# Phy 310 chapter 8

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### Phy 310 chapter 8

1. 1. Dr Ahmad Taufek Abdul Rahman School of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan is defined as the spontaneous disintegration of certain atomic nuclei accompanied by the emission of alpha particles, beta particles or gamma radiation. CHAPTER 8: Radioactivity (3 Hours) 1
2. 2. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Learning Outcome: 8.1 Radioactive decay (2 hours) At the end of this chapter, students should be able to:  Explain α, β+, βˉ and γ decays. State decay law and use  dN  N dt Define activity, A and decay constant, .  Derive and use  N  N 0 e t OR  A  A0 e t Define half-life and use T1/ 2  ln 2  2
3. 3. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1 Radioactive decay    The radioactive decay is a spontaneous reaction that is unplanned, cannot be predicted and independent of physical conditions (such as pressure, temperature) and chemical changes. This reaction is random reaction because the probability of a nucleus decaying at a given instant is the same for all the nuclei in the sample. Radioactive radiations are emitted when an unstable nucleus decays. The radiations are alpha particles, beta particles and gamma-rays. 3
4. 4. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1.1 Alpha particle ()   An alpha particle consists of two protons and two neutrons. It is identical to a helium nucleus and its symbol is 4 4 2 He OR 2 It is positively charged particle and its value is +2e with mass of 4.002603 u. When a nucleus undergoes alpha decay it loses four nucleons, two of which are protons, thus the reaction can be represented by general equation below: α     A 4 Y  4 He  2 Z 2 (Parent) (Daughter) ( particle) Examples of  decay : A ZX 218 Po214Pb 4 He  Q 84 82 2 230 Th226Ra  4 He  Q 90 88 2 Q 226 Ra 222Rn  4 He  Q 88 86 2 238 U234Th 4 He  Q 92 90 2 4
5. 5. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1.2 Beta particle ()   Beta particles are electrons or positrons (sometimes is called beta-minus and beta-plus particles). The symbols represent the beta-minus and beta-plus (positron) are shown below: Beta-minus 0 (electron) : 1    e OR β  Beta-plus (positron) : 0 OR 1e β  Beta-minus particle is negatively charged of 1e and its mass equals to the mass of an electron. Beta-plus (positron) is positively charged of +1e (antiparticle of electron) and it has the same mass as the electron. In beta-minus decay, an electron is emitted, thus the mass number does not charge but the charge of the parent nucleus increases by one as shown below:  A Y  0 e  1 Z 1 (Parent) (Daughter) ( particle) A ZX Q 5
6. 6. DR.ATAR @ UiTM.NS  PHY310 RADIOACTIVITY Examples of  minus decay: 234 Th234Pa 0 e  Q 90 91 1 234 Pa234U 0 e  Q 91 92 1 214 Bi214Po 0 e  Q 83 84 1  In beta-plus decay, a positron is emitted, this time the charge of the parent nucleus decreases by one as shown below: A ZX  (Parent)  For example of A Y  0e  1 Z 1 (Daughter) (Positron) Q  plus decay is 1 1 p0 n  0 e  v  Q 1 1 Neutrino is uncharged particle with negligible mass. 6
7. 7. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1.3 Gamma ray ()       Gamma rays are high energy photons (electromagnetic radiation). Emission of gamma ray does not change the parent nucleus into a different nuclide, since neither the charge nor the nucleon number is changed. A gamma ray photon is emitted when a nucleus in an excited state makes a transition to a ground state. Examples of  decay are : 218 Po 214Pb 4 He  γ 84 82 2 234 Pa 234U 0 e  γ 91 92 1 208  Ti 208Ti  γ 81 81 Gamma ray It is uncharged (neutral) ray and zero mass. The differ between gamma-rays and x-rays of the same wavelength only in the manner in which they are produced; gamma-rays are a result of nuclear processes, whereas x7 rays originate outside the nucleus.
