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Chapter 43

# Chapter 43 - 43.1 a b c 85 37 205 81 28 14 Si has 14...

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43.1: a) Si 28 14 has 14 protons and 14 neutrons. b) Rb 85 37 has 37 protons and 48 neutrons. c) Tl 205 81 has 81 protons and 124 neutrons. 43.2: a) Using , fm) 2 . 1 ( 3 1 A R = the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. b) Using 2 4 R π for each of the radii in part (a), the areas are 163 and fm 353 , fm 2 2 . fm 633 2 c) 3 3 4 R π gives 195 , fm 3 624 3 fm and 1499 . fm 3 d) The density is the same, since the volume and the mass are both proportional to A : 3 17 m kg 10 3 . 2 × (see Example 43.1). e) Dividing the result of part (d) by the mass of a nucleon, the number density is . m 10 40 . 1 fm 14 . 0 3 44 3 × = 43.3: B μ B μ B μ E z z z 2 ) ( = - - = T 533 . 0 T) J 10 051 . 5 )( 7928 . 2 ( 2 Hz) 10 27 . 2 ( s) J 10 63 . 6 ( 2 so , But 27 7 34 = × × × = = = - - B hf B hf E z μ 43.4: a) As in Example 43.2, eV. 10 77 . 2 ) T 30 . 2 )( T eV 10 15245 . 3 )( 9130 . 1 ( 2 7 8 - - × = × = E Since S N and are in opposite directions for a neutron, the antiparallel configuration is lower energy. This result is smaller than but comparable to that found in the example for protons. b) m. .48 4 λ MHz, 9 . 66 = = = = f c h E f

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43.5: a) - = = S N B μ and . B U z μ point in the same direction for a proton. So if the spin magnetic moment of the proton is parallel to the magnetic field, , 0 < U and if they are antiparallel, . 0 U So the parallel case has lower energy. The frequency of an emitted photon has a transition of the protons between the two states given by: Hz. 10 02 . 7 ) s J 10 63 . 6 ( ) T 65 . 1 )( T J 10 051 . 5 )( 7928 . 2 ( 2 2 7 34 27 × = × × = = - = = - - - + h B μ h E E h E f z m. 27 . 4 Hz 10 02 . 7 s m 10 00 . 3 λ 7 8 = × × = = + f c This is a radio wave. b) For electrons, the negative charge means that the argument from part (a) leads to the 2 1 - = s m state (antiparallel) having the lowest energy, since S N and point in opposite directions. So an emitted photon in a transition from one state to the other has a frequency h B h E E h E f z μ 2 2 1 2 1 - = - = = + - But from Eq. (41.22), m. 10 49 . 6 Hz 10 62 . 4 s m 10 00 . 3 λ so Hz 10 62 . 4 kg) 10 11 . 9 ( 4 T) C)(1.65 10 60 . 1 )( 00232 . 2 ( 4 ) 00232 . 2 ( 4 ) 00232 . 2 ( 2 ) 00232 . 2 ( 3 10 8 10 31 19 - - - × = × × = = × = × × = = - = - = f c f π πm eB f m e S m e μ e e z e z R This is a microwave. 43.6: a) %. 0027 . 0 10 66 . 2 ) eV 10 511 . 0 ( ) eV 6 . 13 ( 5 6 = × = × - b) %. 937 . 0 10 37 . 9 ) MeV 3 . 938 ( ) MeV 795 . 8 ( 3 = × = - 43.7: The binding energy of a deuteron is eV. 10 224 . 2 6 × The photon with this energy has wavelength equal to . m 10 576 . 5 eV) J 10 602 . 1 )( eV 10 224 . 2 ( s) m 10 998 . 2 )( s J 10 626 . 6 ( λ 13 19 6 8 34 - - - × = × × × × = = E hc
43.8: a) u, 112 . 0 ) ( 7 N H n = - + m m m which is 105 MeV, or 7.48 MeV per nucleon. b) Similarly, per MeV 7.07 or MeV, 3 . 28 u 03038 . 0 ) ( 2 He n H = = - + m m m nucleon, slightly lower (compare to Fig. (43.2)). 43.9: a) For B 11 5 the mass defect is: ) 5 6 5 11 5 e n p B M( m m m m - + + = u) MeV u)(931.5 081815 . 0 ( energy binding The u 0.081815 u 009305 . 11 ) u 000549 . 0 ( 5 ) u 008665 . 1 ( 6 ) u 007277 . 1 ( 5 B = = - + + = E MeV 76.21 = b) From Eq. (43.11): A Z A C A Z Z C A C A C E 2 4 3 2 1 B ) 2 ( ) 1 ( 3 1 3 2 - - - - - = and there is no fifth term since Z is odd and A is even. 3 1 3 2 11 ) 4 ( 5 ) MeV (0.7100 MeV)(11) 80 . 17 ( 11 ) MeV 75 . 15 ( B - - = E ) 11 ( ) 10 11 ( ) MeV 69 . 23 ( 2 - - MeV. 76.68 B = E So the percentage difference is % 62 . 0 100 MeV 76.21 MeV 76.21 MeV 76.68 = × - Eq. (43.11) has a greater percentage accuracy for Ni.

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