Lecture 16 Notes

# 2 vso rl s we can calculate the dot product d e1 2

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Unformatted text preview: ) Harmonic potential Steeper for neutrons 14 N UCLEAR POTENTIAL Vp = r 2 ✓ ✓ V0 2 R0 V0 (Z 1)e 3 2R0 ◆ 2 3 (Z 1)e 2 R0 2 ◆ Harmonic potential + well depth 15 Vn = r 2 ✓ V0 2 R0 ◆ Steeper and Deeper for neutrons (V0 ) S hell Mode Harmonic oscillator: solve (part of) the radial equation including the angular momentum (centrifugal force term) we obtain the usual principal quantum number n = (N-l)/2+1 16 S pin-Orbit Coupling • The spin-orbit interaction is given by VSO 1 ˆˆ ~· ~ = 2 Vso (r)l s ~ • We can calculate the dot product D E1 ~2 ˆˆ ˆ ˆ ˆ ~ · ~ = (~ 2 ~2 ~2 ) = [j (j + 1) l(l + 1) ls j s l 2 2 1 • Because of the addition rules, j = l ± 2 (2 D E ~ 1 for j=l+ 2 l2 ˆˆ ~· ~ = ls ~2 1 (l + 1) 2 for j=l- 2 17 3 4 S pin-Orbit Coupling • when the spin is aligned with the angular momentum 1 j =l+ 2 the potential becomes more negative, i.e. the well is deeper and the state more tightly bound. • when spin and angular momentum are anti-aligned j = l the system's energy is higher. Vso E= (2l + 1) 2 • The difference in energy is Thus it increases with l . 18 1 2 E xample • 3N level, with l=3 (1f level) j=7/2 or j=5/2 • Level is pushed so down that it forms its own shell 1f 3N 2p 2p1/2 1f5/2 2p3/2 1f7/2 2N 19 20 2f 1h 0 3s 2 2d 1g 1 2p 3 3 ... 3 4 4 3p 5 5 4s 3d 2g 1i 1 6 0 2 4 6 1i11/2 1i13/2 2f5/2 2f7/2 3s1/2 1g7/2 1g9/2 1f 1f5/2 3p1/2 3p3/2 2 1 2s 1d 2s1/2 1 1p 1p1/2 44 126 32 82 22 50 8 28 12 20 6 8 2 2 1h11/2 2d3/2 2d5/2 2p1/2 2p3/2 1d3/2 0 184 1h9/2 1f7/2 2 58 1d5/2 1p3/2 0 0 1s 1s1/2 21 MIT OpenCourseWare http://ocw.mit.edu 22.02 Introduction to Applied Nuclear Physics Spring 2012 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms....
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## This note was uploaded on 11/23/2013 for the course PHYS 22.02 taught by Professor Cappellaro during the Spring '12 term at MIT.

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