8. 8. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1.4 Comparison of the properties between alpha particle, beta particle and gamma ray.  Table 8.1 shows the comparison between the radioactive radiations. Alpha Beta Gamma 1e OR +1e 0 (uncharged) Charge +2e Deflection by electric and magnetic fields Yes Yes No Strong Moderate Weak Penetration power Weak Moderate Strong Ability to affect a photographic plate Yes Yes Yes Yes Yes Yes Ionization power Ability to produce Table 8.1 fluorescence 8
9. 9. DR.ATAR @ UiTM.NS  PHY310 RADIOACTIVITY Figures 8.1 and 8.2 show a deflection of ,  and  in electric and magnetic fields.         α γ    β      E  Figure 8.1  B β γ α Radioactive source Figure 8.2 9
10. 10. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1.5 Decay constant ()  Law of radioactive decay states:  dN   is directly For a radioactive source, the decay rate    dt  proportional to the number of radioactive nuclei N remaining in the source. i.e.  dN  Negative sign means the number of   N  dt  remaining nuclei decreases with time dN  N dt  (8.1) Decay constant Rearranging the eq. (8.1): dN    dt N decay rate  number of remaining radioactiv e nuclei Hence the decay constant is defined as the probability that a radioactive nucleus will decay in one second. Its unit is s1. 10
11. 11. DR.ATAR @ UiTM.NS   PHY310 RADIOACTIVITY The decay constant is a characteristic of the radioactive nuclei. Rearrange the eq. (8.1), we get dN (8.2)  dt N At time t=0, N=N0 (initial number of radioactive nuclei in the sample) and after a time t, the number of remaining nuclei is N. Integration of the eq. (8.2) from t=0 to time t : N dN t   dt N0 N 0  ln N N N    t 0 t 0 N ln   λt N0 N  N 0e  λt (8.3) Exponential law of radioactive decay 11
12. 12. DR.ATAR @ UiTM.NS  PHY310 RADIOACTIVITY From the eq. (8.3), thus the graph of N, the number of remaining radioactive nuclei in a sample, against the time t is shown in Figure 8.3. N Stimulation 8.1 N0 Note: N  N 0e N0 2 N0 8 N0 4 N0 16  t From the graph (decay curve), the life of any radioactive nuclide is infinity, therefore to talk about the life of radioactive nuclide, we refer to its half-life. T1/ 2 : half  life 0 T1/ 2 2T1/ 2 3T1/ 2 4T1/ 2 5T1/ 2 Figure 8.3 time , t 12
13. 13. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1.6 Half-life (T1/2)   is defined as the time taken for a sample of radioactive nuclides disintegrate to half of the initial number of nuclei. From the eq. (8.3), N  N 0 e  t and the definition of half-life, t  T1/ 2 ; N  N 0 , thus 2 1 N0  T1 / 2  e T1 / 2  N 0e 2 2 T1 / 2 when Half-life  2e ln 2  ln eT1 / 2 ln 2 0.693 T1/ 2   λ λ (8.4) The half-life of any given radioactive nuclide is constant, it does not depend on the number of remaining nuclei. 13
14. 14. DR.ATAR @ UiTM.NS   PHY310 RADIOACTIVITY The units of the half-life are second (s), minute (min), hour (hr), day (d) and year (y). Its unit depend on the unit of decay constant. Table 8.2 shows the value of half-life for several isotopes. Isotope Half-life 238 92 U 226 88 Ra 4.5  109 years 210 884 Po 234 90Th 222 86 Rn 138 days 214 83 Bi 20 minutes 1.6  103 years 24 days 3.8 days Table 8.2 14
15. 15. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.1.7 Activity of radioactive sample (A)  dN    of a radioactive sample.  dt   is defined as the decay rate  Its unit is number of decays per second. Other units for activity are curie (Ci) and becquerel (Bq) – S.I. unit. Unit conversion:   1 Ci  3.7 1010 decays per second 1 Bq  1 decay per second  Relation between activity (A) of radioactive sample and time t :  dN From the law of radioactive decay :  N dN dt and definition of activity : A  dt 15
