Physics 111 Fall 2007 Radioactive Decay Problems Solutions PDF

Physics 111 Fall 2007 Radioactive Decay Problems Solutions 1. The 13 H isotope of hydrogen, which is called tritium (because it contains three nucleons), has a half-life of 12.33 yr. It can be used to measure the age of objects up to about 100 yr. It is produced in the upper atmosphere by cosm...

Physics 111 Fall 2007 Radioactive Decay Problems Solutions 1. The 13 H isotope of hydrogen, which is called tritium (because it contains three nucleons), has a half-life of 12.33 yr. It can be used to measure the age of objects up to about 100 yr. It is produced in the upper atmosphere by cosmic rays and brought to Earth by rain. As an application, determine approximately the age of a bottle of wine whose 13 H radiation is about 101 that present in new wine. Because the tritium in water is being replenished, we assume that the amount is constant until the wine is made, and then it decays. We find the number of half-lives from N (); n = 1 2 N0 ( ) , or n log 2 = log (10), which gives n = 3.32. n 0.10 = 1 2 Thus the time is ( )( ) t = nT 1 = 3.32 12.33 yr = 41 yr. 2 2. Strontium-90 is produced as a nuclear fission product of uranium in both reactors and atomic bombs. Look at its location in the periodic table to see what other elements it might be similar to chemically, and tell why you think it might be dangerous to ingest. It has too many neutrons, and it decays with a half-life of about 29 yr. How long will we have to wait for the amount of 90 38 Sr on the Earth’s surface to reach 1% of its current level, assuming no new material is scattered about? Write down the decay reaction, including the daughter nucleus. The daughter is radioactive: write down its decay. We see from the periodic chart that Sr is in the same column as calcium. If strontium is ingested, the body will treat it chemically as if it were calcium, which means it will be stored by the body in bones. We find the number of half-lives to reach a 1% level from N ( ); n = 1 2 N0 ( ) , or n log 2 = log(100), which gives n = 6.64. n 1 0.01= 2  Thus the time is ( )( t = nT 1 = 6.64 29 yr = 193 yr. 2 ) The decay reactions are Sr → 9039Y + 90 38 e + v, 0 −1 90 Y is radioactive; 39 Y → Zr + e + v , 90 39 90 40 0 −1 90 40 Zr is stable. 3. An old wooden tool is found to contain only 6.0% of 14 6C that a sample of fresh wood would. How old is the tool? 14 Since the carbon is being replenished in living trees, we assume that the amount of C is constant until the wood is cut, and then it decays. We find the number of half-lives from N ( ); n = 1 2 N0 0.060 = ( ) , or n log 2 = log (16.7 ), which gives n = 4.06. n 1 2 Thus the time is ( )( ) t = nT 1 = 4.06 5730 yr = 2.3 × 104 yr. 2 4. An amateur archeologist finds a bone that she believes to be from a dinosaur and she sends a chip off to a laboratory for 14C dating. The lab finds that the chip contains 5g of carbon and has an activity of 0.5 Bq. How old is the bone and could it be from a dinosaur? The activity when the bone chip is measured is 0.5 decays/sec. The initial activity when the animal died needs to be determined. In the bone there is found 5g of carbon. Since 1 mole of carbon contains 6.02x1023 atoms and 1 mole of carbon has a mass of 12 g, there are 2.51x1023 carbon nuclei. Further the ration of 14C/12C has remained relatively constant and has a value of 1.3x10-12. Thus the number of 14C nuclei is given as (1.3x10-12)(2.51x1023 nuclei) = 3.26x1011 14C nuclei when the animal died. The initial activity is a product of the decay constant and the number of 14C nuclei present when the animal died. The decay constant is found from the half-life of carbon (5730yrs). 