PC Chapter 13

PC Chapter 13 - 13 Nuclear Structure 13-1 R = R0 A1 3 where...

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203 13 Nuclear Structure 13-1 = 13 0 R RA where = 0 1.2 fm R ; (a) = 4 A so ( 29 (29 == He 1.2 4 fm 1.9 fm R (b) = 238 A so ( 29 U 1.2 238 fm 7.44 fm R (c) U He 7.44 fm 3.92 1.9 fm R R 13-2 ( 29( 29 ρ = = × 14 3 3 15 2.3 10 g cm 10 cm 2.3 10 kg mV !!! 13-3 ρ ρ = NUC NUC NUC ATOMIC ATOMIC ATOMIC MV and approximately; = NUC ATOMIC MM . Therefore ρ ρ  =   3 NUC 0 ATOMIC r R where = 0 0.529 r Å - 11 5.29 10 m and - 15 1.2 10 m R (Equation 13.1 where = 1 A ). So that ρ ρ - - × = × 3 11 13 NUC 15 ATOMIC 5.29 10 m 8.57 10 1.2 10 m . 13-4 (a) ( ( μ - - × = × 27 34 2 1.913 5 5.05 10 J T 1 T 2 29.1 MHz 6.63 10 J s n B f h (b) ( ( - - × × 27 34 2 2.792 8 5.05 10 J T 1 T 42.5 MHz 6.63 10 J s p f (c) In the earth’s magnetic field ( ( ( -- - ×× × 2 76 34 2 2.792 8 5.05 10 J T 50 10 T 2.13 kHz 6.63 10 J s p f .
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204 CHAPTER 13 NUCLEAR STRUCTURE 13-5 (a) The initial kinetic energy of the alpha particle must equal the electrostatic potential energy of the two particle system at the distance of closest approach; α == min kqQ KU r and ( 29 (29 ( ( α - - - ×× = =  ×  2 9 2 2 19 13 min 13 9 10 N m C 2 79 1.6 10 C 4.55 10 m 0.5 MeV 1.6 10 J MeV kqQ r K . (b) Note that α 2 min 1 2 kqQ K mv r , so ( ( ( ( - --  = = 12 2 9 2 2 19 6 27 13 min 2 9 10 N m C 2 79 1.6 10 C 2 6.03 10 m s 4 1.67 10 kg 3 10 m kqQ v mr 13-6 (a) = 13 0 R RA , - 15 0 1.2 10 m R , = 4 10 m R ( ( ( -  = = = ×  ×  = = × × 3 3 4 3 18 56 -15 0 27 56 29 n 10 m 8.33 10 5.8 10 1.2 10 m 1.67 10 kg 5.8 10 9.7 10 kg R A R M mA (about 1/2 the mass of the sun) (b) = 2 GMm mg R so = 2 GM g R ( ( ( 29 - = 11 2 2 29 1 2 4 6.67 10 N m kg 9.7 10 kg 6 .5 10 ms 10 m g (c) ϖ = 2 1 2 KI ϖ π 30 2 rad s ; and ( 29( 29 = = × 2 2 2 9 4 3 72 22 9.7 10 kg 10 m 3.88 10 kg m 55 I MR so ( ( ϖπ = = × × 2 2 37 2 41 11 3.88 10 kg m 30 2 rad s 6.9 10 J . 13-7 μ = -⋅ EB so the energies are μ =+ 1 and μ =- 2 . μμ = 2.792 8 n and μ - 27 5.05 10 J T n μ - = = × × × × = × 27 25 6 2 2 2.792 8 5.05 10 J T 12.5 T 3.53 10 J 2.2 10 eV 13-8 Note: Let the proton’s magnetic moment be = nuc 2.8 p . The neutron’s magnetic moment is μ μ nuc 1.9 n .
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MODERN PHYSICS 205 2.8 μ nuc –2.8 μ nuc B = 0 B > 0 0.9 μ nuc = μ p + μ n B = 0 B > 0 –0.9 μ nuc = – μ p μ n 4.7 μ nuc = μ p μ n –4.7 μ nuc = – μ p + μ n (a) (b) Hydrogen Deuterium (c) For Hydrogen: ( 29 μμ == nuc nuc 2 2.8 5.6 E BB For Deuterium: ( μ ∆= nuc 1.8 2 transitions EB ( 29 μ μ μ nuc nuc nuc 3.8 5.6 2 transitions 9.4 13-9 We need to use the procedure to calculate a “weighted average.” Let the fractional abundances be represented by += 63 65 1 ff ; then ( 29 ( 29 ( + = + 63 65 63 65 Cu 63 65 Cu Cu fm m . We find - = - 65 Cu 63 65 63 Cu Cu Cu mm f , - - 63 64.95 u 63.55 u 0.30 64.95 u 62.95 u f or 30% and = -= 65 63 1 0.70 or 70%. 13-10 (a) - = 13 15 0 1.2 10 A m R RA . When = 12 A , - 15 2.75 10 m R . (b) (29 ( ( - × ×- - 2 9 2 2 19 2 22 9 10 N m C 1.6 10 C 1 1 Z k Ze F RR . When = 6 Z and - 15 2.75 10 m R , = 152 N F . (c) ( ( - × - = 2 9 2 2 19 2 12 9 10 N m C 1.6 10 C 1 1 Z kqq k U R . When = 6 Z and - 15 2.75 10 m R , - = ×= 13 4.19 10 J 2.62 MeV U . (d) = 238 A ; = 92 Z , - 15 7.44 10 m R , = 379 N F and - = 12 2.82 10 J 17.6 MeV U 13-11 ( ( [ ] ( = + 1 1 1.007 276 u 2 1.008 665 u 3.016 05 u 931.5 MeV u 2.657 MeV nucleon 3 b E A 13-12 ( 29 (  = +  56 1 Hn 26 30 Fe 931.5 MeV u 492 MeV b Emm m 492 8.79 MeV nucleon 56 b E A ; agrees with Figure 13.10.
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206 CHAPTER 13 NUCLEAR STRUCTURE 13-13 (a) The neutron to proton ratio, - AZ Z is greatest for 139 55 Cs and is equal to 1.53.
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PC Chapter 13 - 13 Nuclear Structure 13-1 R = R0 A1 3 where...

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