16. 16. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY  Thus A  N  and N  N 0e A   N 0e t  t    N 0 e t and A0  N 0 A  A0 e Activity at time t  λt (8.5) Activity at time, t =0 16
17. 17. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Example 1 : A radioactive nuclide A disintegrates into a stable nuclide B. The half-life of A is 5.0 days. If the initial number of nuclide A is 1.01020, calculate the number of nuclide B after 20 days. 20 Solution : T1/ 2  5.0 days; N 0  1.010 ; t  20 days A BQ The decay constant is given by ln 2  T1/ 2 ln 2  5.0  0.139 days1 The number of remaining nuclide A is N  N 0e t   N  1.0 1020 e 0.13920  6.2 1018 nuclei The number of nuclide A that have decayed is  1.0 1020  6.2 1018  9.381019 nuclei Therefore the number of nuclide B formed is 9.38  1019 nuclei 17
18. 18. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Example 2 : a. Radioactive decay is a random and spontaneous nuclear reaction. Explain the terms random and spontaneous. b. 80% of a radioactive substance decays in 4.0 days. Determine i. the decay constant, ii. the half-life of the substance. Solution : a. Random means that the time of decay for each nucleus cannot be predicted. The probability of decay for each nucleus is the same. Spontaneous means it happen by itself without external stimuli. The decay is not affected by the physical conditions and chemical changes. 18
19. 19. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Solution : b. At time t  4.0 days, The number of remaining nuclei is  80  N  N0   N0   100   0.2 N0 nuclei i. By applying the exponential law of radioactive decay, thus the decay constant is N  N 0e t 0.2 N 0  N 0e  4.0  0.2  e  4.0  ln 0.2  ln e  4.0  ln 0.2   4.0ln e 1   0.402 day ii. The half-life of the substance is T1/ 2  ln 2  T1/ 2 T1/ 2 ln 2  0.402  1.72 days 19
20. 20. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Example 3 : Phosphorus-32 is a beta emitter with a decay constant of 5.6  107 s1. For a particular application, the phosphorus-32 emits 4.0  107 beta particles every second. Determine a. the half-life of the phosphorus-32, b. the mass of pure phosphorus-32 will give this decay rate. (Given the Avogadro constant, NA =6.02  1023 mol1) Solution : dN   5.6 10 s ;  4.0  107 s 1 dt 7 1 a. The half-life of the phosphorus-32 is given by T1/ 2  T1/ 2 ln 2  ln 2  5.6  107  1.24 10 6 s 20
21. 21. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Solution :  dN  5.6  10 s ;  4.0  107 s 1 dt 7 1 b. By using the radioactive decay law, thus dN   N 0 dt  4.0 107   5.6 107 N 0   N 0  7.14 1013 nuclei 6.02  1023 nuclei of P-32 has a mass of 32 g  7.14  1013  7.14  1013 nuclei of P-32 has a mass of    6.02  1023 32    3.80 10 9 g 21
22. 22. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Example 4 : A thorium-228 isotope which has a half-life of 1.913 years decays by emitting alpha particle into radium-224 nucleus. Calculate a. the decay constant. b. the mass of thorium-228 required to decay with activity of 12.0 Ci. c. the number of alpha particles per second for the decay of 8.0 g thorium-228. (Given the Avogadro constant, NA =6.02  1023 mol1) Solution : T1/ 2  1.913 y  1.913 365 24  60  60  6.03107 s   a. The decay constant is given by T1/ 2  ln 2  6.03 10  7 ln 2    1.15  10 8 s 1 22
23. 23. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Solution : b. By using the unit conversion ( Ci decay/second ), 1 Ci  3.7 1010 decays per second the activity is Since 12.0 3.7 1010 A  12.0 Ci   4.44 1011 decays/s A  N then A  4.44 1011  N  N   1.15 10 8 19  3.86 10 nuclei If 6.02  1023 nuclei of Th-228 has a mass of 228 g thus  3.86 1019  3.86  1019 nuclei of Th-228 has a mass of   6.02 10 23 228    2  1.46 10 g 23
