Experiments with the ALPHA/BETAdetector Teacher s Handbook & Student Instruction

Transcription

1 Experiments with the ALPHA/BETAdetector Teacher s Handbook & Student Instruction Guldfolium Laddad partikel Skikt av p-typ Signal R Utarmningsområde Si-kristall av n-typ U Cs y 93.5% % Ba 56 P.O. Box 15120, SE UPPSALA, SWEDEN Phone: , Fax: info@gammadata.se, Internet:

2 Teacher s Handbook for the Alpha & Beta Detector

3 Contents Page 1. Investigation of alpha and beta radiation Purpose Time needed Required theoretical knowledge 1 2. Experimental procedure: alpha radiation 2 3. Experimental procedure: beta radiation Formulas for calculating values for β decay 7

4 1. Investigation of alpha and beta radiation 1.1 Purpose To illustrate the properties of alpha and beta radiation. The energy loss of alpha particles in air and aluminium is studied. The energy difference between the K and L atomic shells of the daughter nuclide is measured with the aid of conversion electrons. 1.2 Time needed 1-2 hours per lab exercise. 1.3 Required theoretical knowledge Basic knowledge of the components of atomic nuclei. Use of the Chart of Nuclides. 1

5 2. Experimental procedure: alpha radiation The experiment is performed with a GDM system equipped with the optional alpha/beta detector, and alpha and beta radiation sources mounted on rods. The alpha/beta detector was designed for Risø's as well as 's alpha and beta sources. The Risø alpha source has, however, one major drawback: It is encapsulated by a thin coating, which decreases the energy of the particles. The energy distribution is also broadened because of the different paths the particles can follow through the coating. Figure 1 compares this distribution with one obtained from an unshielded source. The average energy loss of the alpha particles from the 241 Am Risø source is about 0.8 MeV. This must be considered in the calculations in Experiment 1. The corresponding energy loss in s sources is negligible. Counts Channels Figure 1. Spectrum from the Risø alpha source (1) and from s alpha source (2) 2

6 The alpha experiments give the best results with a source with very thin coating. manufacture such a source. Counts Channels Figure 2. Results from Experiment 1 using 's alpha source Figure 2 shows the spectrum in Experiment 1. The rightmost peak was obtained with the source rod fully inserted, the other peaks with the rod pulled out by 0.5, 1.0, 1.5, 2.0, and 2.5 cm, respectively. The distances should be measured accurately, e.g. with simple gauge blocks. The gain is adjusted with the trim potentiometer so as to put the peak at about channel 480 with the rod fully inserted. 3

7 Diagrams 1 and 2 in the Appendix give the energy loss for charged particles in different materials. The diagrams refer to protons, but can be converted for other charged particles by means of the formulas provided. Diagram A shows what Diagram 1 in Experiment 1 can look like. Notice the two energy axes. First, the axis is scaled in channels. Once the curve has been sketched and extrapolated to the energy axis, it is scaled in MeV. MeV α -energy 5,48 Channels α -energy Distance cm Figure 3. Diagram A 4

8 Diagram B shows the stopping power de/dx as a function of the distance between source and detector (Diagram 2 in Experiment 1). The stopping power was obtained as (the magnitude of) the slope of the graph in Diagram A. The increase at the right-hand end is known as the Bragg peak. MeV/cm Stopping power Distance cm Figure 4. Diagram B 5

9 Counts Channels Figure 5. Results from Experiment 2 using 's alpha source. Figure 5 shows a spectrum from Experiment 2 recorded with an unshielded source. From Diagram A one finds that the energy of the alpha particles is reduced from about 4.9 MeV to 2.4 MeV. (The mean energy is then 3.6 MeV.) Using Diagram 2 in the Appendix for E α = 3.6 MeV (E p = 0.9 MeV) one finds that de/dx is 800 kev/(mg/cm 2 ). The thickness of this aluminium foil is accordingly 3 mg/cm 2. Given the density of aluminium, 2.7 g/cm 3, the thickness of the foil is found to be 0.01 mm. 6

