Ch. 18 Problems, Selected solutions. Sections 18.1
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1 Sections (I) How many ion pairs are created in a Geiger counter by a 5.4-MeV alpha particle if 80% of its energy goes to create ion pairs and 30 ev (average in gases) is required per ion pair? Notice that 80% of the alpha particles energy produces ion pairs in the Geiger counter. Therefore,.80( 5.4 x 0 6 ev) = 4.3 x 0 6 ev of energy is available for ion pair production. Since 30 ev of energy is required for each pair, the Geiger counter will produce: Ans. 4.3 x 0 6 ev ( ion pair 30 ev ) =.44 x 0 5 ion pairs 8.. (I) How many organic molecules can be disrupted by a 4.8-MeV alpha particle absorbed in living tissue? About 5 ev is required to ionize or disrupt an organic molecule. Ans. 4.8 x 0 6 ev ( disruption 5 ev ) = 9.6 x 0 5 disruptions Sections (I) If a model of an atom were created on an enlarged scale with a nucleus 0 cm in diameter, approximately how far away would the outer electrons be? An atom has a diameter of approximately x 0-0 m. Its nucleus has a diameter of x 0-5 m. The entire atom is x 0-0 m x 0-5 m = x 05 times the diameter of its nucleus. If we make a scale model of an atom with a nucleus of.0 m and want to include the electron configuration with the same scale, its diameter will be 0.0 m( x 0 5 ) =0,000 meters = 0 kilometers = 6.3 miles! This problem illustrates the fact that atoms, and things made of atoms, contain mostly empty space (I) If a scale model of a helium atom has a nucleus with a mass of.0 kg, what would the mass of an electron be on that scale? There are four nucleons in the nucleus of a Helium atom: protons and neutrons. Each of the nucleons is approximately 800 times more massive than an electron. The nucleus of a Helium atom is: 4(800) = 700 times more massive than an electron If we have a kilogram model of a Helium atom, an electron in the model would have a mass of:.00 kg( 700 ) =.39 x 0-4 kg= 0.39 grams
2 In each of the following nine problems write the decay equation or equations if the daughter(s) is (are) radioactive. The table of elements in Appendix C can be used to determine atomic numbers. Appendix D gives decay modes of some nuclei. In order to balance the following reactions, you we must conserve charge and mass of both sides of the reaction (I) Write the decay equation for the neutron. Neutrons are produced in large numbers in a nuclear reactor. They are usually stable inside the nucleus but outside the nucleus they undergo beta decay. Ans. Given that a neutron can undergo Beta decay, you should write: n 0 - e + H () Write the decay equation for. A part of radioactive fallout due to weapons tests. Ans. From appendix D we see that Hydrogen-3 undergoes Beta decay 3 H - 0 e + 3 He 8.. (I) Write the decay equation for 40 K, a naturally occurring isotope spread uniformly throughout the ecosphere. Ans. From appendix D we see that Potassium-40 undergoes Beta decay. 40 9K 0 - e Ca 8.. (I) Write the decay equation for 37 I, the isotope most often used in thyroid scans. Ans. From appendix D we see that Iodine-3 undergoes Beta decay 3 53I 0 - e Xe
3 8.3. (I) Write the decay equation for 37 Cs one of the major waste products of nuclear reactors. Ans. From appendix D we see that Cesium-37 undergoes Beta decay Cs 0 - e Ba 8.6. (III) a) Write the decay equation for 38 U, a material contained in pitchblende which undergoes alpha decay. b) Write the decay equation for its daughter which beta decays, c) Write the decay equation for its daughter which also leaves a residual nucleus that beta decays. This is not the end of the decay chain, but you may stop here. Ans. a) 38 9U 4 He Th b) 34 90Th 0 β + 9 c) 34 9Pa 0 β Pa 34 U Section 8.3 Additional data necessary for some of these problems can be found in the tables in this chapter and in the appendices (II) During an archaeological dig an old campfire is discovered. The charcoal in it has one onehundredth the normal amount of 4C. (One part in 0 4 rather than one part in 0.) Calculate the approximate age of the charcoal. Baby step : N = N 0 e -.693T t / ; where t is the time that has elapsed and T the carbon, which is 5,730 years. Dividing both sides of the equation by N o we get: Baby step : N = N o 00 =.0 = e-.693 T Baby step 3: Take the natural log of both sides of the equation: t / ln( 0.0) =ln e T t / is the half-life of Baby step 4: =.693T 5730 yr Baby step 5: Solve for t: T = (5730 years) = 38,00 years
4 8.4. (II) A 50-mCi (millicurie) radioactive source has a half-life of.0 years and is considered safe if its activity is less than.0 μci (microcurie). How much time must pass before the source is safe? Note: The activity of a radioactive substance is often given in millicuries ( 0-3 Curies) and in microcuries( 0-6 Curies) The activity is caused by alpha, beta or gamma decay within the nucleus. The activity A is related to the number of nuclei and the half-life as follows: Baby step : A = A o e.693t t where A o is the original activity. Baby step : Divide both sides by A o and substitute the values..693t t A = e ; A o Baby step 3: Take the natural log of both sides: Baby step 4: Solve for t: ln(4x0 6 ) = ln(e.4 = 0.693T.0yr x t t Ci 50x0-3 Ci = 4x0 6 = e.693t t ) -.4 T=.0yrs = 5 years 8.8. (III) A 48.0 gram sample of carbon is taken from the bones of a skeleton and is found to have a carbon- 4 decay activity of 60 decays/min. It is known that carbon from a living person has a decay rate of 5.0 decays. Given that the half-life of carbon-4 is 5,730 years, how old is the bone? Should we call the min gram police or an archeologist? Baby Step : The activity per gram of the sample is Baby step : Write our equation: 0.693T t A = e A o 60 decays min 48grams = 3.33 decays min g
5 Baby step 3: Substitute values: 3.33 decays min g 5 decays min g 0.693T 5730 yr = 0. = e Baby step 4: Tale the natural log of both sides of the equation: 0.693T 5730 ln( 0.) = ln(e yr ).50 = 0.693T 5730yr Baby step 5: Solve for T Ans. T = yr =,400yr Sections 8.4 and 8.5 Additional data necessary for some of these problems can be found in the tables in this chapter and in the appendices (I) What is the dose in rems due to exposures to (a) 7.0 rad of x-rays in an upper GI series? (b) four dental x-rays of 800 mrad each? (c) 0.50 rad of alpha radiation from an ingested source? Note that the number of REMs equals the relative biological effectiveness(rbe) multiplied by the number of RADs: REM = RBE*RAD a. x-ray 7 rad ( Rem Rad ) = 7 Rem b. x-ray 4 (800 mrad) = 300 mrad = 3. rad 3. rad ( rem ) = 3. rem rad c. alpha.5 rad ( 0 rem ) = 0 rem rad
6 8.3. (I) How many rads produced the following exposures to a human body? (a) 5.0 rem of proton radiation. (b) 5.0 rem of fast neutron radiation. (c) 0 rem of energetic beta radiation. Solving REM= RBE*RAD for rads a. proton 5 rem ( rad ) =.5 rad 0 rem b. fast neutron 5 rem ( rad ) =.5 rad 0 rem c. energetic beta 0 rem ( rad ) = 0 rad rem
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