0.693 0.693 1 yr λ= = × = 3.78 × 10 −12 sec −1. The initial activity is λN = t1 5730 yrs 3.2 × 10 7 sec 2 (3.78x10-12 sec-1)(3.26x1011 14C nuclei) = 1.23 Bq. To calculate the age of the bone we use  the radioactive decay law −120.5sec −1 ) t A = Ao e −λt → 0.5Bq = 1.23Bqe −(3.78×10→ ln( ) = −(3.78 × 10 −12 sec −1 )t 1.23 −12 −1 → −0.902 = −(3.78 × 10 sec )t → t = 2.39 × 10 sec = 7456 yrs. 11 Since the bone is only about 7500 years old and knowing that the dinosaurs disappeared over 65 million years ago, it is probably not the bone of a dinosaur. 5. Calculate the decay energy for the β- decay of 24Na given the following data: m(24Na) = 23.98492 u, m(24Mg) = 23.97845 u, m(24Ne) = 23.98812 u, m(β-) = 5.49x10-4 u. What is the range of the possible energies of the emitted beta particle? For the β-decay reaction of 24Na, ( 11 12 −1 ) ( Q = M 24 Na − M 24 Mg − M 0 e c 2 = 23.98492 − 23.97845 − 5.49 × 10 − 4 uc 2 × ) 931.5MeV uc 2 = 5.52MeV. 6. Show that in alpha decay from a stationary parent nuclide that the conservation of energy and momentum lead to a relation between the decay energy for the nuclear reaction and the kinetic energy gained by the alpha particle, KE, given ⎛ m( 24 He ) ⎞ by Q = KE ⎜⎜1 + ⎟⎟. What is the kinetic energy of the alpha particle emitted ⎝ m (daughter ) ⎠ in the decay of 238U? ( ) For alpha decay: Q = M parent − M daughter − M He c 2. If the parent is at rest when it decays, then from conservation of momentum, the daughter gets a recoil velocity in the direction opposite direction to the velocity of the alpha particle. Conservation of momentum gives: mα 0 = − mdaughter v daughter + mα vα → v daughter = vα. Next we apply conservation of mdaughter  energy and we find: 1 1 m parent c 2 = 2 mdaughter vdaughter + mHe vHe 2 + mdaughter c 2 + mHe c 2. 2 2 Next we can replace the velocity of the recoiling daughter atom in terms of the velocity of the alpha particle which can be measured. 2 1 ⎛ mα ⎞ 1 m parent c = mdaughter ⎜ 2 vα ⎟ + mHe vHe 2 + mdaughter c 2 + mHe c 2. 2 ⎜m ⎟ 2 ⎝ daughter ⎠ Bringing all of the rest energy terms to one side and combinging the terms involving the velocity of the alpha particle, we find that ⎡ 1 mα2 1 ⎤ (mparent c 2 − mdaughter c 2 − m ) He c 2 = Q = ⎢ + mα ⎥ vα. ⎢⎣ 2 mdaughter 2 ⎥⎦ Factoring out the mass of the alpha particle on the LHS we have : 1 ⎡ mα ⎤ Q= mα vα2 ⎢1 + ⎥, which is the desired result. 