Nuclear Physics and Astrophysics 2015 (SPA-5302), Homework 2

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Randolf Higgins
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1 Nuclear Physics and Astrophysics 2015 (SPA-5302), Homework Teppei Katori Name: ID: Problem 1 (4 points) An acient archaeological wooden tool is excavated in China. An experimenter measures the activity from the 4 g sample, and records total 320 counts in a period of 1 hour by a detector with 80% efficiency. From the study they found 10% of measured radiation are natural 14 background radiation, and 90% of radiation come from decays of C which has a decay constant of yr [1] What is the mean lifetime of C? [2] What is the activity of this artefact (in Bq)? [3] Assuming the fraction of C to C in living matter is , how much C was contained when it was buried in this artefact? [4] Using all these information, estimate the age of the artefact. In [1]: la=1.2092*10**(-4.0) tau=1/la print tau Mean life time of 14 C is 8270 years. In [2]: activity=320.0/0.8*0.9/ print activity 0.1 Activity of this artefact is 0.1 Bq now. In [3]: NA=6.022*10**(23) frac=1.0*10**(-12) NumOfAtom=4.0/12.0*NA*frac print NumOfAtom e+11 1 of 6

2 11 14 When this artefact was burried, there was of C atoms. Here, we use N(t) = N o e λ t, N(present)/N(t = 0) = e λ t In [4]: import math OldActivity=la*NumOfAtom/(365.0*24.0*3600) print OldActivity OldTime=math.log(OldActivity/activity)/la print OldTime At the beginning, activity was Bq. Therefore, this artefact is roughly years old. Problem 2 (3 points) Here, a v =15.5 MeV, a s =16.8 MeV, a c =0.72 MeV, a sym =23 MeV, a p =34 MeV. [1] Using Appendix C table of Krane's textbook, estimate the kinetic energy of an alpha particle emitted from the decay of Po to Pb [2] Using semi-empirical mass formula, calculate nuclear masses of Po to Pb, then estimate kinetic energy of an alpha particle emitted by the same process. [3] Why the kinetic energy of alpha estimated from Appendix C and SEMF are different? In [11]: av=15.5;asu=16.8;ac=0.72;asym=23;ap=34 u=931.5;mp=1.007*u;mn=1.009*u M210Po= *u M206Pb= *u M4He= *u Talpha1=M210Po-M206Pb-M4He print M210Po,M206Pb,M4He,Talpha Kinetic energy of α is MeV. In [12]: A=210;Z=84;N=126 B1=av*A-asu*A**(2.0/3.0)-ac*Z*(Z-1)*A**(-1.0/3.0)-asym*(A-2.0*Z)**2.0/A+ print B1 M1=Z*mp+N*mn-B1 print M Total binding energy of Po is MeV. The nuclear mass of Po is MeV. 2 of 6

3 In [13]: A=206;Z=82;N=124 B2=av*A-asu*A**(2.0/3.0)-ac*Z*(Z-1)*A**(-1.0/3.0)-asym*(A-2.0*Z)**2.0/A+ print B2 M2=Z*mp+N*mn-B2 print M Total binding energy of Pb is MeV. The nuclear mass of Pb is MeV. In [14]: A=4;Z=2;N=2 B3=av*A-asu*A**(2.0/3.0)-ac*Z*(Z-1)*A**(-1.0/3.0)-asym*(A-2.0*Z)**2.0/A+ print B3 M3=Z*mp+N*mn-B3 print M Total binding energy of He is MeV. The nuclear mass of He is MeV. In [15]: Talpha2=M1-M2-M3 print Talpha The kinetic energy of alpha particle from SEMF is MeV. Let's take a look the difference of each mass from Appendix C and SEMF. In [16]: print M210Po-M1 print M206Pb-M2 print M4He-M So seems to me SEMF underestimates mass of 210 Po, and that is the origin of lower kinetic energy by SEMF (I don't know why). I will accept any answers. Problem 3 (3 points) 3 of 6

4 Below is a schematic cartoon of the SciBooNE detector at Fermilab (PRD83(2011)012005, ( Neutrino beam, from left, interact and produce muon tracks (arrows) in the detector. Although the figure shows angles of muon tracks, here we assume all muons are streight tracks to z-direction. Also we assume all 3 muons are minimum ionization particles (MIPs) that lose 2 MeV per cm in 1 g/cm. The produced muon tracks are measured in 2 detectors. Scintillation bar detector "SciBar" (2.0m in 3 z-direction) is the detector made by plastic scintillator (1g/cm ), and muon range detector "MRD" (1.6m in z-direction) has alternative layers of plastic scintillator and stainless steel 3 (8g/cm ) in z-direction. Here, we ignore electron catcher "EC". [1] What is the highest energy muon stopped in SciBar? [2] What is the highest energy muon stopped in MRD? [3] How do you know the energy of muon if that does not stop in neither SciBar or MRD? The highest energy muon is the one produced at the biggining of SciBar and stopped at the end of Scibar. Assuming MIP, it drops 2 MeV/cm, so the highest energy muon is 400 MeV. Similarly, highest energy muon is the one produced at the beginning of SciBar and atopped at the end of MRD. It drops 2 MeV/cm in plastic, and 16 MeV/cm in steel, so the highest energy is 1840 MeV. The reason of alternative layers of scintillator and steel is to extend the MRD-stopped muon to higher energy so that they can measure higher energy muons. There are 4 answers, (1) if the MIP particle penetrate MRD, we cannot measure the energy (this is the answer we "cannot"). (2) if there were magnetic field, energy could be estimated from the 2 curvature using mv /r=qvb, (3) if the multiple scattering measured by high resolution detector, energy can be estimated, and (4) if the muon is higher than MIP, energy can be measured assuming radiation energy loss which is a function of muon energy unlike the ionization loss by MIPs. Useful constants: 4 of 6

5 Δν( 133 Cs ) hfs = Hz h = J s c = m/s e = C N A = k B = J/K ħ c = 197 MeV f m m p = 1.007u m n = 1.009u 1u = MeV G N = N m 2 /kg M = kg R = m L = J/s 1pc = 3.26lyr Semi-empirical mass formula of the binding energy B(Z, A) = a v A a s A 2/3 Z(Z 1) (A 2Z) a c a 2 a p ± A 1/3 sym A A 3/4 Orbital angular momentum, L=0, 1, 2, 3... are L=s, p, d, f,... The ordering of nuclear shell 1 s 1/2 : 1 p 3/2 : 1 p 1/2 : 1 d 5/2 : 2 s 1/2 : 1 d 3/2 : 1 f 7/2 : 2 p 3/2 : 1 f 5/2 : 2 p 1/2 : 1 g 9/2 : 2 d 5/2 : 1g Periodic table H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Ga, Ge, As, Se, Br, Kr, Rb... relativistic formula E = m 2 c 4 + p 2 c 2 = mc 2 γ, Δ t = γδt, Δ L = γ 1 ΔL, γ = (1 β 2 ) 1/2, β = v/c Rutherford scattering formula dσ zze = zzħc = dω ( 4πϵ 0 ) 2( 4T α ) ( 137 ) ( 4T α ) sin 4 θ 2 sin 4 θ 2 5 of 6

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Name: Student ID Number: Section Number:

Chem 6A 2011 (Sailor) QUIZ #7 Name: Student ID Number: Section Number: VERSION A KEY Some useful constants and relationships: Specific heat capacities (in J/g. K): H 2 O (l) = 4.184; Al (s) = 0.900; Cu