Study Guide 7: Ionizing Radiation

Transcription

1 Study Guide 7: Ionizing Radiation Text: Chapter 6, sections 1-11 (more than described in Study Guide), plus text 2.5 and lab manual section 7A-1 (on inverse-square law). Upcoming quizzes: Quiz 4 (final day, Friday, March 14, 2008) you should know: Material from SG 5 and SG 6. Quiz 5 (final day, Friday, March 28, 2008) you should know: Material from Computer lab exp. 7 (Radiation) and exp. 8 (Electricity): in particular From semilog and log-log graphs of provided data (exp. 7); 1. How to calculate the linear attenuation coefficient. 2. Determine half-time T 1/2. 3. Study the exponent in power-law decay from a point radioactive source. 1

2 4. Apparatus used in exp. 8 (bottom of page 8-1 of lab manual) and how to arrange this to study parallel and series circuits. 5. Using 4 to calculate current and resistances. NUCLEAR EQUATIONS Nuclear equations are derived from conservation laws: mass/energy charge angular momentum (spin) Radioactive decay is how UNSTABLE NUCLEI can increase their binding energy (i.e., how strongly nuclear particles protons and neutrons stick to each other). 2

3 e.g. The α-decay of Radium into Radon Ra Rn + He + γ A gamma ray, sometimes required for energy conservation Radium (Parent) Radon (Daughter) He-4 nucleus (also called an α particle) The numbers in the above equation have important meanings: The mass number A = (# of protons + # of neutrons). Defines the ISOTOPE Ra Rn + α + γ The CHEMICAL ELEMENT is defined by the atomic number Z = (# of protons) In general A Z X = # of protons + # of neutrons # of protrons X 3

4 Little things you should know 1. THE MASS NUMBER MASS OF 1.0 MOLE ( N A = Avogadro s number) OF THE ISOTOPE IN GRAMS. 2. Symbol X = Fe, H, Au, etc., is redundant (already determined by the atomic number Z). 3. Alternative symbols are occasionally used, e.g., α (instead of He) for alpha particles. 4. A nucleus with extra energy (excited) is designated as follows 222 Rn * 86 This extra energy can be transferred to emitted radiation as kinetic energy in an α or β particle or as a γ photon. 5. Energies can be given in electron volts, 1 ev = Joules. 6. Masses can be given as REST ENERGIES using E=mc 2 law. E.g., E proton =( ) ( ) 2 939MeV. 4

5 TYPES OF RADIATION different types create different decays. we focus on 3 types 1) Alpha (α) Radiation: emission of an α particle (Helium-4 nucleus) from a larger nucleus. e.g Ra Rn + α + γ Rn Po + α + γ e.g. CASCADE, SERIES OR CHAIN Fm Cf + ( K MeV) 55 kev α = + α γ ( ) This denotes the kinetic energy of the α particle The α and γ rays emitted by an isotope have a specific energy spectrum that can be used to identify it. E.g., in the decay above, 87% of the α particles produced have MeV of kinetic energy and 1.5% have MeV. 5

6 GENERAL FORMULA FOR α DECAY A Z X A- 4Y+ 4 Z- 2 2 α + ( γ ) WHY ALPHA PARTICLES? Other nuclei can be emitted (besides 4 He) but is rare. For example, Ra 209Pb+ 14C 82 4 He emission gets rid of more energy per nucleon. All these emissions of smaller nuclei from the parent are really examples of fission because the nucleus splits into two more stable parts. 6 6

7 In the table, the numbers in the upper left corners are the atomic numbers (not mass numbers). You don t have to memorize this table! 7

8 2) Beta (β) Radiation Actually 3 different processes; electron emission, positron emission and electron capture. We only consider electron (β) emission from the nucleus. Can also occur with γ radiation (like α radiation). FOR EXAMPLE Sr Y β + ν Anti-neutrino N O (No γ) + β + ν + γ H He β + ν (No γ) 8

9 Little things you should know You ONLY have to remember that a γ MAY be produced by β and α decays. When a decay will produce a γ is beyond the scope of this course. Remember that ALL electron emissions (the only β decays we study) produce an anti-neutrino. The anti-neutrino is required to conserve momentum. Unlike α s, β s are not emitted only at specific energies but over a range from 0 to a maximum value because the energy is shared with the anti-neutrino. GENERAL FORMULA FOR β DECAY A A 0 X Y Z Z β + ν 9

10 3) Gamma (γ) Radiation: Composed of high energy photons (light) which have no rest mass, therefore parent nucleus = daughter. Often occurs in conjunction with α and β radiation (which do cause the daughter to change from the parent). For example, Excited state 60 Co 60 Ni * 0 β STEP 1 * 60Ni ν Ni + 2γ STEP 2 It is also possible for a nucleus to simply emit a gamma photon. In such cases the general form is, A * X Z A X + Z γ 10

