{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
203 13 Nuclear Structure 13-1 = 1 3 0 R R A where = 0 1.2 fm R ; (a) = 4 A so ( 29( 29 = = 1 3 He 1.2 4 fm 1.9 fm R (b) = 238 A so ( 29( 29 = = 1 3 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 m V !!! 13-3 ρ ρ = NUC NUC NUC ATOMIC ATOMIC ATOMIC M V M V and approximately; = NUC ATOMIC M M . 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) ( 29 ( 29 ( 29 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) ( 29 ( 29 ( 29 - - × = = × 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 ( 29 ( 29( 29 - - - × × = = × 27 6 34 2 2.792 8 5.05 10 J T 50 10 T 2.13 kHz 6.63 10 J s p f .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 K U r and ( 29 ( 29 ( 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 ( 29 ( 29 ( 29 ( 29( 29 - - - × × = = = × × × 1 2 2 9 2 2 19 1 2 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) = 1 3 0 R R A , - = × 15 0 1.2 10 m R , = 4 10 m R ( 29 ( 29( 29 - = = = × = × × = = × × = × 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 m A (about 1/2 the mass of the sun) (b) = 2 GMm mg R so = 2 GM g R ( 29( 29 ( 29 - × × = = × 11 2 2 29 11 2 2 4 6.67 10 N m kg 9.7 10 kg 6.5 10 m s 10 m g (c) ϖ = 2 1 2 K I ( 29 ϖ π = × 30 2 rad s ; and ( 29( 29 = = × = × 2 2 29 4 37 2 2 2 9.7 10 kg 10 m 3.88 10 kg m 5 5 I MR so ( 29 ( 29 ϖ π = = × × = × 2 2 37 2 41 1 1 3.88 10 kg m 30 2 rad s 6.9 10 J 2 2 K I . 13-7 μ = - E B so the energies are μ = + 1 E B and μ = - 2 E B . μ μ = 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 E B 13-8 Note: Let the proton’s magnetic moment be μ μ = nuc 2.8 p . The neutron’s magnetic moment is μ μ = - nuc 1.9 n .
Image of page 2
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 μ μ = = nuc nuc 2 2.8 5.6 E B B For Deuterium: ( 29 μ = nuc 1.8 2 transitions E B ( 29 μ μ μ = = = nuc nuc nuc 3.8 5.6 2 transitions 9.4 E B E B E B 13-9 We need to use the procedure to calculate a “weighted average.” Let the fractional abundances be represented by + = 63 65 1 f f ; then ( 29 ( 29 ( 29 + = + 63 65 63 65 Cu 63 65 Cu Cu f m f m m f f . We find ( 29 ( 29 ( 29 - = - 65 Cu 63 65 63 Cu Cu Cu m m f m m , - = = - 63 64.95 u 63.55 u 0.30 64.95 u 62.95 u f or 30% and = - = 65 63 1 0.70 f f or 70%. 13-10 (a) - = = × 1 3 1 3 15 0 1.2 10 A m R R A . When = 12 A , - = × 15 2.75 10 m R . (b) ( 29 ( 29( 29 ( 29 - × × - - = = 2 9 2 2 19 2 2 2 9 10 N m C 1.6 10 C 1 1 Z k Z e F R R . When = 6 Z and - = × 15 2.75 10 m R , = 152 N F .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern