gamma ray spectroscopy - department of physics at uf€¦ · gamma ray spectroscopy experiment grs...

Gamma Ray Spectroscopy

Experiment GRS

University of Florida — Department of PhysicsPHY4803L — Advanced Physics Laboratory

Objective

Techniques in gamma ray spectroscopy usinga NaI scintillation detector are explored. Youwill learn to use a pulse height analyzer and beable to explain and identify the photopeak, theCompton edge, and the backscatter peak as-sociated with gamma ray interactions. In ad-dition, activity and halflife measurements areexplored.

References

Jerome L. Duggan Laboratory Investigationsin Nuclear Science, 1988, The Nucleus.

EG&G ORTEC Experiments in Nuclear Sci-ence, 1984, EG&G ORTEC.

A. C. Melissinos, Experiments in ModernPhysics, 1966, Academic Press.

Scintillation Detector

A block diagram for a typical scintillation de-tection system is shown in Fig. 1. The scin-tillation detector is illustrated in Fig. 2. Ourdetector has a 4×4 inch cylindrical NaI scintil-lation crystal which is activated with about 1part in 103 thallium impurities. Through var-ious processes, a gamma ray passing into thecrystal may interact with it creating many vis-ible and ultraviolet photons (scintillations).

Oscillo-scope

LinearAmplifierPM Base

NaI (Tl)Detector

Source

HV PowerSupply

PreampMulti-

ChannelAnalyzer

Figure 1: Block diagram for a scintillation de-tector system.

To detect the scintillation photons, the crys-tal is located next to a photomultiplier tube(PMT) and the scintillator/PMT (detector)is enclosed in a reflective, light-tight hous-ing. To minimize the effects of backgroundgamma radiation, the detector is surroundedby a thick lead shielding tube with the desiredgamma rays entering at the scintillator end ofthe tube.

The PMT consists of a photocathode fol-lowed by a series of dynodes (6-10 is typical)followed by and ending with a collection an-ode. Scintillation photons striking the photo-cathode eject electrons via the photoelectriceffect. A high voltage (HV) power supply anda resistor chain (not shown) bias the cathode,dynodes, and anode so as to accelerate elec-

GRS 1

GRS 2 Advanced Physics Laboratory

trons from the cathode into the first dynode,from one dynode to the next, and from thefinal dynode to the anode collector. Each in-cident electron strikes a dynode with enoughenergy to eject around 5-10 (secondary) elec-trons from that dynode. For each initial pho-toelectron, by the end of the chain, there areon the order of 106 electrons reaching the an-ode.

The anode is connected to a charge-sensitive preamplifier which converts the col-lected charge to a proportional voltage pulse.The preamp pulse is then shaped and ampli-fied by a linear amplifier before processing con-tinues.

Because the amount of light (number ofphotons) produced in the scintillation crys-tal is proportional to the amount of gammaray energy initially absorbed in the crystal,so also are the number of photoelectrons fromthe cathode, the final anode charge, and theamplitude of the preamp and amplifier volt-age pulses. The overall effect is that the finalpulse height is proportional to the gamma rayenergy absorbed in crystal.

Pulse Height Analyzer

As the name suggests, a pulse height ana-lyzer (PHA) measures the height of each in-put pulse. Special circuitry, including a sam-ple and hold amplifier and an analog to digitalconverter, determines the maximum positiveheight of the pulse—a peak voltage as mightbe read off an oscilloscope trace. From thepulse height, a corresponding channel numberis calculated. For example, for a PHA hav-ing 1000 channel capability and a pulse heightmeasurement range from 0 to 10 V, a pulseof height 1.00 V would correspond to channel100, one of 2.00 V would correspond to chan-nel 200, one of 8.34 V would correspond tochannel 834, etc. After the correct channel for

a given input pulse has been determined, thePHA then increments the count in that chan-nel.

Our PHA can analyze pulse heights in therange 0-10 V and will be set up to sort theminto 1024 channels.

After many pulses of various sizes have beenprocessed, a plot of the counts in each channelversus the channel number can be displayed toshow the distribution of pulse heights. Withsome caveats to be described shortly, the pulseheight distribution from a scintillation detec-tor can be interpreted as a plot of the numberof gammas versus the energy of the gammasfrom the source, i.e., a gamma ray spectrumof the (radioactive) source. Spectra from pureisotopes can be found in references and com-pared with a source spectra to determine thenuclear composition of the source.

Gamma Interactions

To understand the pulse height distributionassociated with the gamma rays from a ra-dioactive source, it is important to realize thatonly a fraction of the gamma rays interactwith the scintillator; many do not interactat all and simply pass right through. Fur-thermore, when a gamma does interact, thesize of the pulse from the detector depends onwhether all or only part of the gamma ray en-ergy is deposited in the scintillator.

For a given amount of energy deposited inthe scintillator, the output pulse height will bewell-defined but every pulse will not be exactlythe same size. Because of statistical variationsin light production, photon collection, photo-electron production, and electron multiplica-tion, the pulse heights will show a distribu-tion of values with some pulse heights largerand some smaller than the average. Typicalvariations with our detector are in the rangeof 5-10 percent.

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Figure 2: Schematic of a NaI detector and source showing various gamma ray interactions.

The pulse height distribution for a sourceemitting only single energy gamma rays typi-cally appears as in Fig. 3. The large peak atthe far right is called the photopeak and ariseswhen all the gamma ray energy is deposited inthe scintillator. Note the 5-10% width of thispeak due to statistical fluctuations.The most likely interaction to deposit 100%

of the gamma ray energy is the photoelectriceffect. The incident gamma essentially givesup all its energy to eject a bound inner shellelectron from one of the crystal atoms. Theejected electron then has significant kineticenergy (the gamma ray energy less the smallbinding energy of the atomic electron, on theorder of 10 keV) and loses this energy by ex-citing and ionizing more crystal atoms.

Compton scattering is another way gammarays can interact in the crystal.

Exercise 1 Compton scattering is a purelykinematic scattering of an incident gammaphoton of energy Eγ with an electron (massm) in the crystal that is either free or looselybound (Ee ≈ 0, pe ≈ 0). Use conservationof energy and momentum to show that whenthe gamma is scattered through an angle θ, itsfinal energy E ′

γ is reduced to

E ′γ =

Eγ

1 + (Eγ/mc2)(1− cos θ)(1)

The small peak at low voltage (called thebackscatter peak) arises when gamma photonsfirst strike the lead shield and then Compton

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Figure 3: Pulse height spectrum for a monochromatic gamma ray source.

scatter back into the detector. The scatteredphotons, which are greatly reduced in energy,produce the backscatter peak. The Comptonplateau—the relatively flat region extendingfrom the Compton edge to lower energies—occurs when gamma rays Compton scatter inthe scintillator. The recoiling electron’s en-ergy is deposited in the crystal while the scat-tered photon exits the crystal undetected. Therecoil energy varies from a maximum at theCompton edge when the photon backscatters,to zero when the photon is scattered in theforward direction.

Exercise 2 (a) Assume a 0.662 MeV gammafrom a 137Cs source Compton scatters in, andthen escapes from, the scintillation crystal.What is the maximum energy that can be ab-sorbed in the scintillator? (b) Assume thesame energy gamma first Compton scattersfrom the lead shield and is then totally ab-sorbed in the scintillator. What is the mini-mum energy that can be absorbed in the crys-

tal?

The third way gamma rays interact isthrough pair production. In the strong elec-tric fields near crystal nuclei, a gamma raycan create an electron-positron pair as long asthe gamma ray energy exceeds 1.022 MeV (therest mass energy of an electron and positron).Any gamma energy in excess of this becomeskinetic energy of the electron and positron.This kinetic energy is quickly absorbed in thecrystal and when the positron gets to lowenough energy, it annihilates with an electronproducing two 0.511 MeV gammas. The cre-ation, energy loss, and annihilation effectivelyoccur instantaneously. If both annihilationgammas are absorbed, the total energy ab-sorbed will be the original gamma energy andthe event would contribute to the photopeak.However, sometimes either or both of the an-nihilation gammas will escape from the crystalproducing small peaks (called single or doubleescape peaks) 0.511 MeV or 1.022 MeV below

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the photopeak.

Energy Resolution and Calibra-tion

In addition to the gamma ray energy, and theway it interacts in the scintillator, electronicfactors also affect the size of the pulses mea-sured by the PHA. Of course, the preampand amplifier gain will affect the overall sizeof the pulses. In addition, because of themany stages of electron multiplication, thepulse heights are very sensitive to the PMTsupply voltage. Thus, the HV supply mustbe very stable. It is not normally adjusted tochange the overall gain because the PMT typ-ically works best over a fairly narrow range ofvoltages.

To perform an energy calibration on thePHA system, sources of well known gammaray energies are used. The channel number Cγ

for the center of the photopeak correspondingto each gamma ray energy Eγ is recorded and aplot of Eγ vs. Cγ is constructed and analyzed.For non-precision work, the relationship canbe assumed linear with a small zero offset (in-tercept).