11 Some examples: Pu 238 U 92 +? 65 0 Ni? + β + ν Na 24 Mg 12 +? RADIOACTIVITY AND HALF LIVES Over time, a parent sample will gradually turn into a daughter. In this section we will learn how to determine the rate at which a given sample emits radiation and the rate at which radioactive substances are removed from physical and biological systems. Radioactive decay is a random process. Each parent atom has the same probability, λ p, ("p" for physical) of decaying per unit time, Δt. If Δt is short compared to the half life (see below for def n of half life), the number of events that will occur (= the change in the number of parent nuclei, ΔN) is, 11

12 ΔN = Nλ p Δt (1) the "-" indicates that N decreases The parent number change, ΔN, per unit time Δt is then ΔN/Δt = Nλ p (2) For tiny changes, Eq. (1) can be expressed as, dn = - Nλ p dt. Therefore with a little integral calculus, ln N N 0 = λ t p, (3) where N 0 = initial # of parent nuclei and N = final # of parent nuclei. Rearranging Equation (3) a bit we get the: Physical Decay Law N = N 0 e λ p t (4) This equation only applies in the absence of biological decay. 12

13 1. Number of nuclei decreases exponentially over time. 2. Probability/unit time, λ p, (dimension: time -1 ) = the DECAY CONSTANT. 3. The decay constant is related to the half-life, T 1/2. The half life is the time it takes half of the sample to undergo a particular decay. When t = p T 1/2 then N = N 0 /2. Putting this in Eq. (4) gives N ( T ) N λ 0 p p 1/2 = = N 0 e 2 ' ( ) 1 ln = = λ p p T 2 1/2 13

14 Physical Half Life T p 1/2 = λ p (5) Note from the equation N = N 0 e λpt, by differentiating with respect to time t, we get dn dt -λpt = -λpn e = -λ pn 0 The quantity dn/dt is the count-rate, the number of disintegrations per unit time, which is sometimes denoted A (for activity ). We see that A also decreases exponentially with time, i.e. A = A e 0 λ t p = (6) 14

15 Sample problem U α decays with a half life of 72 years into Polonium 228. If a sample of 232 U has an initial count-rate of counts/s (or C/s), what is its count-rate after 50 years? From equation (5), λ p = 1/( p T 1/2 ) = 0.693/72 = year -1 Note, we can keep this unit for λ p if time is also expressed in units of years. From eq. (6), A = A 0 e λ p t A = ( C/s) exp( (50)) = ( C/s) exp( 0.48) = ( C/s) (0.62) = C/s BIOLOGICAL ACTIVITY Foreign materials in living organisms are expelled due to the biological activity (e.g. urination, breathing, sweating) of the organism. Biological activity = # of parent atoms ejected by biological processes/unit time. 15

16 Things to remember 1. Random. Don t know which atoms are excreted. 2. Biological decay depends on BOTH the isotope AND the ORGANISM. 3. Biological decay is governed by exponential laws. If the physical decay = 0, the # of foreign atoms N in a biological system at time t is, Biological Decay Law N = N e 0 λ b t This equation only applies in the absence of radioactive decay. The symbols have the same meaning as for the physical decay equation. The biological decay constant, λ b, replaces the λ p. 16

17 The biological half life b T 1/2 is T b 1/2 = λ b Combining Physical and Biological Activities Radioactive isotopes in living organisms are affected by both physical and biological processes. The effective decay constant, λ e, is given by, λ e = λ p + λ b In terms of half-lives, = + T T T e 1/2 p 1/2 b 1/2 17

18 The number of atoms N present at time t is ( λ λ ) t λe t p + b N = N e = N e 0 0 This equation ALWAYS applies This equation can always be used because if there is no biological decay, then λ b = 0, while if there is no physical decay then λ p = 0. Sample Problem 2: Celtic Remains? THE FACTS: 1) The natural occurrence of carbon-14 on earth (created by cosmic neutron bombardment from nitrogen in the air) is responsible for 15.3 disintegrations per minute per gram of total carbon. 2) Carbon-14 β-decays with a half-life of 5370 years. 3) Charcoal samples from ancient (sacrificial?) fire pits at Stonehenge are found to have an activity of 9.65 disintegrations per minute per gram of carbon. 18

19 THE QUESTIONS: A) What are the end products of this decay?. B) Based on these findings, how old is Stonehenge? (B) Ans. The trees died 3571 years ago. SAMPLE PROBLEM 3 (Similar to killer question 6-8 in text) A radioactive isotope has a physical half-life of 65 min. An initial measurement of it shows a count-rate of 15,500 Cs 1. Thirty-nine minutes later the sample is administered to a mouse. One hour after this, the (unfortunate) mouse has a total body count of 1025 Cs 1. What is the biological half-life of the isotope in the mouse? Ans. b T 1/2 = 25.1 min ABSORPTION OF RADIATION Absorption in living tissue (Bad!): Ionizing radiation damages molecules: can result in cell death. 19

