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## RADIOACTIVITY

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**RADIOACTIVITY**Chapter 28**What is the RADIOACTIVE?**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.**RADIOACTIVITY**Nuclear Radiation Radioactive Decay The Use of Radioisotope ALPHA ( @ He) Law of Radioactive Decay • Radioisotope as • Tracers • Sterilization • Thickness Gauge • Carbon – 14 Dating BETA ( @ ) Decay Constant () Half Time ( ) GAMMA () EXAMPLES QUESTIONS EXAMPLES QUESTIONS**28.1 RADIOACTIVE DECAY**• THE OBJECTIVES: • At the end of this chapter, students should be able to: • Explain α, β+, -andγdecays. • State decay law and use . • Define and determine activity, A and decay constant, . • Derive and use or . • Define and use half-life .**EXPLAIN**, -, +& DECAYS**RADIOACTIVITY**• Is a spontaneous disintegration process of heavy unstable elements accompanied by the emission of alpha particle (), beta particle (), or gamma rays ().**RADIOACTIVE DECAY**• Is a spontaneous reaction that is unplanned, cannot be predicted and independent of physical conditions (such as pressure, temperature) and chemical changes. • It is a 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 DECAY ()**• In decay, the nucleus of heavy radioactive element emits an -particle. • An -particle is a nucleus consists of two protons and two neutrons. • It is positively charged particle and its value is +2e with mass of 4.001502 u. • It is produced when a heavy nucleus (Z > 82) decays. • Alpha particles can penetrate a sheet of paper.**(Daughter)**( particle) (Parent) • When a nucleus undergoes alpha decay it loses 4 • nucleons (2 protons and 2 neutrons). α particle parent daughter • The reaction can be represented by general • equation below :**general equation for -decay:**where X = parent nucleus Y = daughter nucleus Q = energy released or -rays • -decay is followed by the emission of gamma rays for the daughter to be stable • Eamples; parent Daughter particle**The energy released appears in the form of kinetic energy in**the daughter nucleus and the alpha particle which is given by If Q < 0, decay could not occur**BETA DECAY ()**• In decay, the nucleus emits a -particle (+ and -) that has high velocity (v ~ c). • It has the same mass as electron or 0.000549 u. Negatron decay, β- • Negatron (-or ) decay will happen when the number of neutrons are more than the number of protons in a nucleus. • Also called as negatron or electron. • Symbol; β- or • It is produced when one of the neutronsin the parent nucleusdecays into a proton, an electron and anantineutrino.**Massless, neutral**• General equation of negatron decay process: • where is called ‘antineutrino’ (an elementary particle that exist to account the missing energy in negatron decay) parent Daughter particle antineutrino**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 : • Examples of minus decay : (Daughter) ( particle) (Daughter) ( particle) (Parent) (Parent)**Positron decay, β+**Positron (+ or ) decay will happen when the number of protons is more than the number of neutrons in a nucleus. Also called as positron or antielectron. Symbol; β+ or It is produced when one of the protonsin the parent nucleus decays into a neutron, a positronand a neutrino.**Massless, neutral**• General equation of negatron decay process: • where is called ‘neutrino’ (an elementary particle that exist to account the missing energy in positron decay) parent Daughter particle neutrino**In beta-plus decay, a positron is emitted, this time the**charge of the parent nucleus decreases by one as shown below : • Example of plus decay : (Parent) (Parent) (Daughter) (Daughter) (Positron) (Positron)**GAMMA DECAY ()**• In -decay, a photon (-ray) is emitted when the excited nucleus changes from a higher level energy state to a lower level. The wavelengths of the electromagnetic radiation are shorter than 10-10 m. • Gamma rays photon are emitted when an excited nucleus in an excited state makes a transition to a ground state. This will happen when the nucleus decays into alpha or beta particles. • thus gamma-ray emission often associates with other type of decays.**General equation for gamma decay:**• where X* = excited nucleus X = stable nucleus The asterisk (*) indicates that the nucleus is in an excited state. • There is no change in the proton number and the mass number of the nucleus • Examples of -decay; Gamma ray**-ray 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.**Comparison of the properties of the alpha particle, beta**particle and gamma ray. Table 28.1 shows the comparison between the radioactive radiations Yes Yes No Strong Moderate Weak Weak Moderate Strong Yes Yes Yes Yes Yes Yes**Comparison of the properties of the alpha particle, beta**particle and gamma ray.**Comparison of the properties of the alpha particle, beta**particle and gamma ray.