Eγ = κCγ + E0 (2)

where κ, called the energy scale, has units ofenergy/channel. A quadratic term is typicallyadded to account for small nonlinearities.

Once the calibration is known, the energy ofan unknown peak can be obtained through thecenter channel of its PHA photopeak. Becausethe HV and/or amplifier gain may change overtime, it is wise to perform a calibration beforeany critical measurements.

The width of the photopeak is a measureof the energy resolution of the detection sys-tem. The smaller the width, the closer twogammas can become in energy and still be ob-served as separate peaks in the pulse height

spectrum. Consider a photopeak appearing ina pulse height spectrum at channel C and hav-ing a full width at half maximum of ∆C. Theenergy resolution R is commonly expressed bythe ratio of the photopeak’s full width at halfmaximum (in energy units) ∆E = κ∆C toits energy E = κC + E0. If, as is usuallythe case, the calibration offset E0 is small, theenergy resolution is easily obtained from thepulse height spectrum as

R =∆C

C(3)

R decreases with increasing gamma ray energyand is nominally 5-10% for NaI detectors forthe 137Cs 0.662 MeV gamma.The energy resolution will also depend on

the PMT supply voltage. Furthermore, thereis often a “focus” adjustment on the PMT basewhich changes the voltage between the photo-cathode and the first dynode producing smallchanges in the overall gain and energy resolu-tion.

Nuclear Decay

The activity of a sample denotes the rate ofnuclear disintegrations. An activity of 1 curierepresents 3.7 × 1010 nuclear disintegrationsper second. What happens after the disinte-gration depends on the nucleus.For 137Cs, the result is the formation of

137Ba and a β particle (electron). Eight per-cent of the time the β carries away all the ex-cess energy, and 92% of the time the 137Ba isleft in an excited state which decays to theground state by the emission of 0.662 MeVgamma. Thus, the rate of 0.662 MeV gammaemission is 0.92 times the activity.

60Co also decays by β emission to 60Ni. The60Ni is left 99% of the time in an excited state2.507 MeV above the ground state. This ex-cited states decays by emission of a 1.175 MeV,

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followed within a picosecond or so by a second1.332 MeV gamma. Thus each of these gam-mas are emitted at a rate 0.99 times the sourceactivity.

22Na decays by β+ (positron) emission (90%of the time) or by electron capture (10%of the time) to 22Ne. Nearly 100% of thetime the 22Ne nucleus is left in an excitedstate, decaying within a few picoseconds to theground state with the emission of a 1.274 MeVgamma. The high-energy positron thermal-izes in the source (if it is thick enough) orthe surrounding air within a few nanosecondsand then annihilates with an electron produc-ing two 0.511 MeV gammas. Thus the rateof 1.274 MeV gammas is equal to the activ-ity and 0.511 MeV gammas are emitted at1.8 (2× 0.90) times the activity.

54Mn always decays by electron captureto an excited state of 54Cr which then de-cays to the ground state with the emission of0.835 MeV gamma.

Activity Determinations

If one has a source of known activity αk, theunknown activity αu of a source of the sameisotope can be determined from the ratio ofthe count rate of the unknown Ru to that ofthe known source Rk.

αu = αkRu

Rk

(4)

For this method, the unknown and knownsource must be measured under the same con-ditions. For example, the detection systemshould be the same and the sources should beplaced identically.Absolute activity determinations can be

made if various “efficiencies” are known. Forexample, if a photopeak is well-separated inthe source spectrum and the isotope responsi-ble for the photopeak has been identified, the

activity α of the source (from the identifiedisotope) can be based on the observed countrate RE in that photopeak. The activity de-termination would then require that the prob-ability p that a nuclear disintegration resultsin a count in that photopeak be known.

RE = αp (5)

In turn, p will be a product of the fraction fof the nuclear disintegrations which result insuch a gamma, and a probability p′ that thegamma will lead to a count in the photopeak.

p = fp′ (6)

Values for f for various sources is given in thesection on nuclear decay or in reference books.From the spreading out of the gammas we

should expect p′ to decrease as the detector-source distance increases. p′ also decreases forhigher energy gammas which are less likelyto interact in the scintillator. In general, thedependence of p′ (the absolute photopeak ef-ficiency) on the gamma energy and on thedetector-source geometry cannot be separatedin any simple manner. However, the separa-tion of p′ = Gϵ into a geometry-only depen-dent factor G and an energy-only dependentfactor ϵ is often a good approximation. Gwould represent something similar to the frac-tion of the gammas which enter the detector,e.g., G = Ω/4π where Ω is the solid angle sub-tended by the detector at the source. The in-trinsic photopeak efficiency ϵ would then rep-resent the probability that a gamma enteringthe detector produces a count in the photo-peak. Various references show that ϵ typicallydecreases by an order of magnitude or so as thegamma energy increases from 0.5 to 2 MeV.These considerations lead to the approxima-

tionRE = αfGϵ (7)

Note how this equation can also be used for arelative activity determination. By comparing

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counts in the photopeak for a sample of knownand unknown activity, the unknown activitycan be obtained as long as the ratio of thevarious factors is known or can be estimated.