24. 24. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Solution : c. If 228 g of Th-228 contains of 6.02  1023 nuclei thus 8.0 g of Th-228 contains of   15.0  6.02 1023    228   N  3.96 1022 nuclei Therefore the number of emitted alpha particles per second is given by Ignored it. dN A dt  N    1.15 108  3.96 1022 A  4.55 1014 α  particles/ second 24
25. 25. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Learning Outcome: 8.2 Radioisotope as tracers (1 hour) At the end of this chapter, students should be able to:  Explain the application of radioisotopes as tracers. 25
26. 26. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.2 Radioisotope as tracers 8.2.1 Radioisotope    is defined as an isotope of an element that is radioactive. It is produced in a nuclear reactor, where stable nuclei are bombarded by high speed neutrons until they become radioactive nuclei. Examples of radioisotopes: 31 (Radio phosphorus) a. P 1n32 P    Q 15 0 15 P S e  Q 23 1 24 11 Na  0 n  11Na    Q 32 15 b. 32 16 24 11 1 28 28 13 (Radio sodium) Na Mg e  Q c. 13 A l 0 n13 Al   27 0 1 24 12 Q 0 1 (Radio aluminum) 28 Al14 Si  0 e  Q 1 26
27. 27. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 8.2.2 Radioisotope as tracers     Since radioisotope has the same chemical properties as the stable isotopes then they can be used to trace the path made by the stable isotopes. Its method :  A small amount of suitable radioisotope is either swallowed by the patient or injected into the body of the patient.  After a while certain part of the body will have absorbed either a normal amount, or an amount which is larger than normal or less than normal of the radioisotope. A detector (such as Geiger counter ,gamma camera, etc..) then measures the count rate at the part of the body concerned. It is used to investigate organs in human body such as kidney, thyroid gland, heart, brain, and etc.. It also used to monitor the blood flow and measure the blood 27 volume.
28. 28. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Example 5 : A small volume of a solution which contained a radioactive isotope of sodium had an activity of 12000 disintegrations per minute when it was injected into the bloodstream of a patient. After 30 hours the activity of 1.0 cm3 of the blood was found to be 0.50 disintegrations per minute. If the half-life of the sodium isotope is taken as 8 hours, estimate the volume of blood in the patient. Solution : T1/ 2  15 h; A0  12000min1 ; t  30 h The decay constant of the sodium isotope is T1/ 2  ln 2  15  ln 2    4.62  10 2 h 1 The activity of sodium after 30 h is given by A  A0e t  4.62102 30  12000e A  3000min1 28
29. 29. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 1 Solution : T1/ 2  15 h; A0  12000min ; t  30 h In the dilution tracing method, the activity of the sample, A is proportional to the volume of the sample present, V. then A V A1  kV1 and A2  kV2 final initial thus the ratio of activities is given by A1 V1  A2 V2 (8.6) Therefore the volume of the blood is 0.5 1  3000 V2 V2  6000 cm3 29
30. 30. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY 31.2.3 Other uses of radioisotope In medicine  To destroy cancer cells by gamma-ray from a high-activity source of Co-60.  To treat deep-lying tumors by planting radium-226 or caesium137 inside the body close to the tumor. In agriculture  To enable scientists to formulate fertilizers that will increase the production of food.  To develop new strains of food crops that are resistant to diseases, give high yield and are of high quality.  To increase the time for food preservation. 30
