Interactions of Charged Particles, Neutrons, and Photons with Matter, Study notes of Physics

Download Interactions of Charged Particles, Neutrons, and Photons with Matter and more Study notes Physics in PDF only on Docsity! 1 Ph 3504 Radioactive Decay Required background reading Attached are several pages from an appendix on the web for Tipler. You do not have to read them all (unless you want to), but make sure you read the following: • First paragraph after the words “Charged Particles”. The basic idea is that charged particles of a given energy have a reasonably well-defined range when their kinetic energy gets absorbed by matter. • The top paragraph on page 89; the main point there is that the energy loss of a charged particle per unit length in a material is just proportional to the density. So the energy loss in a given material of thickness x is essentially just dE = K x ρ , where K is a constant. So, for convenience, we often express the “thickness” of a material as xρ, which has units of g/cm2. This is a convenient way of comparing the energy of absorbing properties of pieces of material with different compositions. • The section on photons at the end. The main point is that rather than having a well-defined range in materials, the intensity of photons decreases exponentially as the thickness of the shielding material is increased. Read Tipler section 11-3 on radioactivity, which contains the definition of half-life, etc. Prelab Questions 1. Assume you have a monoenergetic source of gamma rays of intensity Io and then you put some lead shielding between it and your detector. The transmitted intensity through the lead as a function of the thickness of the lead absorber is given by: x o x o eIeII ρρµµ )/(−− == where ρ = 11.35 g/cm3 is the density of the lead and µ is the absorption coefficient of the lead for this energy photon. For the 0.662 MeV gamma rays that you will use in this experiment, the absorption coefficient is µ/ρ=0.1137 cm2/g. What thickness of lead, x, is needed to cut down the transmitted intensity of 0.662 MeV photons by a factor of 2 (relative to the case with no shielding present)? 2. In this experiment, you will measure the half-life of a particular short-lived radioactive isotope, 137mBa. You will find that it has a half-life, t1/2, of about 2.55 minutes. Assuming that the original intensity of the source is 1000 Bq = 1000 decays/second, how long do you have to wait until the amount of radioactivity is down to 1 Bq = 1 decay/second? 2 Introduction Radioactivity is produced when the nucleus of an atom decays and emits radiation. The decay products can be electrons, positrons, gamma rays and alpha particles, all of which, due to the ionization they produce, are considered pollutants. They are clearly undesirable, as these rays or particles can penetrate the human body and produce mutation of cells, the results of which are not readily predictable. At very high radiation levels, the effects are well-known and can cause sickness and even death. Fortunately, the radioactive contaminants that surround us are very, very small. Common materials (stone, brick...) contain small concentrations of long-lived radioactive isotopes which produce low levels of radiation. An additional source of radioactivity is the ever-present cosmic radiation which bombards earth from outer space, producing a non-negligible part of the radioactivity in the biosphere. The radioactive sources you will use in this lab also consist of very small amounts of radiation, so they are not harmful to you. The two types of radioactivity you will explore in this lab are those that produce gamma rays (energetic photons with energies ~ 1 MeV) and electrons (from so-called beta decay). We will describe each of these processes in a little more detail as you encounter them in this lab. The equipment you will be using is shown in Figure 1. It consists of the following pieces; please check to be sure you have them all. • Geiger tube mounted in cylindrical holder with shelves below them (please don’t remove this Geiger tube without assistance from your TA; it is easily damaged) • Source set with four sealed radioactive sources - 137Cs, 90Sr, 204Tl, 60Co • Set of absorbers in wood box (plastic, polyethylene, aluminum, lead) • Geiger tube control and counting unit (SPECTECH ST360 Counter) • Minigenerator (at instructor’s table; he or she will prepare this source for you) Figure 1: Experimental equipment for this lab 5 Part 2: Statistics of radioactive decay When you take counts from a radioactive source for a fixed time interval, you get some number N. When you then take another count for the same time interval, you won’t get exactly the same number of counts. If you do it many times you will see a distribution of values. For the case of radioactive decay, this is described well by a distribution called a Poisson distribution, which actually starts to become the same as a Gaussian distribution for a large number of counts. Statistical analysis shows that the standard deviation of this distribution is N=σ , where N is the average number of counts in that time interval. To test this you will take several fixed-time counts for a radioactive source; you will analyze it later and convince yourself that N is a good estimate for the error on a given radioactive count. 