α1

α2

=R1

R2

f2f1

G2

G1

ϵ2ϵ1

(8)

There is an interesting technique for makingan absolute determination of source activity,but it requires that a single nuclear disinte-gration produce two simultaneous gammas ofwell defined energy, such as occurs for 22Na or60Co.We label the two gammas γ1 and γ2. Let p1

(or p2) represent the probability that a nucleardisintegration leads to a γ1 (or γ2) whose fullenergy is deposited in the scintillator, such asmight lead to a count in the photopeak. Letq1 (or q2) represent the probability that a nu-clear disintegration leads to a γ1 (or γ2) hav-ing only partial energy deposited in the scin-tillator, such as might lead to a count in theCompton plateau or the backscatter peak.Because the gammas are emitted simulta-

neously, if both are detected, the result wouldbe a pulse height corresponding to an energyequal to the sum of the energies from eachgamma. For example, sometimes both gam-mas give up their full energy in the scintilla-tor and there appears in the spectrum a sumpeak—a well defined feature that looks like anordinary photopeak and occurs at an energyequal to the sum of the energies of the γ1 andγ2 photopeaks.For some sources, e.g., the 1.274 and

0.511 MeV gammas from 22Na, the two gam-mas are emitted equally likely in all directionsso that p1 and p2 are statistically indepen-dent. For other sources, e.g., the 1.175 andthe 1.332 MeV gammas from 60Co or the two0.511 MeV gammas from 22Na, the outgoingdirections of the two gammas are correlated.For independent gammas, the probability that

both γ1 and γ2 would deposit their full energyin the crystal (i.e., the probability of a sumpeak count) would be p1p2 and the rate of de-tection of sum peak pulses would be

R12 = αp1p2 (9)

The probability of detecting a count in theγ1 photopeak would now be p1(1 − p2 − q2).That is, γ1 must deposit its full energy in thescintillator (hence, the factor p1) and γ2 mustnot deposit any energy (hence, the factor 1−p2−q2). The rate of detection of counts in theγ1 photopeak would then be

R1 = αp1(1− p2 − q2) (10)

Similarly, the rate of detection in the γ2 pho-topeak would be

R2 = αp2(1− p1 − q1) (11)

In the experiment, you will measure R1, R2

and R12. Since there are five unknowns (α,p1, p2, q1, and q2), with only three measure-ments, two more relations are needed. If theratios δi = qi/pi are assumed known, (the pro-cedure section will describe a measurement fora rough determination of these ratios) then theactivity α can be determined from the threepreceding equations.

α =(R1 +R12(1 + δ2))(R2 +R12(1 + δ1))

R12

(12)Note that p1 and p2 can also be determined

p1 =R12

R2 +R12(1 + δ1)(13)

p2 =R12

R1 +R12(1 + δ2)

Procedure137Cs Spectrum

A 137Cs source is known to emit a singlegamma ray of energy 0.662 MeV and should

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produce a pulse height spectrum like thatof Fig. 3 except that the backscatter peak,which is affected by the significant lead shield-ing around our detector, will be bigger andbroader.

1. Turn on the NIM electronics supply rack.

2. Check that the PMT HV power supplyis set to positive polarity. The switch ison the back. Set the PMT HV supply to800 V. Turn on the supply.

3. Set the preamplifier on the ×1 setting.Set the amplifier input to negative andoutput to unipolar. Set the amplifiercoarse gain at 4.

4. Place a 1 µC 137Cs disk source about 10centimeters from the front of the detec-tor. Look at the preamp output with theoscilloscope—sensitivity 1 V/division, 1-20 µs/division time base. Describe whatyou see on the oscilloscope.

5. Look at the amplifier output with theoscilloscope. Describe what you see.Switch the amplifier to bipolar output.How do the pulse shapes change? Thezero-crossing of bipolar pulses will bemore precisely correlated with the timeat which the original gamma interactedin the detector, while the amplitude ofunipolar pulses will typically be more pre-cisely correlated with the energy of theoriginal gamma. Thus, the bipolar settingis normally used when one is interested inthe timing of the gamma and the unipolarsetting is normally used for spectroscopicanalysis. Set the amplifier for unipolaroutput.

