Physics Solution Manual for 1100 and 2101

# 1 yr ln ln0000 100 387 yr 0693 n0 0693 t1 2 36

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Unformatted text preview: and 31.5). Since the half-life is inversely proportional to the decay constant (see Equation 31.6), the half-life also remains constant when the mass of the substance increases. 12. N/N0 = 0.63 13. (b) The activity of an isotope depends on the number of nuclei present and the decay constant of the isotope (see Section 31.6). The decay constant, on the other hand, is inversely proportional to the half-life (see Equation 31.6). Thus, it is possible for the two samples to have the same activity, provided the ratio of the number of nuclei to the half-life is the same for both. 14. (e) Since the half-life is 1 day, one-half of the sample remains after 1 day, and one-half of that amount (or one-quarter of the original) remains after 2 days. 15. (a) The activity of a radioactive sample is equal to the number N of nuclei present times the decay constant λ of the isotope (see Section 31.6). The decay constant is inversely proportional to the half-life T1/2 (see Equation 31.6). Thus, the activity is proportional to the ratio N/T1/2, and this ratio is greatest for Sample 1. 16. (d) The half-life is the time required for one-half of the original nuclei to disintegrate (see Section 31.6). 17. (b) The 14 C dating technique assumes that the isotope 14 C was ingested by a living 6 6 organism (see Section 31.7). At one time, all of these items were part of a living animal. 4 18. (e) The 16 C activity when the animal died was 0.23 Bq per gram of carbon (see Section 31.7). The present activity is 0.10 Bq per gram of carbon, which is less than half of the original activity. Since the half-life of 14 C is 5730 yr, the bones must be more than 5000 6 years old. 19. Age = 6.9 × 103 years Chapter 31 Problems 1585 CHAPTER 31 NUCLEAR PHYSICS AND RADIOACTIVITY PROBLEMS 1. SSM REASONING For an element whose chemical symbol is X, the symbol for the A nucleus is Z X , where A represents the total number of protons and neutrons (the nucleon number) and Z represents the number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by Equation 31.1: A = Z + N . SOLUTION For the nucleus 208 82 Pb , we have Z = 82 and A = 208 . a. The net electrical charge of the nucleus is equal to the total number of protons multiplied by the charge on a single proton. Since the 208 Pb nucleus contains 82 protons, the net 82 electrical charge of the 208 82 Pb nucleus is bg q net = 82 ( +1.60 × 10 –19 C) = +1.31 × 10 –17 C b. The number of neutrons is N = A – Z = 208 – 82 = 126 . c. By inspection, the number of nucleons is A = 208 . d. The approximate radius of the nucleus can be found from Equation 31.2, namely r = (1.2 × 10 –15 m) A 1/ 3 = (1.2 × 10 –15 m)(208) 1/3 = 7.1 × 10 –15 m e. The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be found by multiplying the mass m nucleon of a single nucleon by the total number A of nucleons in the nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is ρ= mtotal V = mnucleon A 4 3 πr 3 = mnucleon A 4 3 π (1.2 × 10 –15 m) A 1/ 3 3 = mnucleon 4 3 π ( 1.2 × 10 –15 m) 3 Therefore, 1.67 × 10 –27 kg ρ=4 = –15 m) 3 3 π ( 1. 2 × 10 2.3 × 10 17 kg / m 3 1586 NUCLEAR PHYSICS AND RADIOACTIVITY 2. REASONING a. The number of protons in a given nucleus A ZX is specified by its atomic number Z. A b. The number N of neutrons in a given nucleus Z X is equal to the nucleon number A (the number of protons and neutrons) minus the atomic number Z (the number of protons): N = A − Z (Equation 31.1). c. In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to the number of protons in the nucleus. SOLUTION a. The number of protons in the uranium b. The number N of neutrons in the 238 92 U nucleus 202 80 Hg c. The number of electrons that orbit the is Z = 92 . nucleus is N = A − Z = 202 − 80 = 122 . 93 41 Nb nucleus in the neutral niobium atom is equal to the number of protons in the nucleus, or 41 . 3. REASONING For an element whose chemical symbol is X, the symbol for the nucleus is A X where A represents the number of protons and neutrons (the nucleon number) and Z Z represents the number of protons (the atomic number) in the nucleus. SOLUTION a. The symbol 195 78 X indicates that the nucleus in question contains Z = 78 protons, and N = A − Z = 195 – 78 = 117 neutrons . From the periodic table, we see that Z = 78 corresponds to platinum, Pt . b. Similar reasoning indicates that the nucleus in question is sulfur, S , and the nucleus contains N = A − Z = 32 – 16 = 16 neutrons . c. Similar reasoning indicates that the nucleus in question is copper, Cu , and the nucleus contains N = A − Z = 63 – 29 = 34 neutrons . d. Similar reasoning indicates that the nucleus in question is boron, B , and the nucleus contains N = A − Z = 11 – 5 = 6 neutrons . Chapter 31 Problems e. Similar reasoning indicates that the nucleus in question is 1587 plutonium, Pu , and the nucleus contains N =...
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