pset5sol - 22.02 Problem Set 5 Solution Krane 5.1 Solution...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
22.02 Problem Set 5 Solution Krane 5.1 Solution Let’s take extreme limit of the shell model: assume only the single unpaired nucleon determines the properties of the nucleus. (a) 7 Li: A=7, Z=3, N=4 There is an unpaired proton at 1p 3/2 . Spin: 3/2, Parity: (-1) 1 =-1. Thus I π =3/2 - (b) 11 B: A=11, Z=5, N=6 There is an unpaired proton at 1p 3/2 . Spin: 3/2, Parity: (-1) 1 =-1. Thus I π =3/2 - (c) 15 C: A=15, Z=6, N=9 There is an unpaired neutron at 1d 5/2 . Spin: 5/2, Parity: (-1) 2 =1. Thus I π =5/2 + (d) 17 F: A=17, Z=9, N=8 There is an unpaired proton at 1d 5/2 . Spin: 5/2, Parity: (-1) 2 =1. Thus I π =5/2 + (e) 31 P: A=31, Z=15, N=16 There is an unpaired proton at 2s 1/2 . Spin: 1/2, Parity: (-1) 0 =1. Thus I π =1/2 + (f) 141 Pr: A=141, Z=59, N=82 There is an unpaired proton at 2d 5/2 . Spin: 5/2, Parity: (-1) 2 =1. Thus I π =5/2 + Krane 5.2 Solution 13 C: A=13, Z=6, N=7 In the ground state, all nucleons stay at the lowest possible energy level as follows: 1d 5/2 1d 5/2 1p 1/2 1p 1/2 1p 3/2 1s 1/2 1p 3/2 1s 1/2 6 protons 7 neutrons The unpaired single neutron stays at 1p 1/2 , which gives I π =1/2 - First excited state: I π =1/2 +
Background image of page 1

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

View Full DocumentRight Arrow Icon
22.02 Problem Set 5 Solution 1d 5/2 1d 5/2 1p 1/2 1p 1/2 1p 3/2 1s 1/2 1p 3/2 1s 1/2 6 protons 7 neutrons The unpaired single neutron stays at 1s 1/2 , which gives I π =1/2 + Second excited state: I π =3/2 + 1d 5/2 1d 5/2 1p 1/2 1p 1/2 1p 3/2 1s 1/2 1p 3/2 1s 1/2 6 protons 7 neutrons The unpaired single neutron stays at 1p 3/2 , which gives I π =3/2 - Third excited state: I π =5/2 + 1d 5/2 1d 5/2 1p 1/2 1p 1/2 1p 3/2 1s 1/2 1p 3/2 1s 1/2 6 protons 7 neutrons The unpaired single neutron stays at 1d 5/2 , which gives I π =5/2 + Krane 5.3 Solution 203 Tl: A=203, Z=81, N=122 We only consider the single unpaired nucleon in this problem. Check figure 5.6, we find the 81 st proton should stay at 1h 11/2 , which means I π =11/2 - for 203 Tl. If the pairing force increases strongly with bigger l, it is more favorable in terms of lowering energy for a proton to stay at 1h
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/09/2009 for the course PHY 4604 taught by Professor Mucciolo during the Spring '09 term at University of Central Florida.

Page1 / 7

pset5sol - 22.02 Problem Set 5 Solution Krane 5.1 Solution...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online