20 Damaged DNA causes genetic mutations (AND superheros!). REDUCTION OF RADIATION EXPOSURE: 1) DISTANCE AND TIME In the absence of an absorptive barrier, activity A (C in lab manual) obeys an inverse square law: A K 2 r r = distance from source A person three times further away from the source, receives 1/9 as much exposure. This does not include absorption by air (or other material) that intervenes. 2) ABSORPTION BY MATTER Alpha, beta and gamma radiation each behave differently in the presence of barriers. Let's assume the incident 20

21 radiation has intensity I 0 or activity A 0 (text and lab manual use N). GAMMA RAY (OR X-RAY) absorption is similar to radioactive decay or absorption of light. Each γ- photon can be absorbed at any point along the path through the barrier. The probability of absorption/unit length/photon is μ (linear attenuation coefficient). For a beam of intensity, I, traversing a distance dx along its path, the change in intensity will be (compare to the expression for decay) di = - μidx With a little calculus, we see or ln I I I 0 = μx = e I 0 μx μ depends on: 1) Energy of the photon - The more energetic the photon is, the more deeply it penetrates. 21

22 2) Nature of the absorbing material. However, if we define the mass attenuation coefficient, μ m, μ = μ m ρ where ρ is the density (mass per volume), we find that μ m is largely independent of the material. It still depends on the gamma ray (or x-ray) energy. For a 100 kev x-ray, 3.0 cm of iron cuts the intensity by a factor of two. Given that human tissue is about 1/10 as dense as iron, these x-rays would penetrate about 30 cm of human tissue. As one might expect, this is about the energy of a medical x-ray. ABSORPTION OF CHARGED PARTICLES (α AND β) occurs everywhere along the path (rather than at just one point as for γ-rays). The mean distance traveled by the particles is represented by the penetration depth, defined below. The activity drops sharply beyond this distance for particle beams of a single energy. Alpha and beta rays are much less penetrative than γ-rays. 22

23 ABSORPTION OF α PARTICLES 1) High mass means very little deflection with each collision. 2) Short intense tracks in target material. 3) Intense ionization (damage) along path. 4) A 5.3 MeV alpha has a penetration depth of 30 mm in air and about 30 micrometers in water or human tissue. 5) A major hazard if adjacent to the skin (α emitter was swallowed), but can be easily shielded. ABSORPTION OF β PARTICLES 1) Low mass means beta particles are strongly deflected with each collision. 2) Leave long and tortuous (twisted) tracks in target material. 3) Mean penetration depth of β particles of energy, E > 0.6 MeV is given by: Penetration depth = [5.42 E(MeV) 1.33]/ρ where ρ = density (kg/m 3 ) 23

24 4) Note that that higher energy beams penetrate more deeply. 5) Can penetrate several meters of air (see text example 6-2) and are hazardous at a distance. Sample Problem 4: Lead Shielding: A researcher working at some distance from a γ- emitting radioisotope would be exposed to 100 times the allowable yearly limit of radiation over one hour (lethal!). What thickness of lead shielding is required to reduce the exposure to one tenth the yearly limit for an hour-long exposure at the same distance? The mass attenuation coefficient of lead is m 2 /kg for the γ-photons involved and the density of lead is kg/m 3. Ans. 5.0 cm Sample Problem 5: Beta radiation in air and with shielding: What is the penetration depth in air of β radiation from Bismuth-210? The E for 210 Bi β s is 1.17 MeV and the density of air is 1.3 kg/m 3. Would 1.0 mm 24

25 of Pb (density kg/m 3 ) stop this radiation? How about 1.0 mm of Al (density kg/m 3 )? Ans. D = 3.88 m Yes No Some final things Radiation damage at the molecular level: most radiation causes ionization of molecules, disrupting chemical bonds (such as in DNA and proteins), producing free radicals and high-speed electrons which can cause other ionizations. Qualitative only: (text Sections 6.10 and 6.11 are not very clear): Effects of radiation on an organism can be described by the following concepts: Exposure, Absorbed Dose and Equivalent Dose. (1) Exposure = amount of radiation energy which is incident per unit mass of victim. (DOES NOT DEPEND ON ABSORBER) (Ignore mysterious question 1, especially 1(c), on Self- Test IV. This mixes up exposure, absorbed dose and equivalent dose. Also skip text exercise 6-4) 25

26 (2) Absorbed Dose = amount of radiation energy which is absorbed per unit mass of victim. (DOES DEPEND ON THE PROPERTIES OF THE ABSORBER E.G., IONIZATION ENERGIES) (3) Equivalent Dose: measures damage produced by absorbed dose. Depends on the interaction of the radiation type (e.g., α) with the victim material. 26