1. Use the exact same setup as part 1: • 10 second counting interval • 90Sr source on the top shelf (yellow label down) • GM tube set at your chosen operating voltage 2. Take 10 separate 10 second counts and note down your numbers. You will analyze these data later for your report. Part 3: Absorption of Gamma Rays in Matter This part of the experiment will examine the means by which nuclear radiation can be attenuated. First you will consider gamma rays (energetic photons in the MeV range that come from the decay of excited nuclear states). When a beam of gamma rays strikes a slab of material, reactions with the electrons and nuclei in the material (photoelectric effect, Compton scattering, and pair production) will remove some of the gamma rays from the beam. If N gamma rays are incident on a given slab of thickness dx in a given time interval, then that number is reduced by some amount dN through interactions in the material. The number dN is given by the relation dxNdN µ−= where µ is a proportionality constant, called the absorption coefficient, which is characteristic of the material and also depends on the energy of the gamma rays. A negative sign appears because dN represents a removal of gammas from the beam. Solving this relation gives ))(/( 00)( xx eNeNxN ρρµµ −− == Here, N(x) is the number of gammas remaining in the beam after the beam has traversed a distance x into the material and N is the number or gammas entering the material at x =0. 6 In the second part of the expression we have expressed the “thickness” of the material as ρx where ρ is the density of the material. This has units of g/cm2. In that scheme we represent the absorption coefficient as µ/ρ with units of cm2/g. Remove your 90Sr source and get out the 137Cs source. We are interested in the 0.662 MeV gamma rays that 137Cs emits. You will study absorption of this gamma ray in two different types of absorbers – aluminum and lead. Take a look at the absorber set (in the wooden box). Note that for each different type of absorber, there are two numbers given for it – the thickness in inches (or in mils, which is 1/1000 of an inch) and the thickness given as ρx (in units of mg/cm2). Also, there is a code for each one. 1. Start by setting up the counter for 30 second fixed counts and leave the high voltage of the GM tube at your desired operating voltage. Initially, take a “background” count by moving the other radioactive sources (including the 137Cs) to the edge of the table. Take three 30 second counts and note them down. These are your counts of the background from radioactive material in the walls of the room and from cosmic rays. 2. Put the 137Cs source onto the plastic tray with its yellow label UP (this is because 137Cs emits an electron as part of a beta decay, but we are not interested in it here, so we orient the 137Cs source with the most absorber in the way of the electron). Put the source on the fifth shelf from the bottom. This will leave room for you to put in absorbers. Prepare a data table that has columns for absorbers (and thicknesses) used and counts per 30 seconds. Take data for aluminum absorbers in the thickness range .040 inches up to a maximum of 0.395 inches (which can be obtained by putting the four thickest aluminum absorbers in the shelves above the 137Cs source. Take data with at least seven different absorber combinations (for a total of 7 different thicknesses). 3. Now do the same as step 2, but for the lead absorbers. Here, you can put the 137Cs source on the third shelf from the top. Take data for seven different combinations of lead absorbers for a thickness range of .032 inches up to .375 inches. 