6. Turn on the Ortec PHA-equipped com-puter. Start the MAESTRO software.Select Acquire|MCB Properties|ADC and

set the Gain to 1024 and the Gating toOff. Select Presets in this menu and clearall values or set them to zero. SelectCalculate|Calibration and, if not grayedout, select Destroy Calibration. In theright-most item of the tool bar, selectfrom the drop-down menu the detector(2 L1249-F MCB 1, and not the buffer.Select Display|Full View so you will dis-play the entire 1024 channel spectrum onthe main display. There are several waysto zoom in. If needed, learn these onyour own. Select Acquire|Clear to erasethe current detector spectrum. SelectAcquire|Start to begin data acquisition.

7. Do the pulses viewed on the oscilloscopehave a range of amplitudes? How do thescope traces demonstrate that all pulseamplitudes are not equally likely andthere are more pulses at some amplitudesthan at others. How does this correlatewith what is observed on the PHA dis-play?

8. Adjust the amplifier gain so that the pho-topeak pulse heights are around 6 V. Se-lect Acquire|Start to begin data acquisi-tion. After acquiring a spectrum, se-lect Acquire|Stop. Set a region of interest(ROI) over the 0.662 MeV photopeak andthen click inside this region. The soft-ware performs a fit to a Gaussian profileand supplies the peak center and the fullwidth at half maxima (FWHM). Recordthese and determine the energy resolu-tion.

9. Check how things change with count rate.The 137Ba generator (≈ 10 µC 137Cssource) is good for this part of the exper-iment. Move it close to the detector forhigh count rates, farther away for lowercount rates. You may want to switch to

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the 1 µC disk source to get to the low-est count rates. Check how the followingchange and explain what you find:

(a) Resolution. Hint: look at thepreamp pulses. The amplifier worksoff the derivative of the leading edgeof the preamp pulse.

(b) Gain. What happens to the positionof the photopeak? Below whatcount rate or dead time doesthe gain become constant?

10. In the next few steps, you will look forthe dependence of the gain and resolu-tion on the PMT high voltage setting.Retake a 137Cs spectrum (at the current800 V setting) with the source far enoughaway from the detector that the photo-peak position and resolution should beunaffected by the counting rate. Leave itsposition fixed throughout these measure-ments. Measure and record the peak po-sition and FWHM and calculate the res-olution. Also record the amplifier coarseand fine settings.

11. Being careful not to touch the 500 V se-lector switch (leave this at 500 V at alltimes), increase the high voltage to 850 V.Adjust the amplifier gain to get the posi-tion of the photopeak to roughly the sameposition as in the previous step. Measureand record the peak position and FWHMand calculate the resolution. Also recordthe amplifier coarse and fine settings.

12. Repeat the previous step at a PMT highvoltage of 750 V.

13. Discuss the results.

14. Set the PMT high voltage back to 800 V.

Dead Time

Note how the dead time percentage increasesfor high count rate conditions. Dead time istime the PHA is busy processing a pulse andnot able to look at additional pulses comingin from the detector. Were you to measurespectra for a fixed time (called the real time)for both low and high count rate conditions,the effective time would be lower for the highercount rate because the acquisition would be“dead” for a greater fraction of the time. TheMaestro PHA software can largely correct forthis by providing the live time. The softwareeffective runs the live time clock only when itis not busy processing pulses.

15. Take a spectrum for a minute or so withthe 137Ba generator up against the detec-tor noting the dead time percentage. Setone ROI over the entire spectrum. Clickin it and record the total counts. Alsorecord the live time and the real time.The difference between the real time andthe live time is the time the PHA wasbusy processing pulses for this spectrum.Determine how long the PHA is dead foreach pulse it analyzes.

16. Repeat the determination of this pulseprocessing time with the source backedaway from the detector such that the deadtime is about 10%.

Exercise 3 Show that, because of dead time,the measured count rate Rm would be lowerthan the true rate R (the rate that would bemeasured were there no dead time) accordingto the relation

Rm =R

1 +RT(14)

where T is the dead time associated witheach measured pulse, (i.e., as determined inSteps 15 and 16).