2 ⎢⎣ mdaughter ⎥⎦ Looking up the decay energy, Q = 4.28 MeV, mα = 4.0015u, mdaughter = mthorium = 233.99409u, and we find the velocity of the alpha particle to be 1.42x107 m/s. 7. Calculate the binding energies of radium-226 (m = 225.97709 u), radium-228 (m = 227.98275 u), and thorium-232 (m = 231.98864 u). The nuclear binding energy is given through: NBE = Zm p c 2 + NmN c 2 − matom c 2 ; where m p = 1.00727u and mn = 1.00867u. For radium-226: 931.5MeV NBE = [88(1.00727u ) + 138(1.00867u ) − 225.97709u ]c 2 × = 1731.8MeV. 1uc 2 NBE 1731.8MeV The = = 7.66 nucleon MeV. nucleon 226 For radium-228: 931.5MeV NBE = [88(1.00727u ) + 140(1.00867u ) − 227.98275u ]c 2 × = 1742.7 MeV. 1uc 2 NBE 1742.7 MeV The = = 7.64 nucleon MeV. nucleon 228  For thorium-232: 931.5MeV NBE = [90(1.00727u ) + 142(1.00867u ) − 231.98864u ]c 2 × = 1766.9MeV. 1uc 2 NBE 1766.9MeV The = = 7.62 nucleon MeV. nucleon 232 8. What is the activity of 1gram of radium-226? The activity is given as the product λN, where λ is the decay constant and N is the number of nuclei that decay. Given that are sample is radium-226, with a half-life of 1600 years, we can 0.693 0.693 1 yr calculate the decay constant. λ = = × = 1.35 × 10 −11 sec −1. t1 1600 yrs 3.2 × 10 sec 7 2 To calculate the number of nuclei present we use the mass given, 1g. There are 226 g of radium per mole and in 1 mole there are 6.02x1023 nuclei. Thus in 1 g there are 2.66x1021 nuclei. The activity is therefore λN = (1.35x10-11 sec-1)(2.66x1021 nuclei) = 3.6x1010 decays/sec = 3.6x1010 Bq. 9. Strontium is chemically similar to calcium and can replace calcium in bones. The radiation from 90Sr can damage bone marrow where blood cells are produced, and lead to serious health problems. How long would it take for all but 0.01% of a sample of 90Sr to decay? Here we take the initial mass M0 = 1 and the final mass is 0.01% of the initial mass. Thus Mf = 1x10-4 M0. The decay constant for 90Sr is λ = 0.024 yr-1. From the radioactive decay law: M f = M o e − λt → 1× 10 −4 M 0 = M o e − (0.024 yr )t → t = 383.4 yrs. −1 10. After the sudden release of radioactivity from the Chernobyl nuclear reactor accident in 1986, the radioactivity of milk in Poland rose to 2 000 Bq/L due to iodine-131 present in the grass eaten by dairy cattle. Radioactive iodine, with half-life 8.04 days, is particularly hazardous because the thyroid gland concentrates iodine. The Chernobyl accident caused a measurable increase in thyroid cancers among children in Belarus. (a) For comparison, find the activity of milk due to potassium. Assume that one liter of milk contains 2.00 g of potassium, of which 0.011 7% is the isotope 40 K with half-life 1.28 × 109 yr. (b) After what time interval would the activity due to iodine fall below that due to potassium? 40 (a) One liter of milk contains this many K nuclei:  ⎛ 6.02 × 10 23 nuclei mol ⎞ ⎛ 0.011 7 ⎞ N = (2.00 g )⎜ 18 ⎟ ⎜⎝ 100 ⎟⎠ = 3.60 × 10 nuclei ⎝ 39.1 g mol ⎠ ln 2 ln 2 ⎛ 1 yr ⎞ −17 −1 λ= = ⎜⎝ ⎟ = 1.72 × 10 s T1 2 1.28 × 10 yr 3.156 × 107 s ⎠ 9 ( )( R = λ N = 1.72 × 10−17 s −1 3.60 × 1018 = 61.8 Bq ) ln 2 (b) For the iodine, R = R0 e− λ t with λ= 8.04 d 1 ⎛ R ⎞ 8.04 d ⎛ 2 000 ⎞ t= ln ⎜ 0 ⎟ = ln ⎜ = 40.3 d λ ⎝ R⎠ ln 2 ⎝ 61.8 ⎟⎠ 11. A small building has become accidentally contaminated with radioactivity. The longest-lived material in the building is strontium-90. ( 90 38 Sr has an atomic mass 89.907 7 u, and its half-life is 29.1 yr. It is particularly dangerous because it substitutes for calcium in bones.) Assume that the building initially contained 5.00 kg of this substance uniformly distributed throughout the building and that the safe level is defined as less than 10.0 decays/min (to be small in comparison to background radiation). How long will the