chapter 43 solutions

# chapter 43 solutions - N UCLEAR P HYSICS 43.1(a 28 14 Si...

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Unformatted text preview: N UCLEAR P HYSICS 43.1. (a) 28 14 Si has 14 protons and 14 neutrons. (b) 85 37 Rb has 37 protons and 48 neutrons. (c) 205 81 Tl has 81 protons and 124 neutrons. 43.2. (a) Using 1 3 (1.2 fm) , R A = 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 2 2 fm ,353fm and 2 633 fm . (c) 3 4 3 R π gives 195 3 fm , 624 3 fm and 1499 3 fm . (d) The density is the same, since the volume and the mass are both proportional to A : 17 3 2.3 10 kg m 2 (see Example 43.1). (e) Dividing the result of part (d) by the mass of a nucleon, the number density is 3 3 44 0.14 fm 1.40 10 m . = 43.3. I DENTIFY : Calculate the spin magnetic energy shift for each spin state of the 1 s level. Calculate the energy splitting between these states and relate this to the frequency of the photons. S ET U P : When the spin component is parallel to the field the interaction energy is . z U B μ = - When the spin component is antiparallel to the field the interaction energy is . z U B μ = + The transition energy for a transition between these two states is 2 , z E B μ ∆ = where n 2.7928 . z μ μ = The transition energy is related to the photon frequency by , E hf ∆ = so 2 . z B hf μ = E XECUTE : 34 6 27 (6.626 10 J s)(22.7 10 Hz) 0.533 T 2 2(2.7928)(5.051 10 J/T) z hf B μ-- ״ = = = E VALUATE : This magnetic field is easily achievable. Photons of this frequency have wavelength / 13.2 m. c f λ = = These are radio waves. 43.4. (a) As in Example 43.2, 8 7 2(1.9130)(3.15245 10 eV T)(2.30 T) 2.77 10 eV. E-- ∆ = = Since and S r r μ 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) 66.9 MHz,λ 4.48 m. E c f h f ∆ = = = = 43.5. I DENTIFY : Calculate the spin magnetic energy shift for each spin component. Calculate the energy splitting between these states and relate this to the frequency of the photons. (a) S ET U P : From Example 43.2, when the z-component of (and ) S r r μ is parallel to , | | z B U B μ = - = r n 2.7928 . B μ- When the z-component of (and ) S r r μ is antiparallel to B r , n | 2.7928 . z U B B μ μ = - = + The state with the proton spin component parallel to the field lies lower in energy. The energy difference between these two states is n 2(2.7928 ). E B μ ∆ = E XECUTE : 27 n 34 2(2.7928 ) 2(2.7928)(5.051 10 J/T)(1.65 T) so 6.626 10 J s E B E hf f h h μ-- ∆ ∆ = = = = 7 7.03 10 Hz 7.03 MHz f = = And then 8 7 2.998 10 m/s 4.26 m 7.03 10 Hz c f λ = = = E VALUATE : From Figure 32.4 in the textbook, these are radio waves. 43-1 43 43-2 Chapter 43 (b) S ET U P : From Eqs. (27.27) and (41.22) and Fig.41.14 in the textbook, the state with the z- component of r μ parallel to B r has lower energy. But, since the charge of the electron is negative, this is the state with the electron spin component antiparallel to . B r That is, for the 1 2 state s m = - lies lower in energy.lies lower in energy....
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chapter 43 solutions - N UCLEAR P HYSICS 43.1(a 28 14 Si...

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