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Background Correction

The detected gamma rays come not only fromthe source one is trying to measure, but fromother sources as well. Since you will be work-ing with several sources, it is important tomove those not being studied far from the de-tector. Even with such precautions, there willstill be detected gammas from cosmic rays,naturally occurring radioactivity in the build-ing materials and from other sources. Forsome measurements it is wise to take into ac-count these “background” gammas.One of the simplest ways to do this with

our system is via a background subtraction.To subtract out a background, first save yourspectrum to the buffer using Acquire|Copy toBuffer. Next leaving all else unchanged, movethe source away from the detector, and takea background spectrum. You want to take along background to improve the precision ofthe correction, but you do not have to worryabout differences in the collection times forthe background and source spectrum. Anytime difference will be scaled out in the sub-traction. Save the background spectrum todisk and remember the name you give it. Se-lect the buffer from the drop-down menu onthe tool bar. You should be back to yoursource spectrum. Select Calculate|Strip andthen select the previously saved backgroundspectrum. The buffer should now contain thebackground-subtracted spectrum.When in doubt, take a new background

spectrum. It would change, for example, ifyou change the amplifier gain or move sourcesaround.

Energy Calibration

As we will now be concerned with the relation-ship between the gamma energies and theirphotopeak positions, all spectra should betaken with the sources placed far enough

away from the detector that the gain re-duction at high count rates is negligible. (SeeStep 9b.) Furthermore, while peak positionsare generally insensitive to smoothly varyingbackgrounds, you should at least check one ortwo spectra to see if the photopeak positionschange after a background subtraction.

17. Go back to using the 137Cs disk source.Take a new, background-corrected spec-trum. Record the channels at the photo-peak center, the Compton edge and thecenter of the backscatter peak.

18. If the photopeak center channel is Cγ,a “quick and dirty” one-point calibra-tion can then be performed assuminglinearity and a zero offset. E0 = 0,κ = 0.662Mev/Cγ. This step can bedone manually or by the Maestro pro-gram by creating an ROI over the pho-topeak, clicking the cursor in the pho-topeak ROI, selecting Calculate|Calibrateand then entering the 137Cs gamma en-ergy of 0.662 MeV. From then on, thechannel and energy will be displayedaccording to the one-point calibration.Print out the 137Cs spectrum markingand labeling the channel value and en-ergy at the photopeak, Compton edge andbackscatter peak. Also supply on thegraph, the value of κ. Roughly comparethe Compton edge and backscatter peakenergies with the predictions based on theCompton formula, Eq. 1.

19. Select Calculate|Calibration and destroythe one-point calibration.

20. Adjust the amplifier gain to make a3 MeV gamma produce about a 10 Vpeak. That is, make the 0.662 MeV137Cs photopeak occur around channel220. This step will make sure that the

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highest energy peak (about 2.5 MeV) ob-served in this part of the experiment willbe “in range.”

21. Take spectra for a 22Na, 54Mn, 60Co, and137Cs sources. For each source, set thenecessary ROI’s and obtain the channelat the photopeak center. Make sure thecount rate/dead time are low enough notto affect the gain. If in doubt, move thesource farther away and check whetherthe peak position changes. Make an en-ergy calibration table including the sourceisotope, the known gamma ray energy,and the photopeak center.

22. Make a plot of gamma ray energy vs. themeasured photopeak center and performa regression. Comment on the fit pa-rameters and the fitting model (linear, orquadratic) and the “goodness of fit.”

23. Measure the spectrum of the unknown—a jar of white crystals that looks a bitlike salt or sugar, but is neither. Justplace the whole jar close to the detector.It is a naturally occurring mineral witha low natural abundance of a radioactiveisotope. The spectrum will be weak andmay need to be collected for over 30 min-utes and stripped of the background. Useyour calibration to determine the energyat the photopeak(s) and determine theisotope(s) present. Appendix A SectionII of Duggan has a sorted list of gammaray energies and isotopes.

Activity Measurements

These measurements are concerned with theoverall count rates for the various sources.Thus, careful background subtractions willtypically be important. All spectra in this sec-tion must be background-corrected.

24. Make the following measurements withthe 22Na, 54Mn, and 137Cs sources. Takebackground-corrected spectra for eachsource placing it along the centerline ofthe detector at distances of 0, 5, 15, and30 cm.

25. Use the ROI feature to make tables of thenet counts in each photopeak for all pho-topeaks, including the sum peak for the22Na source. For the 22Na source makesure the acquisition time is long enoughto get several thousand counts in the sumpeak (except at the 30 cm source distancewhere the sum peak is too weak). Forthe 54Mn and 137Cs sources, also makea single ROI to cover both the Comptonplateau and the backscatter peak for usein activity determinations described later.Record the gross counts for these ROI’s.For each spectrum, make sure youalso record the live time.

26. Check if the count rates behave (roughly)as expected with regard to distance. Arethe ratio of the rates at the different dis-tances consistent from source to source?What does this imply about the separa-bility of the efficiency into a geometricand energy factor? Is the ratio of thecounts at different distances reasonablyrelated to an average detector solid an-gle at these distances? Keep in mind thedetector is 4 inches long.