31. 31. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY In industry  To measure the wear and tear of machine part and the effectiveness of lubricants.  To detect flaws in underground pipes e.g. pipes use to carry natural petroleum gas.  To monitor the thickness of metal sheet during manufacture by passing it between gamma-ray and a suitable detector. In archaeology and geology  To estimate the age of an archaeological object found by referring to carbon-14 dating.  To estimate the geological age of a rock by referring to potassium-40 dating. 31
32. 32. DR.ATAR @ UiTM.NS Example 6 : PHY310 RADIOACTIVITY   Radioactive iodine isotope 131I of half-life 8.0 days is used for 53 the treatment of thyroid gland cancer. A certain sample is required to have an activity of 8.0  105 Bq at the time it is injected into the patient. a. Calculate the mass of the iodine-131 present in the sample to produce the required activity. b. If it takes 24 hours to deliver the sample to the hospital, what should be the initial mass of the sample? c. What is the activity of the sample after 24 hours in the body of the patient? (Given the Avogadro constant, NA =6.02  1023 mol1) 32
33. 33. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Solution : T1/ 2  8.024  60  60  6.91105 s; A0  8.0 105 Bq The decay constant of the iodine isotope is T1/ 2  ln 2  6.9110  5 ln 2    1.00 106 s 1 a. From the relation between the decay rate and activity,  dN  A0     dt  0 A0  N 0 8.0 105  1.00 106 N0 N0  8.0 1011 nuclei If 6.02  1023 nuclei of I-131 has a mass of 131 g thus  8.0 1011  8.0  1011 nuclei of I-131 has a mass of   6.02 10 23 131    10  1.74 10 g 33
34. 34. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY T1/ 2  8.024  60  60  6.91105 s; A0  8.0 105 Bq 4 b. Given t  24 hr  24  3600  8.64 10 s 10 Let N : mass of I-131 after 24 hours  1.74 10 g Solution : N0 : initial mass of I-131 By applying the exponential law of radioactive decay, thus N  N 0e t   1.74 10  N0e  10 1.00106 8.64104  N0  1.74 10 e 10 N 0  1.90 10 g  1.00106 8.64104 10 c. Given t  24 hr  24  3600  8.64 10 The activity of the sample is A  A0e  t   4 A  8.0 10 e  A  7.34 105 Bq 5 s   1.00106 8.64104  34
35. 35. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Example 7 : An archeologist on a dig finds a fragment of an ancient basket woven from grass. Later, it is determined that the carbon-14 content of the grass in the basket is 9.25% that of an equal carbon sample from the present day grass. If the half-life of the carbon-14 is 5730 years, determine the age of the basket. Solution :  9.25  2 N   N 0  9.25  10 N 0 ;T1/2  5730 years  100  The decay constant of carbon-14 is T1/ 2  ln 2 5730     1.21 10 4 y 1 The age of the basket is given by N  N 0e  t ln 2 9.25 10 N 0  N0e     1.21104 t 2    ln 9.25102   1.21104 t ln e t  19674 years 35
36. 36. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Exercise 8.1 : Given NA =6.021023 mol1 1. Living wood takes in radioactive carbon-14 from the atmosphere during the process of photosynthesis, the ratio of carbon-14 to carbon-12 atoms being 1.25 to 1012. When the wood dies the carbon-14 decays, its half-life being 5600 years. 4 g of carbon from a piece of dead wood gave a total count rate of 20.0 disintegrations per minute. Determine the age of the piece of wood. ANS. : 8754 years 2. A drug prepared for a patient is tagged with Tc-99 which has a half-life of 6.05 h. a. What is the decay constant of this isotope? b. How many Tc-99 nuclei are required to give an activity of 1.50 Ci? c. If the drug of activity in (b) is injected into the patient 2.05 h after it is prepared, determine the drug’s activity. (Physics, 3rd edition, James S. Walker, Q27&28, p.1107) ANS. : 0.115 h1; 1.7109 nuclei; 1.19 Ci 36
37. 37. DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY Next Chapter… CHAPTER 9 : Nuclear Reactor 37