# chap43 - 43.1: a) Si 28 14 has 14 protons and 14 neutrons....

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Unformatted text preview: 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 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 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 . 3 44 3 = 43.3: B B B E z z z 2 ) ( =-- = T 533 . 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 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 MHz, 9 . 66 = = = = f c h E f 43.5: a) - = = S N B and . B U z point in the same direction for a proton. So if the spin magnetic moment of the proton is parallel to the magnetic field, , &lt; U and if they are antiparallel, . 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 h E E h E f z m. 27 . 4 Hz 10 02 . 7 s m 10 00 . 3 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 2 2 1 2 1- =- = = +- But from Eq. (41.22), m. 10 49 . 6 Hz 10 62 . 4 s m 10 00 . 3 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 m eB f m e S m e e e z e z R This is a microwave. 43.6: a) %. 0027 . 10 66 . 2 ) eV 10 511 . ( ) eV 6 . 13 ( 5 6 = = - b) %. 937 . 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 ( 13 19 6 8 34--- = = = E hc 43.8: a) u, 112 . ) ( 7 N H n =- + m m m which is 105 MeV, or 7.48 MeV per nucleon....
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## This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap43 - 43.1: a) Si 28 14 has 14 protons and 14 neutrons....

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