building be unsafe? m ass present 5.00 kg N0 = = = 3.35 × 10 25 nuclei ( m ass of nucleus (89.907 7 u ) 1.66 × 10−27 kg u ) ln 2 ln 2 λ= = = 2.38 × 10−2 yr −1 = 4.53 × 10−8 min −1 T1 2 29.1 yr ( )( ) R 0 = λ N 0 = 4.53 × 10−8 min −1 3.35 × 10 25 = 1.52 × 1018 counts min R 10.0 counts min = e− λ t = = 6.59 × 10−18 R0 1.52 × 1018 counts min and ( ) λ t = − ln 6.59 × 10−18 = 39.6 39.6 39.6 giving t = = = 1.66 × 103 yr. λ 2.38 × 10−2 yr −1 12. Natural uranium must be processed to produce uranium enriched in 235U for bombs and power plants. The processing yields a large quantity of nearly pure 238U as a by- product, called “depleted uranium.” Because of its high mass density, it is used in armor-piercing artillery shells. (a) Find the edge dimension of a 70.0-kg cube of 238U. The density of uranium is 18.7 × 103 kg/m3. (b) The isotope 238U has a long half-life of 4.47 × 109 yr. As soon as one nucleus decays, it begins a relatively rapid series of 14 steps that together constitute the net reaction  238 92 U → 8( 42 He ) + 6( ‐01 e ) + 206 82 Pb + 6 v + Qnet Find the net decay energy. (Refer to Table A.3.) (c) Argue that a radioactive sample with decay rate R and decay energy Q has power output ℘ = QR. (d) Consider an artillery shell with a jacket of 70.0 kg of 238U. Find its power output due to the radioactivity of the uranium and its daughters. Assume that the shell is old enough that the daughters have reached steady-state amounts. Express the power in joules per year. (e) A 17-year-old soldier of mass 70.0 kg works in an arsenal where many such artillery shells are stored. Assume that his radiation exposure is limited to absorbing 45.5 mJ per year per kilogram of body mass. Find the net rate at which he can absorb energy of radiation, in joules per year. 1 3 13 m ⎛ m⎞ ⎛ 70.0 kg ⎞ (a) V = l3 = , so l = ⎜ ⎟ =⎜ 3⎟ = 0.155 m ρ ⎝ ρ⎠ 3 ⎝ 18.7 × 10 kg m ⎠ (b) Add 92 electrons to both sides of the given nuclear reaction. Then it becomes 238 92 U atom → 8 42 He atom + 206 82 Pb atom + Qnet. Q net = ⎡ M 238 U − 8 M 4 H e − M 206 Pb ⎤ c 2 = [238.050 783 − 8 (4.002 603 ) − 205.974 449 ]u (931.5 MeV u ) ⎣ 92 2 82 ⎦ Q net = 51.7 MeV (c) If there is a single step of decay, the number of decays per time is the decay rate R and the energy released in each decay is Q. Then the energy released per time is P = QR. If there is a series of decays in steady state, the equation is still true, with Q representing the net decay energy. (d) The decay rate for all steps in the radioactive series in steady state is set by the parent uranium: ⎛ 7.00 × 10 4 g ⎞ N =⎜ ⎝ 238 g m ol ⎠ ( 23 ) 26 ⎟ 6.02 × 10 nuclei m ol = 1.77 × 10 nuclei ln 2 ln 2 1 λ= = = 1.55 × 10−10 T1 2 4.47 × 109 yr yr ⎛ 1⎞ R = λ N = ⎜ 1.55 × 10− 10 ⎝ yr ⎟⎠ ( ) 1.77 × 10 26 nuclei = 2.75 × 1016 d ecays yr , ⎛ 1⎞ so P = QR = (51.7 MeV )⎜ 2.75 × 1016 ⎝ yr ⎟⎠ ( ) 1.60 × 10−13 J MeV = 2.27 × 105 J yr. ⎛ 4.55 × 10−3 J ⎞ (e) The allowed whole-body dose is then (70.0 kg )⎜ ⎟⎠ = 3.18 J yr. ⎝ kg ⋅ yr  − 13. A sealed capsule containing the radiopharmaceutical phosphorus-32 ( 32 15 P ), an e emitter, is implanted into a patient’s tumor. The average kinetic energy of the beta particles is 700 keV. The initial activity is 5.22 MBq. Determine the energy absorbed during a 10.0-day period. Assume that the beta particles are completely absorbed within the tumor. 