4. For your report, you will analyze the data you took above to determine the absorption coefficient for gamma rays of 0.662 MeV energy in aluminum and lead. Part 4: Absorption of Energetic Electrons in Matter Now you will consider how electrons are attenuated in matter. There is a fundamental difference between the absorption of gammas and electrons. Whereas a photon is completely removed from the beam in a single collision, the electron loses its energy gradually in many collisions and is characterized by a definite range. The absorption of electrons depends on the fact that they have charge as compared to the gammas which do not. As an electron passes through matter, the probability of a direct collision with an electron or nucleus is very small. But its electric field interacts with other electrons in the material exciting atomic electrons or removing (ionizing) them. Since the incident particle does work on these atomic electrons in the material, its energy decreases, and it slows down. The electron comes to rest when all of its kinetic energy has been 7 transferred to atoms of the material. For a given electron kinetic energy that means that electrons make it through until the thickness of material is enough to stop the electrons of that energy. Electrons of higher kinetic energy require more material to be stopped. As pointed out in the introduction, the amount of energy lost in a given thickness of material is just proportional to ρx, independent of the type of materials (for materials with low Z atoms), where ρ is the density and x is the thickness. The actual source of electrons you will use in this part is a 204Tl source which emits a spectrum of electrons with kinetic energies ranging from 0 to 0.765 MeV. So what you will actually see in this part is similar to Figure 3, which shows a typical “number of counts” versus “absorber thickness” plot. As the absorber thickness is increased, you will see the count rate decrease as the lower energy electrons are stopped. Eventually, there is enough absorber that even the electrons with the maximum kinetic energy are stopped. At that point you will essentially be seeing only the background count rate, independent of how much more absorber you add. Figure 3: Number of electron counts versus absorber thickness (plotted on a semi-log graph, where the y-axis is the logarithmic axis). 1. Set the counting time interval to 20 seconds. Before inserting the source, take three 20 second measurements so you have a determination of the background count rate. 2. Mount the 204Tl source on the third shelf from the top. For this part of the experiment you should use the absorbers with the codes A – F. You don’t have to worry about which type of material it is, just note down the total ρx thickness in mg/cm2 for each combination of absorbers you use. Take counts for 20 seconds. Record the absorber thickness and count rate in a table. Gradually increase the absorber thickness, you should see the count rate decrease until it levels off at some thickness (when it should then be down near to the background count rate). (Note: make sure to handle the thin absorbers by their frame on the edges; don’t grab them in the middle of the absorber.) 3. You will analyze your data for your report to determine the range in matter of the electrons with the highest kinetic energy (0.765 MeV) from 204Tl. 10 paper available on the lab website). On a semi-log plot, the results should give a line. Do a least squares fit in each case to determine the absorption coefficient in units of cm2/g (refer to the section VI on page A-18 of the error analysis appendix for how to do a least squares fit in a case like this where the errors on each point are different). Compare your results to the following accepted values: 0.0750 cm2/g for aluminum and 0.1137 cm2/g for lead for the 0.662 MeV gamma ray. The density of aluminum is 2.7 g/cm3 and the density of lead is 11.35 g/cm3. 5. Part 4: Absorption of Energetic Electrons in Matter: Correct each of your measurements for the background counts and assign an error using N=σ . Plot your results on a semi-log plot. What range (in units of g/cm2) corresponds to the point indicated in Figure 3 where the slope definitely changes? This is the range for the most energetic electrons (0.765 MeV) in your 204Tl radioactive sample. 6. Part 5: Measurement of a radioactive half-life: Correct each of your measurements for the background counts and assign an error using N=σ . Plot your results on a semi-log plot and do a least squares fit to determine the decay constant and the half-life. Refer to section VI on page A-18 of the error analysis appendix for details of how to do the least squares fit. 7. Conclusion In this section, we will discuss briefly the main interactions of charged particles, neu- trons, and photons with matter. Understanding these interactions is important in the development of nuclear detectors, the design of radiation shielding, and the study of effects of radiation on living organisms. For the important