**Example 28.1:**• Write equations to represent the following radioactive decay. • decays by emitting an alpha-particle and a gamma photon. • decays by beta-emission. • decays by positron-emission. • You may use X, Y and Z to represent the daughter nuclides. • Solution 28.1: • a) • b) • c)**Example 28.2:**decays through a series of transformations to a final stable nuclide. The particles emitted in the first five successive transformations are alpha-particle, negatron-particle, electron-particle, alpha-particle and alpha-particle. Write an equation to represent each sequence of the nuclear transformations. You may use A, B, C, D and E to represent the ‘daughter’ nuclides for the successive transformations. Solution 28.2:**Example 28.3:**• decays to Po according to the equation • where is a -particle and is a -ray photon. What is the proton number (atomic number) of Po? • 82 B. 84 C. 207 D. 212 • Example 28.4 • Complete the following radioactive decays and identify the radiations emitted. • a) • b) • c)**QUESTION 1:**Determine the energy released when a Uranium U decays by emitting an -particle to form a Thorium nucleus = Th. (mass of U = 238.0508 u; mass of Th = 234.0436 u; mass of He = 4.0026 u; 1 u = 934 MeV). (4.3 MeV) QUESTION 2: Find the energy released during a -decay in which a Thorium nucleus Th is converted to a protactinium nucleus Pa. (mass of Th = 234.0436 u; mass of Pa = 234.04330 u; 1 u = 934 MeV). (0.27 MeV) QUESTION 3 Polonium Po undergoes an -decay to produce a daughter nucleus that itself undergoes of -decay. What is the atomic number and mass number of the final nuclide? 238 92 234 90 238 92 234 90 4 2 234 90 234 91 234 91 234 90 216 84**State Decay Law & Use**Define & Determine Activity, A & Decay Constant, **DECAY CONSTANT ()**• Decay constant, • Decay law states that the rate of disintegration of a given nuclide at any time (rate of decay) is directly proportional to the number of nuclei N present at that time. OR • For a radioactive source, the decay rate is directly proportional to the number of radioactive nuclei N remaining in the source.**Negative signmeans the number of nuclei present decreases**with time Decay rate known as Activity, A • Unit is the Becquerel, (Bq) • 1 Bq is the rate of decay of 1 nucleus per second = the activity = the decays per second = the disintegrations per second by the radioactive nucleus (1 Bq = 1 decay/s) (1 Ci (curie) =3.7 × 1010 Bq) N = the number of nuclei remain (present).**λ = decay constant**• Hence, theDecay constantof a nuclide is the probability that a radioactive atom will decay in one second. • Its unit is s-1. • It has different values for different nuclides. • Decay constant is the characteristic of the • radioactive nuclide. • The larger the decay constant, the greater is the • rate of decay.**Derive & Use**N = Noe-t or A = Aoe-t**DERIVATION**• From the equation , • At time t=0, N=N0 (initial number of radioactive nuclei in the sample) and after a time t, the number of radioactive nuclei present is N. • By integrating the equation from t = 0 to time t : Exponential law of radioactive decay**Graph of N (number of remaining nucleus) versus t (decay**time) N No N = Noe-t No / 2 T½ = Half-life No / 4 No / 8 time, t 0 T½ 2T½ 3T½ 4T½**From the law of radioactive decay,**and definition of Activity, Thus, and and Activity at time, t =0 Activity at time t**Define & Use**Half-life,**Half-life,**• time required for the number of radioactive nuclei to decrease to half of the original number of nuclei • At t = and N = N0 / 2**The half-life of any given radioactive nuclide is constant,**it does not depend on the number of nuclei present. • The units of the half-life are second (s), minute (min), hour (hr), day and year (yr). • Its depend on the unit of the decay constant. Table 28.2 shows the value of half-life for several isotopes.**Example 28.5**A sample of of mass 4.0 × 10-12 kg emits 4.2 × 107-particles per second. What is the decay constant of ? Solution 28.5: Mass of 1 mol of is 0.032 kg. Hence 0.032 kg contains 6.02 × 1023 atoms.**Example 28.6:**Initially, a radioactive sample contains 1.0 106 of radioactive nucleus. Half-life of the sample is T½. Find the number of nucleus that still remains after 0.5 T½.**Example 28.7:**Thorium-234 has = 24 days. Initial activity of this particular isotope source is 10 Ci. a) How much is the activity of this source after 72 days? b) How long does it take for the activity to become 2.5 Ci?**Example 28.8:**• The activity of a sample of Radon-222 contains 3.0 × 107 radon atoms is 120 Bq. The half-life of Radon-222 is 3.8 days. • What is the decay constant of Radon-222? • Calculate the number of Radon-222 atoms in the sample. • How many atoms of Radon-222 remain in the sample when the activity is 40 Bq? • How many Radon-222 atoms present after 19 days? • Find the activity of the Radon after 19 days. • Solution 28.8: