HW4Ans

HW4Ans - MP 200 Radiation Physics Assignment 04 Solution...

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Unformatted text preview: MP 200 Radiation Physics Assignment 04 Solution Hao Li Oct. 2010. 1. A. 2. B. The grand unified theory refers to any of several similar models in particle physics that three gauge interactions in the standard model, which define EM, weak and strong interactions, are merged into one single interaction characterized by a larger gauge symmetry and one unified coupling constant rather than three independent ones. Therefore the answer is two. 3. B. 4. C. 5. C. ρ nuclear r 5.29 ×10−11 m 3 = ( atomic )3 = ( ) = 8.6 ×1013 −15 1.2 × 10 m ρ atomic rnuclear 6. 7. 8. 9. 10. A. C. A. A. Q = (m 1 H + m 2 H − m 3 He )c 2 = (1.007825 + 2.014102 − 3.0160293) × 931.5MeV = 5.494 MeV 1 1 2 11. (m13C + m 4 He − m16O − mn )c 2 6 2 8 = (13.003355 + 4.002603 − 15.994915 − 1.008665) × 931.5MeV = 2.215MeV 12. The semi‐empirical equation: 3 2 Mc = [ Zm p + ( A − Z )mn ]c − C1 A + C2 A + 2 2 C3 Z ( Z − 1) 1 A3 ⎛1 ⎞ C4 ( A − 2 Z ) 2 C5 ⎜ ⎟ + − 0 A A⎜ ⎟ ⎜ −1⎟ ⎝⎠ Differentiate the equation with A constant: C C dm = m p − mn + 3 (2 Z − 1) − 4 4 ( A − 2 Z ) 1 dZ A A3 dm = 0 gives a minimum Z, which gives: dZ (mn − m p + Z min = C3 1 3 + 4C4 ) . A 2C3 8C4 + 1 A A3 Substitue the value gives Z=44. 13. a) 11 6 C : Z=6 N=5 J = 3/2, parity = odd 44 20 Ca : Z=20 N=24 Even‐even combination. J = 0, parity = even 73 32 Ge : Z=32 N=41 61 28 J=9/2, parity = even Ni : Z=28 N=33 J=5/2, parity = odd; b) 61 The discrepancy is observed for 28 Ni . It is possible the neutron in 1 f 1 stays in 2 p 3 , thus giving 2 2 J = 3/2, parity = odd. 14. m( 235U )c 2 = [92m p + 143mn ]c 2 − C1 × 235 + C2 × 235 3 + 92 = 218883.8MeV 87 m( 35 Br )c 2 = [35m p + 52mn ]c 2 − C1 × 87 + C2 × 87 3 + 2 2 C3 × 92 × 91 C4 (235 − 2 × 92) 2 + 1 235 235 2 C4 (87 − 2 × 35) 2 87 C3 × 35 × 34 87 1 2 + = 80946.4 MeV m( 145 La )c 2 = [57m p + 88mn ]c 2 − C1 ×145 + C2 ×145 3 + 57 = 134960.7 MeV 3m(n)c 2 = 3 × 939.566 MeV = 2818.7 MeV 2 C3 × 57 × 56 145 1 2 + C4 (145 − 2 × 57) 2 145 Q = 218883.8 − 80946.4 − 134960.7 − 2818.7 MeV = 158MeV 15. Q= 269 269 × 9.23MeV = 9.369 MeV Kα = 265 265 Q = (m 269 Hs − m 265 Sg − m 4 He )c 2 = (m 269 Hs − 265.1211 − 4.002603) × 931.5MeV = 9.369MeV 108 106 2 108 m 269 Hs = 269.1338u 108 m 269 Hs − nucleus = 269.1338 − 108 × 5.485799 ×10−4 = 269.075 108 16. a) ΔBE = [(5mH + 6mn − mB ) − (6mH + 5mn − mC )]c 2 = (− mH + mn + mC − mB )c 2 = (−1.007825 + 1.008665 + 11.011432 − 11.009305) × 931.5MeV = 2.764 MeV b) 3 e2 (62 − 52 ) = 2.78MeV 5 4πε 0 r r = 3.42 ×10−15 m 17. Add three reactions together, we will have: 2 31 29 30 29 30 31 1 4 3 1 H + 15 P + 14 Si + 14 Si → 14 Si + 14 Si + 14 Si + 2 1 H + 2 He Some particles on both sides cancel out, we have 2 31 31 1 4 3 1 H + 15 P → 14 Si + 2 1 H + 2 He The Q value of this action is: Q = Q1 + Q2 + Q3 = 8.158 + 8.388 + 4.364MeV = 20.910 MeV or 2 31 31 1 4 Q = [3m( 1 H ) + m( 15 P ) − m( 14 Si ) − 2m( 1 H ) − m( 2 He)]c 2 31 31 = [m( 15 P ) − m( 14 Si )]c 2 + [3 × 2.014102 − 2 × 1.007825 − 4.002603] × 931.5MeV 31 31 = [m( 15 P ) − m( 14 Si )]c 2 + 22.4054 MeV Therefore 31 31 [m( 15 P) − m( 14 Si )]c 2 = 20.910 − 22.405MeV = −1.495MeV 31 The beta decay of 14 Si is: 31 14 31 0 Si → 15 P + −1 e +ν e The Q value of this beta decay is 31 31 31 31 Q = [m( 14 Si ) − m( 15 P)]c 2 = −[m( 15 P) − m( 14 Si )]c 2 = 1.495MeV 18. a) β − : 126 I → 126 Xe + e− + ve 53 54 Q = (m( 126 I ) − m( 126 Xe))c 2 = (125.905624 − 125.904274) × 931.5MeV = 1.258MeV 53 54 β + : 126 I → 126Te + e+ + ve 53 52 Q = (m( 126 I ) − m( 126Te) − 2me )c 2 = (125.905624 − 125.903312) × 931.5MeV − 0.511× 2 MeV 53 52 = 1.132 MeV EC: 126 I + e − → 126Te + ve 53 52 Q = (m( 126 I ) − m( 126Te))c 2 = (125.905624 − 125.903312) × 931.5MeV 53 52 = 2.154 MeV b) 19. 20. 224 Ra → 220 4 Rn + 2 He Q1 = [m( Ra ) − m( Rn) − m( He)]c 2 = [224.020212 − 220.011394 − 4.002603] × 931.5MeV = 5.788MeV 220 220 × 5.788MeV = 5.68MeV Kα1 = Q1 = 224 224 Q2 = 5.788MeV − 0.241MeV = 5.547 MeV Kα 2 = 220 220 × 5.547 MeV = 5.45MeV Q2 = 224 224 ...
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This note was uploaded on 10/30/2010 for the course MP 200 taught by Professor Guna during the Fall '10 term at Duke.

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