10 3. Experimental procedure: beta radiation The Risø gamma-ray source ( 137 Cs) could be used as described in the beta-ray exercise. However, this source also seems to be covered with a coating that stops electrons so that the conversion peaks cannot be seen in the spectrum. Risø s beta-ray source contains 90 Sr, which does not emit any conversion electrons. Nevertheless, it can be used to illustrate the appearance of a beta-ray spectrum. Counts Eβ 0 0,50 1,00 1,50 MeV Figure 6. Beta spectrum from 137 Cs The beta spectrum from 137 Cs is shown in Figure 6. The conversion-electron peaks can be seen on the right-hand side of the spectrum. The larger peak is due to K electrons, while L electrons give rise to a small peak to the right of the K peak. As was mentioned above, Risø s cesium source cannot be used in this lab exercise. Mätteknik AB have manufactured a source. (Permit issued by the Swedish Radiation Protection Institute, October 10, 1991, document no. 6541/11.) It serves as both a beta source and a gamma source. 3.1 Formulas for calculating values for β decay Q β - = m x c 2 - m y c 2 Q β + = m x c 2 - m y c 2-2m β c 2 NOTE: m x and m y are the atomic masses of the mother and daughter nuclide, respectively. Q β - = E b (y) - E b (x) + Δmc 2 Q β + = E b (y) - E b (x) - Δmc 2-2m e c 2 where: E b (x) and E b (y) is the binding energy of the mother and daughter nucleus, respectively. Δmc 2 = (m n - m p - m e )c 2 = MeV m e c 2 = MeV 7

11 Student Instruction for the Alpha & Beta Detector

12 Contents Page 1. Investigation of alpha radiation Purpose Equipment Theory Alpha decay The detector Experiment Stopping power of air for alpha particles Diagram A Stopping power of air for alpha particles Diagram B Stopping power of air for alpha particles Experiment Determination of the thickness of an aluminium foil 6 2. Investigation of beta radiation Purpose Equipment Theory Beta decay The detector Experiment Study of beta particles and conversion electrons 9 3. Exercises 10

13 1. Investigation of alpha radiation 1.1 Purpose To investigate alpha radiation with the aid of a surface-barrier detector. The energy loss of alpha particles in air is studied. Then, alpha particles are used to measure the thickness of an aluminium foil. 1.2 Equipment A GDM system with a surface-barrier detector; an alpha-radiating source. 1.3 Theory Alpha decay Ionising radiation is a way for unstable atomic nuclei to get rid of excess energy. For some (usually heavy) nuclei it can be advantageous to emit alpha particles, i.e. 4 He nuclei. The resulting daughter nucleus then contains two protons and two neutrons less than the original nucleus. The alpha particles are ejected with specific energies, i.e. only a limited number of alpha energies will occur. Figure 1 shows a so-called decay scheme, in this case for 235 U. The daughter nucleus is usually excited, so that the alpha particles get different energies. The percentage distribution of the occurring alpha energies is shown on the right-hand side U % 5.7% 0.6% 3.4% 18% 57% 4% % 3.7% 4.6% Th Figure 1. Example of a decay scheme. 1

