Physics Solution Manual for 1100 and 2101

02 1023 nucleimol 256 1021 nuclei each nucleus yields

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Unformatted text preview: y is released during β decay, so the combined mass of the bismuth 211 211 − 83 Bi daughter nucleus and the β particle is less than the mass of the lead 82 Pb parent nucleus. The difference in mass is equivalent to the energy released. To obtain the energy in MeV, we will use the fact that 1 u is equivalent to 931.5 MeV. SOLUTION The mass difference that accompanies the β − decay is given by 210.988 735 u − 210. 987 255 u = 1.48 × 10 energy released is ( −3 u. Since 1 u is equivalent to 931.5 MeV, the ) 931.5 MeV Energy released = 1.48 ×10−3 u = 1.38 MeV 1u 58. REASONING In the radiocarbon method, the number of radioactive nuclei remaining at a given instant is related to the number present initially, the time that has passed since the individual died, and the decay constant for 14 C . Thus, to determine how long ago the 6 individual died, we will need information about the number of nuclei present in the material discovered with the mummy and the number present initially, which can be related to the activity of the material found with the body and the initial activity. We will also need the decay constant, which can be obtained from the half-life of 14 C . 6 1620 NUCLEAR PHYSICS AND RADIOACTIVITY SOLUTION The number N of radioactive nuclei present at a time t is N = N 0 e −λ t (31.5) where N0 is the number present initially at t = 0 s and λ is the decay constant for Solving this equation for t, we find that N = e −λ t N0 or N = −λ t ln N 0 or 14 6C . 1 N t = − ln λ N0 Since the activity A is proportional to the number N of radioactive nuclei, this expression for t becomes 1 N 1 A (1) t = − ln = − ln λ N0 λ A0 The decay constant is related to the half-life T1/2 according to λ= 0.693 T1/2 (31.6) Substituting this expression into Equation (1) reveals that A 1 1 A t = − ln = − ln λ A0 0.693 / T1/2 A 0 Noting that A/A0 = 0.785 and that T1 / 2 = 5730 yr , we find that T A 5730 yr 3 t = − 1/2 ln = − ln ( 0.785 ) = 2.00 ×10 yr A 0.693 0.693 0 59. SSM WWW REASONING AND SOLUTION As shown in Figure 31.18, if the first dynode produces 3 electrons, the second produces 9 electrons ( 3 2 ), the third produces 27 electrons ( 3 3 ), so the Nth produces 3N electrons. The number of electrons that leaves the 14th dynode and strikes the 15th dynode is 314 = 4 782 969 electrons Chapter 31 Problems 1621 60. REASONING The mass m of a substance is equal to the number n of moles times the mass per mole (see Section 14.1). The mass per mole (in grams per mole) has the same numerical value as the atomic mass of the substance. Since the atomic mass of gold is 197.968 u, its mass per mole is 197.968 g/mol. Thus, m = n (197.968 g/mol). Recall also that the number n of moles is equal to the number N of gold atoms divided by Avogadro’s number NA; n = N /NA. Therefore, the mass of gold can be written as N m = n (197.968 g/mol ) = (197.968 g/mol ) NA (1) The activity ∆N / ∆t of a radioactive sample is the number of disintegrations per second that occurs. The activity is related to the decay constant λ and the number N of atoms by Activity = ∆N = λN ∆t (31.4) The minus sign has been removed from Equation 31.4, since we have taken the absolute value of both sides of this equation. Solving Equation 31.4 for N and substituting the result into Equation (1), we find ∆N (2) m = ∆t (197.968 g/mol ) λN A Finally, we recognize that the decay constant λ is related to the half-life T1/2 by λ = 0.693/ T1/2 (Equation 31.6). Substituting this expression for λ into Equation (2) gives ∆N T (197.968 g/mol ) ∆t 1/ 2 m= ( 0.693) NA SOLUTION Expressing the activity of radioactive gold in becquerels (1 Ci = 3.70 × 1010 4 Bq) and the half-life in seconds (1 day = 8.64 × 10 s), we find that the mass of gold required to produce an activity of 315 Ci is 1622 NUCLEAR PHYSICS AND RADIOACTIVITY ∆N T (197.968 g/mol ) ∆t 1/ 2 m= ( 0.693) NA 10 4 ( 315 Ci ) 3.70 × 10 Bq ( 2.69 days ) 8.64 × 10 s (197.968 g/mol ) 1 Ci 1 day = 23 −1 ) ( 0.693) ( 6.02 × 10 mol = 1.29 × 10−3 g 61. REASONING AND SOLUTION reaction gives The conservation of linear momentum applied to the mαvα + mΤvΤ = 0 (1) The energy released in the decay is assumed to be kinetic, so (1/2)mαvα2 + (1/2)mΤvΤ2 = 4.3 MeV (2) Solving Equation (1) for vα, substituting into Equation (2) and rearranging, gives the kinetic energy of the thorium atom. 1 2 2 mT v T = 4 .3 MeV 4 .3 MeV = = mT 234 .0436 u 1+ 1+ 4 .0026 u mα 0.072 MeV The kinetic energy of the α particle is, then, 4.3 MeV – 0.072 MeV = 4.2 MeV . 62. REASONING a. Each α decay process takes the general form A A− 4 4 ZP → { + { Z −2 D 2 He { Parent nucleus Daughter nucleus α particle Therefore, each α decay reduces the atomic mass number A by 4 and the atomic number Z by 2. Writing out the overall decay process for four α decays will allow us to determine the atomic mass number and atomic number of the “great-great-granddaughter” of the thorium 228 90Th nucleus. Chapter 31 Problems 1623 b. The total amount of ene...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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