Natural Abundance of K40and Detector EfficiencyYour goal in this experiment is to determine the natural abundance of the iso-tope K40in a KCl sample. The natural abundance is the fractional amount of theisotope compared to the non-radioactive isotopes found on earth. For K40, the nat-ural abundance is the ratio of the number of K40nuclei divided by total number ofnon-radioactive potassium nuclei (mostly K39) found on earth. Since isotopes arenot separated by chemical means, the natural abundance of K40is the same for anysample of potassium on earth. In this experiment, our sample will be KCl.To find the natural abundance, we need examine a sample containing potassium,and measureA) the total number of potassium (mostly K39) nuclei, andB) the total number of K40nuclei.A) We can determine the total number of potassium nuclei in our sample by measuringthe mass of the KCl sample. For this class, we will use a container of salt substitute:pure KCl. We will find the mass by subtracting the mass of the container. Since weknow the sample is pure KCl, the mass measurement gives us the number of moles.Using Avagadro’s number, 6.02 × 1023, you can determine the number of potassiumnuclei in the sample.B) Since K40is radioactive, we can determine the number of K40nuclei in the sampleby measuring its activity. The number of radioactive nuclei, N, is related to theactivity, A (decays/sec), byA =Nln(2)τ(1)where τ is the half-life of the isotope. For K40the half-life τ = 1.277 × 109years.Once A is measured, the number of K40nuclei N can be determined from the aboveequation. To obtain an accurate measurement of the activity, it is important to knowthe efficiency of the detector. The most important part of the experiment will be todetermine the detector’s efficiency. In the next section, we discuss what efficiency isand how to measure it.Efficiency Calibration of Solid Scintillation DetectorsThe efficiency ε of a detector is defined as (the number of particles detected)/ (thenumber of particles emitted):1ε =the number of particles detectedthe number of particles emitted(2)The efficiency is a number between zero and one. If we know the efficiency of ourdetector, then measuring the number of particles detected will allow us to determinethe number of particles emitted in our sample. The efficiency of a detector will dependon a few factors, the most important are:1.The source-detector geometry: The number of particles detected will dependon how close the source is to the detector. The closer the source is to the detector,the larger the efficiency will be.2.The size of the detector: Larger detectors will usually be more efficient, sincethey have a larger volume for the particles to absorbed in.3.The energy of the gamma (or X-ray) radiation: The photopeak is producedby photo-absorption. The photo-absorption process has a strong energy dependence.For high energy photons, photo-absorption has a lower probability to occur than pho-tons of low energy.For solid scintillation detectors, NaI and Ge, the dependence of ε on energy, num-ber 3 above, is quite large. For example, NaI detectors can detect 100 KeV gammasabout 4-5 times more efficiently than 1200 KeV gammas. This means that although aphotopeak at 1200 KeV is small compared to one at 100 KeV in a particular spectrum,there might be more 1200 KeV gamma emitted than 100 KeV gammas.Since the efficiency depends on the three factors listed above, one often keeps thesource-detector geometry fixed during a series of experiments. That is, for a series ofexperiments one places all the samples in the exact location relative to the detector.Also, one uses samples that are all the same size and shape. If this is done, thenfactors 1 and 2 above are the same for all the samples in a particular experiment.In this case, the only efficiency calibration necessary is the energy dependence ofε. The energy dependence for a particular source-detector geometry is measured byusing standardized sources. One can purchase sources in which the activity has beencalibrated by the manufacture. If the activity of the source is known, then the numberof gamma particles emitted can be calculated. By measuring the number of gammas(of a particular energy) detected during a specific time interval, the efficiency ε canbe determined.2Efficiency Calculation including Source-Detector GeometryThe distance from the sample to the detector and the size of the detector areimportant factors that affect the detector’s efficiency. We would like to include theseaffects in our measurements. The method that seems to work best (i.e. is simple andfairly accurate) with our NaI detectors is to use the ansatz:Counts Detectedtime=γ′s emittedtime(πr24π(x + d)2) ǫ (3)where πr2is the cross sectional area of the detector, x is the distance from the sourceto the side of the detector, and d is the distance from the side of the detector to the”effective center” of the detector. The geometry factor (πr2)/(4π(x + d)2) representsthe fraction of gammas emitted that go through the detector. This geometry factoris just an approximation, but usually gives consistent results in our experiments. Inour laboratory, we have NaI detectors of two sizes: with a diameter of 1 1/2 inches,and with a diameter of 2 inches. One can measure x, so once the ”effective distance”d is known, the efficiency ǫ can be determined.The number of γ’s emitted can be determined from the activity, A, of the sample.For a particular γ energy, the number of γ’s emitted per second equals the activitytimes the Yield, Y . The Yield is the probability that a γ is emitted during the nucleardecay. In terms of the activity and yield, the above equation becomesCounts Detectedsec= AY (πr24π(x + d)2) ǫ (4)In the experiment, you will first determine d by collecting data from one sourcelocated at different distances x from the detector. A computer program that does aχ2fit of the data to determine the best estimate of d is available on the lab computers.The program was written by Sue Hoppe (2003), a Cal Poly Pomona physics major.Once you have determined d for your detector, you can measure how the effi-ciency ǫ depends on the energy of the gamma. For standards, we will use Cs137(Eγ= 662KeV),60Co (Eγ= 1173.237 and 1332.501 KeV), and Na22(Eγ= 511 and1275 KeV). Once the efficiencies for the energies of the calibration sources has beendetermined, a graph of efficiency vs. energy, ǫ(E) can be plotted. As you will see, ǫhas