43 1 Model The nucleus is composed of Z protons and neutrons a 3H has Z 1 proton and 3 1 2 neutrons Solve b 40Ar has Z 18 protons and 40 18 22 neutrons c 40Ca has Z 20 protons and 40 20 20 neutrons d 239Pu has Z 94 protons and 239 94 145 neutrons a 6Li has Z 3 protons and 3 3 3 neutrons 43 2 Model The nucleus is composed of Z protons and neutrons Solve b 54Cr has Z 24 protons and 54 24 30 neutrons c 54Fe has Z 26 protons and 54 26 28 neutrons d 220Rn has Z 86 protons and 220 86 134 neutrons 43 3 Solve a The radius and diameter of the nucleus of 4He are r r0A1 3 1 2 fm 4 1 3 1 90 fm d 3 8 fm b For 40Ar r 1 2 fm 40 1 3 4 10 fm and d 8 2 fm c For 220Rn r 1 2 fm 220 1 3 7 24 fm and d 14 5 fm 43 4 Solve Using Equation 43 2 r r A 0 1 3 7 46 fm 2 1 2 fm A 1 3 A 7 46 fm 2 4 fm 3 30 Only silicon has a stable isotope of A 30 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 m 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 The radii are computed from r 1 3 r A 0 r where 0 1 2 fm The densities are computed from Solve a For 6 Li m r 4 3 3 b For 207 Pb m 6 0135 u 9 988 10 27 kg 27 kg 1 661 10 1 u r 1 3 r A 0 1 2 10 15 m 6 1 3 2 18 10 15 m 2 2 10 15 m m r 3 4 3 4 3 27 9 988 10 2 18 10 kg 15 m 3 17 2 3 10 kg m 3 m 206 9309 u 3 437 10 25 kg 27 kg 1 661 10 1 u r 1 3 r A 0 1 2 10 15 m 207 1 3 7 10 10 15 m 7 1 10 15 m m r 3 4 3 4 3 25 3 437 10 7 10 10 kg 15 m 3 17 2 3 10 kg m 3 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 43 6 Model Assume that air is 21 O2 and 79 N2 and that its density is 1 2 kg m3 Solve The masses of oxygen and nitrogen molecules are Moxygen 0 21 airV 0 21 1 2 kg m3 1 m3 0 252 kg Mnitrogen 0 79 airV 0 79 1 2 kg m3 1 m3 0 948 kg The number of oxygen molecules is N oxygen M oxygen M A N A 0 252 kg 0 032 kg mol 6 02 10 mol 23 1 4 74 10 24 The number of protons from oxygen molecules is Likewise the number of protons from nitrogen molecules is N oxygen protons 4 743 10 24 16 7 59 10 25 N nitrogen protons 0 948 kg 0 028 kg mol 6 02 10 mol 23 1 26 14 2 85 10 The total number of protons is 7 59 1025 2 85 1026 3 6 1026 Because nitrogen and oxygen each have an equal number of protons and neutrons this is also the total number of neutrons 43 7 Solve The nuclear density was found to be 2 3 1017 kg m3 Thus M nuclear V 17 2 3 10 kg m 3 0 5 10 m 2 1 2 10 kg 11 3 4 3 Assess The nuclear density is tremendously large compared to the density of familiar liquids and solids 43 8 Solve The chart shows stable and unstable nuclei for all nuclei with Z 8 A black dot represents stable isotopes a dark gray dot represents isotopes that undergo beta minus decay and a light gray dot represents isotopes that undergo beta plus decay or electron capture decay a A 36 for which 36S and 36Ar are stable 43 9 Solve b Nuclei with A 8 and A 5 have no stable nuclei 43 10 Solve From Equation 43 6 the binding energy for 3 H is B Zm H Nm m n atom 1 1 00783 u 2 1 00866 u 3 01605 u The binding energy per nucleon is For 3He the binding energy is 0 00910 u 931 49 MeV u 8 48 MeV 3 8 48 MeV 2 83 MeV 1 B Zm H Nm m n atom 2 1 00783 u 1 1 00866 u 3 01603 u 0 00829 u 931 49 MeV u 7 72 MeV The binding energy per nucleon is 1 3 7 72 MeV 2 57 MeV 43 11 Solve From Equation 43 6 the binding energy for 54Fe is B Zm H Nm m n atom 26 1 00783 u 32 1 00866 u 57 933278 u 0 54742 u 931 49 MeV u 58 509 92 MeV 8 7917 MeV 509 92 MeV 1 The binding energy per nucleon is For 58Ni the binding energy is B Zm H Nm m n atom 28 1 00783 u 30 1 00866 u 57 935346 u The binding energy per nucleon is 0 54369 u 931 49 MeV u 58 506 45 MeV 8 7318 MeV 506 45 MeV 1 Assess The binding energy per nucleon is higher for iron but the binding energy per nucleon for nickel isn t much different we expect both of these results from Figure 43 6 43 12 Solve From Equation 43 6 the binding energy for 3He is B Zm H Nm m n atom 2 1 00783 u 1 1 00866 u 3 01603 u The binding energy per nucleon is For 4He the binding energy is 0 00829 u 931 49 MeV u 7 722 MeV 1 3 7 722 MeV 2 1 00783 u Nm m n atom 2 57 MeV B Zm H 0 03038 u 931 49 MeV u 28 30 MeV 2 1 00866 u 4 00260 u The binding energy per nucleon is for 4He than for 3He 4He is more tightly bound 1 4 28 30 MeV 7 07 MeV Because the binding energy per nucleon is more 43 13 Solve From Equation 43 6 the binding energy for 12C is B Zm H Nm m n atom 6 1 00783 u 6 1 00866 u 12 00000 u The binding energy per nucleon is For 13C the binding energy is 0 09894 u 931 49 MeV u 92 162 MeV 1 12 92 162 MeV 7 68 MeV B …
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