CHECKPOINT: The procedure shouldbe completed to the previous step, in-cluding the regression analysis for theenergy scale calibration.

27. Here you will determine the activities ofthe 22Na source using the technique forsimultaneous gammas. You will need anindependent estimate of δ1 and δ2. From

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Step 24 you should already have the nec-essary data to determine δ’s for the 137Csand 54Mn sources which produce singleenergy gammas (0.662 MeV for 137Cs,0.835 MeV for 54Mn). These δ’s are sim-ply the ratio of the counts in the ROI cov-ering the Compton plateau and backscat-ter peak to the counts in the ROI coveringthe photopeak. Check how δ varies for thefour source-detector distances measuredand how it varies for the two sources, i.e.,how it varies with gamma energy.

28. Use the two δ’s from the previous stepto estimate the δ’s at the 22Na gammaenergies of 0.511 and 1.274 MeV

29. Use the appropriate ROI data for 22Naand the live time to determine the countrates in the 0.511 and 1.274 MeV photo-peaks and in the sum peak. Use Eq. 12,with these three rates R1, R2 and R12 andthe estimates of δ1 and δ2 of the previousstep to determine the sample activity forthe three source distances 0, 5 and 15 cm.

30. Compare the activities obtained in theprevious step with each other and withthe activity as determined by the originalsource activity, its age, and its halflife.

31. Most photopeak efficiency curves in thereferences show that in the energy rangefrom about 0.2-2.0 MeV, p′ falls off withenergy E as E−q with q around 0.9 for a3× 3 inch detector, increasing somewhatfor smaller detectors. Determine q fromthe count rates for 0.511 and 1.274 MeVpeaks for the 22Na source. Use what youhave learned to determine the activityof the 54Mn source assuming the 137Cssource activity was correctly labeled atthe manufacture date.

Halflife of 137Ba

The 137Ba used for this experiment is ametastable excited state of a nucleus left be-hind when the 137Cs isotope emits a beta parti-cle. The 137Ba is said to be the daughter of theparent nucleus 137Cs. The 0.662 MeV gammais emitted when the excited 137Ba nucleus de-cays to the ground state. The excited 137Badecays with a short halflife while the 137Csdecays with a much longer halflife. The bar-ium will be chemically separated (“milked”)from the cesium by forcing a mild acid solutionthrough a container of powdered 137Cs. Onlythe barium dissolves in the acid which will becollected in a dish and the detected gammaswill be monitored over time to determine thehalflife.Nuclear decay is a homogeneous Poisson

process1 where Γ, the probability per unit timefor the nucleus to decay (the decay rate) de-pends on properties of the nucleus and the par-ticular excited nuclear state. For a nucleus ina given excited state at t = 0, the probabilitydP that it will decay between t and t+ dt fol-lows the exponential probability distributiondP = Γe−Γtdt. The mean or average lifetimefor that excited state is then given by τ = 1/Γand the halflife is given by τ ln 2.Multiplying dP by the initial number of ex-

cited nuclei in a sample and by the probabilitythat a decay will lead to a detected event givesthe expected number of decay events in the in-terval from t to t + dt. Dividing this numberby dt gives the rate of decay events at time t.Adding in the constant average background

rate occurring from other processes gives thepredicted overall event rate as a function oftime (uncorrected for detector dead time)

R = RB +R0e−t/τ (15)

1See, for example, the addendum on Poisson Vari-ables in the Statistics section of the lab homepage foradditional information.

April 19, 2013


where R0 is the event rate due to nuclear decayat t = 0 and RB is the background rate.

The dead-time corrected rate is given byEq. 14, Rm = R/(1 + RT ), with R given byEq. 15. The predicted count Cp for the inter-val from ti to tf would then be the integralof the dead-time corrected count rate over themeasurement interval

Cp =∫ tf

tiRm(t)dt

=∫ tf

ti

R(t)

1 +R(t)Tdt

=∫ tf

ti

RB +R0e−t/τ

1 + (RB +R0e−t/τ )Tdt

=RB(tf − ti)

1 +RBT+

(τ

T− RBτ

1 +RBT

)·

ln1 +RBT +R0Te

−ti/τ

1 +RBT +R0Te−tf/τ(16)

Equation 16 is cumbersome and prone tonumerical error. The formulation that followsis much simpler and gives quite reasonable re-sults.