14 1.3.2 The detector The detector used here is of the semiconductor type. Often simply called a semiconductor detector, the more proper term is surface-barrier detector. It is used to detect charged particles and measure their energy. These could be alpha or beta particles from radioactive sources, or other particles, such as protons, produced in nuclear reactions. The detector consists of an n-doped silicon crystal. By an etching procedure, a very thin (1 μm) p-doped layer has been created at one surface of the crystal. This layer is covered by a thin (40 μg/cm 2 ) gold layer used for electrical contact. When a voltage is applied with the positive potential on the n-layer, the crystal can be regarded as a reverse-biased diode. A depletion layer with very few charge carriers will form in the n-layer. The thickness of this layer will depend on the applied voltage and the electrical resistivity of the crystal. A thin depletion layer is sufficient to detect alpha particles owing to their short range, whereas the detection of beta particles requires a much thicker layer (on the order of mm). The depletion layer of the beta detector used in the present experiment is 0.5 mm thick. The working principle of the surface-barrier detector is shown in figure 2. While a charged particle is being stopped in the depletion layer, it ionises the atoms of the crystal, creating electron-hole pairs. These are collected in the applied electric field, thereby causing a charge pulse. Gold layer Charged particle Layer of p-type Signal R Depletion region U Si-crystal of n-type + - Figure 2. The working principles of the surface-barrier detector. If the depletion layer is thick enough, all the kinetic energy of the particle will be deposited, producing a corresponding number of electron-hole pairs. The size of the charge pulse is then proportional to the energy of the particle. 2

15 1.4 Experiment Stopping power of air for alpha particles The radioactive nuclide used in this experiment is 241 Am. Write down the reaction formula (use the Chart of Nuclides):... Use the mass data in the Chart of Nuclides to calculate the energy released in the decay:... How many (and which) alpha energies are expected to appear in this reaction, according to the Chart of Nuclides?... (Compare with the more detailed decay scheme and the table given in the Appendix.) Place the alpha source in the detector box, as far in as possible. Put the switch on the amplifier in the alpha position. Start the data aquisition. Stop it after a few seconds. Pull out the source 0.5 cm. Start the data aquisition and stop it when the peak in the spectrum is about as big as the first one. Repeat this procedure with the source pulled out in a few more steps of 0.5 cm. Determine the centroid of each peak and enter the result into the table, together with the detector-source distance. Remember to add the distance between the source and the detector with the source pushed as far into the detector box as possible, as an offset to the measured distances. This offset distance is 1.1 cm for both s alpha source and the Risø source. Peak Distance (cm) Centroid (channels) 3

16 1.4.1 Diagram A Stopping power of air for alpha particles Make a diagram of the centroid channel number vs. the distance. Draw a graph that fits the points as well as possible, and extrapolate it to the y axis. The crossing point corresponds to the centroid at zero distance. We can now put an energy scale along the y axis. (The crossing point corresponds to the alpha energy listed in the Chart of Nuclides, and we assume that channel 0 corresponds to zero energy. For the Risø source, which is encapsulated, the energy must be corrected for the energy loss in the coating material.) Calculate the range in air of these alpha particles by extrapolating the graph down to the x axis. Answer:... The slope of the curve gives the energy loss per unit length in air. The energy loss per unit length (de/dx) is called the stopping power. Calculate the stopping power of air for 3.0-MeV alpha particles. Answer:... Compare with diagram 1 in the Appendix:... By calculating the stopping power, i.e. the slope of the graph in diagram 1, at different distances from the source, one can construct the so-called Bragg curve, which shows the stopping power as a function of the distance that the particle has travelled in the substance. Distance (cm) de/dx (MeV/cm)

17 1.4.2 Diagram B Stopping power of air for alpha particles Draw a diagram of the stopping power vs. distance. As you can see, the stopping power varies with the distance that the alpha particles have travelled. The lower the kinetic energy of the alpha particle, the larger is the stopping power. The maximum stopping power occurs at the end of the particle track. This appears in the diagram as a peak, the Bragg peak. The Bragg peak means that large amounts of energy are deposited per unit length towards the end of the particle track. This is used in radiation therapy to damage a tumour as much as possible while sparing the surrounding tissue. 5

18 1.5 Experiment Determination of the thickness of an aluminium foil Place the alpha source as far inside the detector box as possible. Start the data acquisition and stop after a few seconds. Insert an aluminium foil (mounted in the absorber holder) between the source and the detector. Acquire data again for a few seconds. Determine the centroid channels for the two peaks in the spectrum. Peak 1 2 Centroid Energy (MeV) From diagram 2 in the Appendix, the following energy loss per unit length is obtained for alpha particles in aluminium: de/dx =... for E α =... (the average of the energies above) Calculate the thickness of the foil. (Assume that the thickness is x mg/cm 2. Notice the unit! The energy difference E 2 - E 1 can be calculated as (de/dx) x. Finally calculate the thickness in mm.)