latter case, we will exam- ine the principal factors involved in stopping or attenuating a beam of particles. Charged Particles When a charged particle traverses matter, it loses energy mainly through collisions with electrons. This often leads to the ionization of the atoms in the matter, in which case the particle leaves a trail of ionized atoms in its path. If the energy of the parti- cles is large compared with the ionization energies of the atoms, the energy loss in each encounter with an electron will be only a small fraction of the particle’s energy. (A heavy particle cannot lose a large fraction of its energy to a free electron because of conservation of momentum, as we saw in Section 4-2. For example, when a bil- liard ball collides with a marble, only a very small fraction of the energy of the billiard ball can be lost.) Since the number of electrons in matter is so large, we can treat the loss of energy as continuous. After a fairly well-defined distance, called the range, the particle will have lost all its kinetic energy and will come to a stop. Near the end of the range, the view of energy loss as continuous is not valid because the kinetic energy is then small and individual encounters are important. For electrons, this can lead to a significant statistical variation in path length, but for protons and other heavy particles with energies of several MeV or more, the path lengths vary by only a few percent or less, for identical monoenergetic particles. The statistical varia- tion of the path lengths is called straggling. We can get an idea of the important factors in the stopping of a heavy charged particle by considering a simple model. Let ze be the charge and M the mass of a par- ticle moving with speed v past an electron of mass me and charge e. Let b be the impact parameter. We can estimate the momentum imparted to the electron by assum- ing that the force has the constant value F  kze2/b2 for the time it takes the particles to pass the electron, which is of the order of t  2b/v (see Figure 12-18). The momen- tum given to the electron is equal to the impulse, which is of the order of magnitude 12-16p  Ft  kze2 b2 2b v  2kze2 bv More Interaction of Particles and Matter (Continued) 88 More where k is the Coulomb constant. (The same result is obtained by integration of the variable impulse, assuming the particle moves in a straight line and the electron remains at rest.) The energy given to the electron is then 12-17 This is the kinetic energy lost by the particle in one encounter. To find how many such encounters there are, consider a cylindrical shell of thickness db and length dx (see Figure 12-19). There are Z (NA/A)2b db dx elec- trons in the shell, where Z is the atomic number, A the atomic weight, NA Avogadro’s number, and  the mass density. The energy lost to these electrons is then If we integrate from some minimum b to some maximum b, we obtain 12-18 where 12-19 The range of the values of b can be estimated from general considerations. For exam- ple, this model is certainly not valid if the collision time is longer than the period the electron is in orbit. The requirement that 2b/v be less than this time sets an upper limit on b. The lower limit on b can be obtained from the requirement that the maxi- mum velocity the electron can receive from a collision is 2v (obtained from the L  lnbmaxbmin  dK dx  4k2z2e4(Z /A)NA mev2 L dK  2k2z2e4 mev 2b2 Z NA A 2b db dx Ke  p2 2me  2k2z2e4 mev2b2 M ze v b 2b e– Impulse ≈ Ft = kze 2 –––– b2 F = kze 2 –––– b2 2b–– v Fig. 12-18 Model for calculating the energy lost by a charged particle in a col- lision with an electron. The impulse given to the electron is of the order Ft, where F  kze2/b2 is the maximum force and t  2b/v is the time for the particle to pass the electron. M ze v x dx b db Volume of shell is 2πb db dx Number in shell is n2πb db dx Fig. 12-19 In path length dx, the charged par- ticle collides with n2b db dx electrons with impact parameters in db, where n  Z(NA/A) is the number of electrons per unit volume in the material. (Continued) Interaction of Particles and Matter The chance of a neutron’s being removed from a beam within a given path dis- tance is proportional to the number of neutrons in the beam and to the path distance. Let  be the total cross section for the scattering plus the absorption of a neutron. If I is the incident intensity of the neutron beam (the number of particles per unit time per unit area), the number of neutrons removed from the beam per unit time will be R  I per nucleus (Equation 12-5). If n is the number density of the nuclei (the nuclei per unit volume) and A is the area of the incident beam, the