect the module.) Set the high voltage value in the box in the upper left corner to “1000”. Turn on the high voltage by clicking the “OFF” button which should now read “ON”. (These PMT tubes can safely run up to +1000 V. Do not set the value in the box to higher than 1000.)  - Attach the 50 Ω terminator to an input of the scope, and attach a coaxial cable from the PMT anode output to the terminated scope input. - Adjust the scope’s gain, trigger slope and trigger level until the pulses are clearly visible.

> Question 5: Sketch the pulses in your lab notebook. What is the pulse polarity? Does this polarity make sense, given how the PM works? What is the duration of the pulses? What is the range of pulse heights? What are the rise and fall times of the pulses? (See Fig. 5.)

{ ${/download/attachments/164078984/fig_5.png?version=1&modificationDate=1499973046000&api=v2}$ Figure 5: Defining the rise and fall times for a positive and negative pulse.

Remove the terminator and re-connect the cable directly to the scope. Look at the pulses again and try to measure their duration and height with no terminator at the scope input.

The RG58-U coaxial cable you are using has a characteristic impedance of 50 Ω (independent of its length!). On the other hand, the input impedance of the scope is 1 MΩ. This change in impedance from low to high at the cable-scope interface will cause a reflection of the pulse without a change in polarity. The pulses observed on the scope are the sum of the incident and reflected pulses. Using the terminator places 50 Ω in parallel with the scope’s 1 MΩ input impedance.

Question 6:  What is the equivalent resistance of this parallel combination? How will the reflection be affected by the use of the 50 Ω terminator?

We would like to perform a quantitative analysis of the pulse heights. To do so, a histogram of heights may be created by using a pulse height analyzer (PHA). On the PHA, the horizontal axis – channel number – is proportional to pulse height. The vertical axis gives number of pulses having each height. Such a histogram is usually called a spectrum. The UCS-30 spectrometer may be used as a PHA.

1. Plug the coaxial cable from the PMT anode into the input on the rear panel of the UCS-30. This input is designed to take negative pulses from a PMT tube without a 50 Ω termination. - Under the Mode menu, select PHA - PreAmp In. In this mode, the PHA takes the small, negative input and amplifies it into a larger, positive signal. - Set the Coarse Gain to 1 and the Fine Gain to 1. - Under the Settings menu, select _Energy Calibration: Uncalibrate. _This clears any existing calibration from previous groups. If the option is greyed-out, then the calibration has already been cleared. - Under the Settings menu, select High Voltage/Amp/ADC and set Conversion Gain to 2048_._ This setting divides the range of the amplified pulses from the PMT into channels 0 to 2047. The PHA is set up to  histogram pulses with voltages from 0 to 8 V, so an amplified signal of (for example) 4 V should appear in the middle of the screen at approximately channel 1023. - Drag the lower level discriminator (the triangle near left-most channel edge) to approximately the 5% position. This setting serves to eliminate low-level electronic noise pulses. - Drag the upper level discriminator (the triangle near right-most channel edge) all the way to the 100% position. - Place a Na-22 source near the scintillator. Click the green diamond icon to start taking data. You should see a spectrum begin to appear on screen. - Adjust the spectrum so that it fills the full channel range. You may click the “X” to erase the spectrum without stopping collection, or you may click the red stop sign to stop collection. - Each source has a maximum energy gamma, so there should be a natural fall-off to zero activity for channels above the full photopeak corresponding to this energy. If you observe activity all the way up to the last channel on the PHA, you may need to lower the voltage to the PMT in order to reduce the pulse heights and bring the spectrum back into range.
   1. Conversely, if the pulse heights are too small, your spectrum will appear smooshed into only the lowest channels. To expand the spectrum, increase the Coarse and/or Fine Gains to amplify the pulses into higher channels.
   2. You have adjusted the spectrum correctly when the highest energy peak from Na-22 is fully on-screen but near the right-most edge. When this is complete, do not change either the high voltage or gain settings for the rest of the experiment.

| - Refer to the nuclear decay schemes here to help identify the origin of each peak in this spectrum. |

We wish to re-scale the horizontal axis of the PHA from channel number to energy in MeV (million electron-volts). To do so, we will use two known gamma energies: the 1.27 MeV of Na-22 and the 356 keV of Ba-133. Be sure to leave the high voltage and gain fixed for the rest of the experiment in order to preserve your calibration.

1. The higher energy gamma of Na-22 is 1.27 MeV. Using your spectrum of Na-22, determine which peak is due to the 1.27 MeV gamma. - Use the mouse or arrows on the keyboard to move the PHA cursor to the 1.27 MeV peak position. Record the channel number for this peak. - Erase the data and acquire a spectrum for Ba-133. Identify its 0.356 MeV peak. - Move the cursor to the Ba-133 peak you selected and record this channel number. - Under the _Settings _menu, select Energy Calibration: 2 Point Calibration. - For _Units _enter MeV and click Set. - On the next screen, enter the channel and energy of the 1.27 MeV gamma from Na-22. - On the next screen, enter the channel and energy of the 0.356 MeV gamma from Ba-133. - The horizontal axis should now be re-scaled and labeled in units of MeV.

For the Ba-133, Cs-137 and Na-22 sources, acquire spectra and sketch them into your notebook. Measure the energies of the gamma peaks you observe and compare them to the decay schemes for each source.

Question 7: For each of the gammas, plot the energy vs. PHA channel number. How are they related? Does this relation make sense in light of the physical processes involved in the detection? Please explain.

1. Acquire a spectrum of Cs-137 with sufficient data to give a fairly smooth spectrum. Sketch the spectrum. Measure the gamma energy on the PHA.  - Using the Compton formula, calculate the predicted energy of the Compton edge. Note that this feature represents the maximum energy given by the gamma to an electron due to a single, head-on collision. Move the cursor to the calculated energy of the Compton edge. Identify this feature on your sketch. - Calculate the energy of the _backscatter _peak. This feature represents gammas which have undergone a large angle scatter outside the scintillator and then entered the scintillator, losing all of their remaining energy in the scintillator. Move the cursor to the calculated energy. Identify this feature on your sketch of the spectrum.

> Question 8: What is the energy relationship among the Compton edge, the backscatter peak and the full-energy peak?