19 2. Investigation of beta radiation 2.1 Purpose User s Guide for he Alpha/ & Beta Detector version 3.2 To investigate beta radiation with the aid of a semiconductor detector. The spectrum of conversion electron energies is used to determine the energy difference between the atomic K and L shells. 2.2 Equipment A GDM system with a semiconductor detector; s beta-radiating source. 2.3 Theory Beta decay There is a type of radioactivity in which a nucleus is transformed by emitting beta radiation. For instance, 14 C decays according to 14 C 14 N + β - + ν This type of instability is due to the fact that the composition of protons and neutrons in the nucleus is not the energetically most favourable one. By radioactive decay the nucleus can reach a state of lower energy. In this process a neutron in the nucleus is transformed to a proton according to n p + β - + ν The total electric charge stays the same, because the positive charge of the proton is balanced by the creation of a negative particle, β -, which is in fact an electron. An additional particle is also created, a neutrino (in this case an antineutrino, ν). Another type of beta decay is the transformation of a proton into a neutron: p n + β + + ν β + is a positive electron, called a positron. It is the anti-particle of the electron. The released energy becomes the kinetic energy of the created β and ν particles. The energy is arbitrarily distributed between the two particles. The beta energy can get all values from 0 to the maximum energy, which equals the energy released in the decay. If the daughter nucleus is formed in its ground state, the released energy equals the value of the decay. 7

20 Figure 3 shows how 137 Cs undergoes β - decay to 137 Ba. After the decay, the daughter nucleus may be in its ground state, but is usually in an excited state with an excitation energy of MeV. This excess energy is emitted either as gamma radiation or through so-called internal conversion. In the latter case, the nucleus transfers its excess energy to one of the electrons of the atom, which is ejected with a kinetic energy of MeV minus the electron binding energy. Another example of beta decay ( 90 Sr) is illustrated in the Appendix. 90 Sr produces no conversion electrons. 137 Cs 55 β 30.0 y 93.5% % Figure 3. Decay scheme for 137 Cs Ba The detector The beta particles are detected with the semiconductor detector described in chapter Investigation of alpha radiation. 8

21 2.4 Experiment Study of beta particles and conversion electrons Place the 137 Cs source in the detector box. The switch should be in the beta position. Collect a spectrum for at least 5 minutes. A beta-ray spectrum is characterised by the continuous distribution of beta-particle energies. The spectrum from 137 Cs is dominated by beta particles emitted in decay to the excited state, since 93.5% of the decays go to that state. The far right end of the spectrum shows a pair of discrete peaks, a large one to the left of a small one. These are due to the conversion electrons that are emitted when the MeV excited state deexcites. The two peaks show that the conversion electrons come from two different electron shells, the K and L shells in the daughter nuclide 137 Ba. 9

22 3. Exercises 1. Write down the reaction formula for the decay of 137 Cs Calculate the energy released in the decay, the so-called Q value Calculate the maximum beta-particle energy when the daughter nucleus is left in a) the ground state:... b) the excited state: Calculate the energy of a conversion electron originating from the K shell. The binding energy of K electrons is 37.4 kev Make an energy calibration of the beta spectrum by means of the K peak, assuming that channel 0 = 0 MeV. Then calculate the energy of the conversion electrons from the L shell and the binding energy of L electrons Calculate the maximum beta-ray energy from the decay to the excited state by extrapolating the right-hand edge of the spectrum distribution down to the energy axis. Result:... According to exercise 3b:... 10