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CHAPTER 28: Radioactivity (2 Hours)

is defined as the spontaneous disintegration of certain atomic nuclei accompanied by the emission of alpha particles, beta particles or gamma radiation. CHAPTER 28: Radioactivity (2 Hours). Learning Outcome:. 28.1 Radioactive decay (1 1/2 hours).

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CHAPTER 28: Radioactivity (2 Hours)

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  1. is defined as the spontaneous disintegration of certain atomic nuclei accompanied by the emission of alpha particles, beta particles or gamma radiation. CHAPTER 28: Radioactivity(2 Hours)

  2. Learning Outcome: 28.1 Radioactive decay (1 1/2 hours) At the end of this chapter, students should be able to: • Explainα, β+, βˉ and γ decays. • Statedecay law and use • Defineactivity, A and decay constant, . • Derive and use • Definehalf-life and use OR

  3. 28.1 Radioactive decay Radioactivity is a phenomenon in which an unstable nuclei undergoes spontaneous decay as a result of which a new nucleus is formed and energy in the form of radiation is released • 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.

  4. 28.1.1 Alpha particle () • An alpha particle consists of two protons and two neutrons. • It is identical to a helium nucleus and its symbol is • 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: OR ( particle) (Parent) (Daughter) • Alpha particles can penetrate a sheet of paper.

  5. α particle parent Examples of  decay : daughter

  6. 28.1.2 Beta particle (β) • Two types : • a) Beta minus , β- • b) Beta plus , β+ • A beta particle has the same mass and charge as anelectron. • Beta particles can penetrate a few mm of Al and their velocity is high (v ~ c).

  7. Beta minus (β)-negatively charge. • Also called as negatron or electron. • Symbol; • β- or • It is produced when one of the neutrons in the parent nucleus decays into a proton, an electron and an antineutrino. massless, neutral

  8. In beta-minus decay, an electron is emitted, thus the mass number does not charge but the chargeof the parent nucleus increases by one as shownbelow : ( particle) (Parent) (Daughter) • Examples of  minus decay :

  9. Beta plus (β+)- positively charge. • Also called as positron or antielectron. • Symbol; • β+ or • It is produced when one of the protons in the parent nucleus decays into a neutron, a positron and • a neutrino. massless,neutral

  10. In beta-plus decay, a positron is emitted, this time the charge of the parent nucleus decreases by one as shown below : (Positron) (Daughter) (Parent) • Example of  plus decay :

  11. 28.1.3 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 : • 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 x-rays originate outside the nucleus. Gamma ray

  12. 28.1.4 Comparison of the properties between alpha particle, beta particle and gamma ray. • Table 28.1 shows the comparison between the radioactive radiations. 1e OR +1e 0 (uncharged) +2e Yes Yes No Strong Moderate Weak Weak Moderate Strong Yes Yes Yes Yes Yes Yes Table 28.1

  13. (28.1) 28.1.5 Decay constant () • Law of radioactive decay states: For a radioactive source, the decay rate is directly proportional to the number of radioactive nuclei N remaining in the source. i.e. • Rearranging the eq. (28.1): Hence the decay constant is defined as the probability that a radioactive nucleus will decay in one second. Its unit is s1. Negative sign means the number of remaining nuclei decreases with time Decay constant

  14. (28.2) (28.3) • The decay constant is a characteristic of the radioactive nuclei. • Rearrange the eq. (28.1), we get 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. (28.2) from t=0 to time t : Exponential law of radioactive decay

  15. The number of nuclei N as function of time t Half-life is the time required for the number of radioactive nuclei to decrease to half the original number (No)

  16. From Hence

  17. 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 28.2 shows the value of half-life for several isotopes. Table 28.2

  18. 28.1.6 Activity of radioactive sample (A) • is defined as the decay rate of a radioactive sample. • Its unit is number of decays per second. • Other units for activity are curie (Ci) and becquerel (Bq) – S.I. unit. • Unit conversion: • Relation between activity (A) of radioactive sample and time t : • From the law of radioactive decay : and definition of activity :

  19. (28.4) and • Thus and Activity at time, t=0 Activity at time t

  20. Example 28.1.1 : 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. Solution : The decay constant is given by The number of remaining nuclide A is The number of nuclide A that have decayed is Therefore the number of nuclide B formed is 20

  21. Example 28.1.2 : 80% of a radioactive substance decays in 4.0 days. Determine i. the decay constant, ii. the half-life of the substance. 21

  22. Solution : At time The number of remaining nuclei is i. By applying the exponential law of radioactive decay, thus the decay constant is ii. The half-life of the substance is 22

  23. Example 28.1.3 : 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 15.0 g thorium-228. (Given the Avogadro constant, NA =6.02  1023 mol1) Solution : a. The decay constant is given by 23

  24. Solution : b. By using the unit conversion ( Cidecay/second ), the activity is Since then If 6.02  1023 nuclei of Th-228 has a mass of 228 g thus 3.86  1019 nuclei of Th-228 has a mass of 24

  25. Solution : c. If 228 g of Th-228 contains of 6.02  1023 nuclei thus 15.0 g of Th-228 contains of Therefore the number of emitted alpha particles per second is given by Ignored it. 25

  26. Example 28.1.4 : A sample of radioactive material has an activity of 9.00 x 1012 Bq. The material has a half-life of 80.0 s. How long will it take for the activity to fall to 2.00 x 1012 Bq ? Solution

  27. Example 28.1.5 : N = 25% , No = 100% t =34.6 min (1.72 h)

  28. Learning Outcome: 28.2 Radioisotope as tracers (1/2 hour) At the end of this chapter, students should be able to: • Explainthe application of radioisotopes as tracers.

  29. 28.2 Radioisotope as tracers • Radioisotope (unstable isotope) is an isotope • which is exhibits radioactivity (known as radioactive • isotope). a) Blood volume • The volume of blood in the bloodstream, V2 can • be determined by using dilution method as given • below.

  30. where A1 = activity of the blood drawn from the patient A2 = activity of the blood in the bloodstream V1 = volume of the blood drawn from the patient V2 = volume of blood in the bloodstream of the patient

  31. Example 28.2.1 A small volume of a solution which contains a radioactive isotope of sodium Na-24 has an activity of 1.5 x 104 Bq. The solution is injected into the bloodstream of a patient. The half-life of the sodium isotope is 15 hours. After 30 hours, the activity of 1.0 cm3 of blood is measured and found to be 0.50 Bq. Estimate the volume of blood in the patient.

  32. Solution 28.2.1

  33. b) Detecting leaks in underground pipes. • The exact position of an underground pipe can be located if a small quantity of radioactive liquid is added to the liquid being carried by the pipe. • Geiger counter can be used to detect the leaks. • Any leaks would be detected by an increase in radiation reading . • The soil close to the leak becomes radioactive. • The short-lived radioisotope is used to avoid from the permanent contamination of the soil.

  34. c) Detecting brain tumors. • Technitium-99 is a gamma emitter (half-life 6hours) and is used as a medical tracer. • When injected into the blood stream, 99 Tc will not be absorbed by the brain, because of the blood-brain barrier. • However, tumors do not have this barrier. • Thus, brain tumors readily absorb the 99 Tc. • These tumors then show as gamma-ray • emitters on detectors external to the body. • The short-lived radioisotope is used so that it • can quickly eliminate from the body.

  35. Good luck For 2nd semester examination

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