32 The half-life of P is 14.26 d. Thus, the decay constant is ln 2 ln 2 λ= = = 0.048 6 d −1 = 5.63 × 10−7 s −1. T1 2 14.26 d R0 5.22 × 106 d ecay s N0 = = = 9.28 × 1012 nuclei λ 5.63 × 10−7 s −1 At t = 10.0 days , the number remaining is ( ) ( N = N 0 e− λ t = 9.28 × 1012 nuclei e ) − 0.048 6 d −1 (10. 0 d ) = 5.71 × 1012 nuclei so the number of decays has been N 0 − N = 3.57 × 1012 and the energy released is ( ) ( E = 3.57 × 1012 (700 keV ) 1.60 × 10−16 J keV = 0.400 J. ) 14. To destroy a cancerous tumor, a dose of gamma radiation totaling an energy of 2.12 J is to be delivered in 30.0 days from implanted sealed capsules containing palladium- 103. Assume that this isotope has half-life 17.0 d and emits gamma rays of energy 21.0 keV, which are entirely absorbed within the tumor. (a) Find the initial activity of the set of capsules. (b) Find the total mass of radioactive palladium that these “seeds” should contain. ln 2 ln 2 The decay constant is λ = = = 0.040 8 d. The number of nuclei remaining after 30 d T1 2 17 d is N = N 0 e− λT = N 0 e−0.040 8(30) = 0.294N 0. The number decayed is N 0 − N = N 0 (1 − 0.294 ) = 0.706 N 0. Then the energy release is ⎛ 1.6 × 10−19 J ⎞ ( 2.12 J = 0.706N 0 21.0 × 103 eV ⎜ ⎝ 1 eV ) ⎟⎠ 2.12 J N0 = −15 = 8.94 × 1014 2.37 × 10 J ⎛ 1d ⎞ (a) R0 = λ N 0 = 0.040 8 d ( 8.94 × 1014 ⎜ ) ⎝ 86 400 s ⎟⎠ = 4.22 × 108 Bq ⎛ 1.66 × 10−27 kg ⎞ original sample mass = m = N original mone atom = 8.94 × 1014 (103 u )⎜ ⎟ (b) ⎝ 1u ⎠ = 1.53 × 10−10 kg = 1.53 × 10−7 g = 153 ng 15. A living specimen in equilibrium with the atmosphere contains one atom of 14C (half- life = 5 730 yr) for every 7.7 × 1011 stable carbon atoms. An archeological sample of wood (cellulose, C12H22O11) contains 21.0 mg of carbon. When the sample is placed inside a shielded beta counter with 88.0% counting efficiency, 837 counts are accumulated in one week. Assuming that the cosmic-ray flux and the Earth’s atmosphere have not changed appreciably since the sample was formed, find the age of the sample. ⎛ 0.021 0 g ⎞ NC = ⎜ ⎝ 12.0 g mol ⎟⎠ ( 6.02 × 10 23 molecules mol ) (N C ) = 1.05 × 1021 carbon atoms of which 1 in 7.70 × 1011 is a 14 C atom (N 0 )C-14 = 1.37 × 109 , ln 2 λC-14 = = 1.21 × 10−4 yr −1 = 3.83 × 10−12 s −1 5 730 yr R = λ N = λ N 0 e− λ t At t = 0 , ⎡ 7 (86 400 s ) ⎤ ( )( R0 = λ N 0 = 3.83 × 10−12 s −1 1.37 × 109 ⎢ ) ⎣ 1 week ⎥⎦ = 3.17 × 103 decays week. 837 At time t, R= = 951 d ecays w eek. 0.88 R −1 ⎛ R⎞ Taking logarithms, ln = −λ t so t= ln ⎜ ⎟ R0 λ ⎝ R0 ⎠ −1 ⎛ 951 ⎞ t= ln ⎜ = 9.96 × 103 yr. 1.21 × 10 −4 yr −1 ⎝ 3.17 × 103 ⎟⎠ 16. In an experiment on the transport of nutrients in the root structure of a plant, two radioactive nuclides X and Y are used. Initially 2.50 times more nuclei of type X are present than of type Y. Just three days later there are 4.20 times more nuclei of type X than of type Y. Isotope Y has a half-life of 1.60 d. What is the half-life of isotope X? We have all this information: N x (0) = 2.50N y (0) N x (3d ) = 4.20N y (3d ) − λ y 3d N x (0 ) − λ y 3d N x (0 )e− λx 3d = 4.20N y (0 )e = 4.20 e 2.50 2.5 3d λ y e3d λx = e 4.2 ⎛ 2.5 ⎞ 3d λx = ln ⎜ + 3d λ y ⎝ 4.2 ⎟⎠ 0.693 ⎛ 2.5 ⎞ 0.693 3d = ln ⎜ + 3d = 0.781 T1 2 x ⎝ 4.2 ⎟⎠ 1.60 d T1 2 x = 2.66 d