Although the source decays exponentiallyover the full course of the measurements, overthe much shorter time scale of a single mea-surement interval, the detection rate can beassumed to change linearly. With this as-sumption, the average uncorrected event rateover an interval from ti to tf will be the valueof R at the midtime of the interval: tmid =(ti + tf )/2. Then, with every time interval setto give the same live time tL, the count Cp inany interval will be that average rate multi-plied by the live time.

Cp = CB + C0e−tmid/τ (17)

where CB = RBtL and C0 = R0tL.

After milking the 137Cs you will have towork quickly, so doing a dry run is definitelyworthwhile. The basic principal is to take a

spectrum for some fixed live time and deter-mine the total counts either over the photo-peak and Compton plateau or just over thephotopeak. These counts are obtained repeat-edly as the activity of the sample decays.Choose the Services|Job Control and select

the JOB control program Decay.JOB. At thispoint just examine the program—don’t run it.Here’s what it looks like:

SET_DETECTOR 1

SET_PRESET_CLEAR

SET_PRESET_LIVE 10

LOOP 150

CLEAR

START

WAIT

FILL_BUFFER

SET_DETECTOR 0

REPORT "DECAY???.RPT"

SET_DETECTOR 1

END_LOOP

The SET PRESET LIVE 10 command instructsthe software to count for a live time of 10 sec-onds. The LOOP 150 command repeats thefollowing commands 150 times. The CLEAR,START and WAIT acquire another spectrum andthe REPORT command creates a text file DE-CAY???.RPT (with the loop number for ???)in the Maestro user directory. This file con-tain information about the acquired spectrumand the ROIs, in particular, the start time forthe spectrum and the gross count within theROI.An Excel spreadsheet program will read the

full set of report files made during execu-tion of the Decay.JOB program and will pro-duce columns containing the starting time, thegross count, and the net count and its uncer-tainty for each file.For everything to work properly:

1. Make sure there is only one ROI setto include the photopeak and Compton

April 19, 2013


plateau and that it is in the detector spec-trum, not the buffer.

2. Milk the 137Cs cow with 1 cc of the acidsolution letting the 137Ba containing fluidfall into the aluminum dish. Place thedish next to the detector and leave itundisturbed through the rest of the mea-surements.

3. Chose the Tools|Job Control and selectthe Decay.JOB program. It will take 150spectra, each for a live time of 10 seconds,and it will make reports on the ROI foreach, taking about 30 minutes total. Overthis time the sample activity will have de-cayed substantially.

4. Move the source away from the detec-tor and take a long background count.You may have to reset the presets in theAcquire|MCB Properties tab list. Recordthe counts in the ROI and the live time.

5. When finished, dump the solution downthe drain and wash out the dish. Thisviolates no laws and is perfectly safe.

6. Activate the Excel spreadsheet programand click on the Read .RPT files (happyface icon) in the tool bar. First, click inthe Data Table Location and select the topright cell address where the ROI data ta-ble will be written. Then click on the Get-Data button. From the file selection dia-log box, find and select the first of the.RPT files—DECAY000.RPT. The pro-gram then reads all the files, and createscolumns for the start time of the spec-trum (as an Excel time stamp), the grosscount, the net count, and the net countuncertainty.

7. Make a column for the starting timefor the count—in seconds, starting at

zero for the first count. This step canbe performed with the spreadsheet for-mula (B10-$B$10)*86400 and copyingdown (assuming the times start in cellB10). (The factor 86400 is the numberof seconds in one day because Excel timestamps are real numbers in days sincemidnight, Jan. 1 1904.) Make a secondcolumn for the midtime of the data acqui-sition interval. You can take tmid as themidtime between the start of one intervaland the start of the next interval. Thereis a fixed delay of about 2-seconds whilethe Decay.JOB program writes the ROIinformation to the hard drive and startstaking data for the next interval. Thus,the delay would shift every tmid by about1 second, but any constant offset to allvalues of tmid in Eq. 17 can be factoredinto the constant C0 and would not affectany other parameters.

8. Make a graph of the gross count vs. themidtime of the counting period. Performa properly weighted, least squares fit toyour data using Eq. 17. Use “countingstatistics” with the uncertainty in eachmeasured count Cm taken as its squareroot

√Cm. If significant, the starting

time of an interval has an approximateuncertainty of ±0.5 s.

9. Check if your results (χ2, and halflife)vary with the fitting region chosen. Inparticular, there may be significant errorsat the beginning of the data set when thecount rate is high and dead time correc-tions are most important. Discuss howthe χ2 and the lifetime vary as variousamounts of this beginning data are leftout of the fit. Compare the fitted halflifewith the accepted value of 2.55 minutes.

April 19, 2013

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