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Chapter 43 - 43.1 Model The nucleus is composed of Z...

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43.1. Model: The nucleus is composed of Z protons and Α Ζ neutrons. Solve: (a) 3 H has Z = 1 proton and 3 1 = 2 neutrons. (b) 40 Ar has Z = 18 protons and 40 18 = 22 neutrons. (c) 40 Ca has Z = 20 protons and 40 20 = 20 neutrons. (d) 239 Pu has Z = 94 protons and 239 94 = 145 neutrons.
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43.2. Model: The nucleus is composed of Z protons and Α Ζ neutrons. Solve: (a) 6 Li has Z = 3 protons and 3 3 = 3 neutrons. (b) 54 Cr has Z = 24 protons and 54 24 = 30 neutrons. (c) 54 Fe has Z = 26 protons and 54 26 = 28 neutrons. (d) 220 Rn has Z = 86 protons and 220 86 = 134 neutrons.
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43.3. Solve: (a) The radius and diameter of the nucleus of 4 He are r = r 0 A 1/3 = (1.2 fm)(4) 1/3 = 1.90 fm d = 3.8 fm (b) For 40 Ar, r = (1.2 fm)(40) 1/3 = 4.10 fm and d = 8.2 fm. (c) For 220 Rn, r = (1.2 fm)(220) 1/3 = 7.24 fm and d = 14.5 fm.
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43.4. Solve: Using Equation 43.2, 3 1/ 3 1/ 3 0 7.46 fm 7.46 fm 2(1.2 fm) 30 2.4 fm r r A A A = = = = Only silicon has a stable isotope of A = 30.
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43.5. Visualize: The masses of the nuclei are found by subtracting the mass of Z electrons from the atomic masses in Appendix C, and then converting to kg. 6 ( Li) 6 015121 u 3(0 000548) u 6 0135 u m = . . = . 207 ( Pb) 206 975871 u 82(0 000548) u 206 9309 u m = . . = . The radii are computed from 1 3 0 r r A / = where 0 1 2 fm. r = . The densities are computed from 3 4 3 m r ρ π = Solve: (a) For 6 Li: 27 27 1 661 10 kg 6 0135 u 9 988 10 kg 1 u m . × = . = . × 1 3 15 1 3 15 15 0 (1 2 10 m)(6) 2 18 10 m 2 2 10 m r r A / / = = . × = . × . × 27 17 3 3 15 3 4 4 3 3 9 988 10 kg 2 3 10 kg/m (2 18 10 m) m r ρ π π . × = = = . × . × (b) For 207 Pb: 27 25 1 661 10 kg 206 9309 u 3 437 10 kg 1 u m . × = . = . × 1 3 15 1 3 15 15 0 (1 2 10 m)(207) 7 10 10 m 7 1 10 m r r A / / = = . × = . × . × 25 17 3 3 15 3 4 4 3 3 3 437 10 kg 2 3 10 kg/m (7 10 10 m) m r ρ π π . × = = = . × . × Assess: The densities are very similar because the nucleons are tightly packed in both cases. All nuclei have similar densities. This is also a typical density for a neutron star.
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43.6. Model: Assume that air is 21% O 2 and 79% N 2 , and that its density is 1.2 kg/m 3 . Solve: The masses of oxygen and nitrogen molecules are M oxygen = (0.21) ρ air V = (0.21)(1.2 kg/m 3 )(1 m 3 ) = 0.252 kg M nitrogen = (0.79) ρ air V = 0.79(1.2 kg/m 3 )(1 m 3 ) = 0.948 kg The number of oxygen molecules is oxygen 23 1 24 oxygen 0.252 kg 6.02 10 mol 4.74 10 0.032 kg/mol A A M N N M = = × × = × The number of protons from oxygen molecules is 24 25 oxygen protons 4.743 10 16 7.59 10 N = × × = × Likewise, the number of protons from nitrogen molecules is 23 1 26 nitrogen protons 0.948 kg 6.02 10 mol 14 2.85 10 0.028 kg/mol N = × × × = × The total number of protons is 7.59 × 10 25 + 2.85 × 10 26 = 3.6 × 10 26 . Because nitrogen and oxygen each have an equal number of protons and neutrons, this is also the total number of neutrons.
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43.7. Solve: The nuclear density was found to be 2.3 × 10 17 kg / m 3 . Thus ( ) ( ) 3 17 3 2 11 nuclear 4 2.3 10 kg/m 0.5 10 m 1.2 10 kg 3 M V π ρ = = × × = × Assess: The nuclear density is tremendously large compared to the density of familiar liquids and solids.
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43.8.
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