number of nuclei encountered in a distance dx is nA dx. The number of neutrons removed from the beam in a distance dx is thus 12-21 where N  IA is the total number of neutrons per unit time in the beam. Solving Equation 12-21 for N, we obtain 12-22 If we divide by each side of Equation 12-22 by the area of the beam, we obtain a similar equation for the intensity of the beam: 12-23 We thus have an exponential decrease in the neutron intensity with penetration. After a certain characteristic distance x1/2, half the neutrons in a beam are removed. After a second equal distance, half of the remaining neutrons are removed, and so on. Thus, there is no well-defined range. At the half-penetration distance x1/2, the number of neutrons will be (1/2)N0. From Equation 12-22, 12-24 The main source of energy loss for a neutron is usually elastic scattering. (In materi- als of intermediate weight, such as iron and silicon, inelastic scattering is also impor- tant. We shall neglect inelastic scattering here.) The maximum energy loss possible in one elastic collision occurs when the collision is head-on. This can be calculated by considering a neutron of mass m with speed vL making a head-on collision with a nucleus of mass M at rest in the laboratory frame (see Problem 12-24). The result is that the fractional energy lost by a neutron in one such collision is 12-25 This fraction has a maximum value of 1 when M  m and approaches 4(m/M) for M  m. EXAMPLE 12-10 Penetration of Neutrons in Copper The total cross section for the scattering and absorption of neutrons of a certain energy is 0.3 barns for cop- per. (a) Find the fraction of neutrons of that energy that penetrates 10 cm in  E E  4mM (M  m)2  4(m/M) [1  (m/M)]2 x 1/2  ln 2 n enx1/2  2 1 2 N0  N0enx1/2 I  I0e nx N  N0enx dN  I(nA dx)  nN dx 91 (Continued) 92 More copper. (b) At what distance will the neutron intensity drop to one-half its initial value? Solution (a) Using n  8.47  1028 nuclei/m3 for copper, we have According to Equation 12-22, if we have N0 neutrons at x  0, the number at x  0.10 m is 12-26 The fraction that penetrates 10 cm is thus 0.776, or 77.6 percent. (b) For n  8.47  1028 nuclei/m3 and   0.3  1028 m2, we have from Equation 12-24 Photons The intensity of a photon beam, like that of a neutron beam, decreases exponentially with distance in an absorbing material. The intensity versus penetration is given by Equation 12-23, where  is the total cross section for absorption and scattering. The important processes that remove photons from a beam are the photoelectric effect, x 1/2  ln 2 (0.3  1028 m2)(8.47  1028/m3)  0.693 2.54 m  0.273 m  27.3 cm N  N0enx  N0e0.254  0.776N0 nx  (3.0  1028 m2)(8.47  1028/m3)(0.10 m)  0.254 1 40 50 60 70 20 30 10 0.2 1 20.5 5 20 Pair production Compton scattering Total Photoelectric 10 50 100 σ, b ar ns Ephoton, MeV Fig. 12-23 Photon interac- tion cross sections vs. energy for lead. The total cross sec- tion is the sum of the cross sections for the photoelectric effect, Compton scattering, and pair production. (Continued) Interaction of Particles and Matter Compton scattering, and pair production. The total cross section for absorption and scattering is the sum of the partial cross sections for these three processes: pe, cs, pp. These partial cross sections and the total cross section are shown as functions of energy in Figure 12-23. The cross section for the photoelectric effect dominates at very low energies, but it decreases rapidly with increasing energy. It is proportional to Z4 or Z5, depending on the energy region. If the photon energy is large compared with the binding energy of the electrons (a few keV), the electrons can be considered to be free, and Compton scattering is the principal mechanism for the removal of photons from the beam. The cross section for Compton scattering is proportional to Z. If the photon energy is greater than 2mec2  1.02 MeV, the photon can disappear, with the creation of an electron-positron pair. This process, called pair production, was described in Section 2-4. The cross section for pair production increases rapidly with the photon energy and is the dominant component of the total cross section at high energies. As was discussed in Section 2-4, pair production cannot occur in free space. If we consider the reaction  : e  e, there is some reference frame in which the total momentum of the electron-positron pair is zero; however, there is no reference frame in which the photon’s momentum is zero. Thus, momentum conservation requires that a nucleus be nearby to absorb momentum by recoil. The cross section for pair production is proportional to Z2 of the absorbing material. 93