physics 2203, fall 2021 modern physics monday, nov. 12 ...physics 2203, fall 2021 modern physics ....
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Physics2203,Fall2021ModernPhysics
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Monday,Nov.12th,2012.‐‐‐Magne<sm‐‐‐Ch.16:SubatomicPhysics‐‐‐Quiztoday—FermiEnergy
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Announcements‐‐‐TermpapersdueNov.21‐‐‐First‐everphysics‐majoricecreamsocial:Wed.Nov.14th4:30pmlibrary.
WeknowhowtodothisbecauseweknowtheshiPinenergywithappliedmagne<cfield. ΔE ≈ µBB0a)Showsthedensityofstatesbutnowfortwospinalignments.ThedashedlineistheFermienergy.
b)APerafieldisapplied.
c)APerthespinshavehad<metorelax.Nowwehaveamagne<cmaterial.
Manymaterialshavespontaneousmagne<cordering.ThefamiliaroneisFerromagne<sm—themagnetsonyourrefrigerator.
Butthespinsmayaligninanan<‐parallelfashion—an<ferromagne<sm.
Ferrimagne<smisamixtureofthetwo.
Thedreamoftheelectronicsindustryishalfmetallicmetal.ALLONESPIN!
Basicbuildingblock
CaMnO3
OK,letsfigureouttheformalvalenceofeachelement.Chemistry.OTakesuptwoelectronstofillthepshell—O2‐Calosestwoelectrons{Ar}—Ca2+
WhatisthevalenceofMn?
ThenucleusofanatomcontainsZprotonswhichcreatesthecoulombfieldtobindtheZelectrons.
ThenucleuscontainsAnucleons(protonsandneutrons).Aisreferredtoasthemassnumber.
Thenucleusprovides~99.9%ofthemassofanatom,butonly10‐15ofthevolume.Thesizeismeasuredinfm(10‐15m).
Whatkeepsthenucleusfromflyingapart?Apointchargeonthesurfaceofatypicalnucleusexperiences~20MeVpotenBalenergy!NuclearForcemustbelargerthanthis!
ErnestRutherford(1871‐1973)1908NobelPrize“Chemistry”fordiscoveringthatatomscanbebrokenapartbyalphapar<cles‐‐
(1891‐1974)astudentofRutherfordwonthe1935NobelPrize“forthediscoveryoftheneutron.”
NobelLecture:TheseideasthusexplainthegeneralfeaturesofthestructureofatomicnucleianditcanbeconfidentlyexpectedthatfurtherworkontheselinesmayrevealtheelementarylawswhichgovernthestructureofmaGer.
Therearetwomajordifferencesbetweenthestudyoftheatom(electrons)andthenucleus.1) Fortheelectronswecouldtreatthemasquasi‐
independentpar<cleswithsomeZeff.Notwiththenucleus—atruemany‐bodyphysicsproblem.
2) Thereisnoclosedformexpressionforthenuclearforce.
Historically(before1932)believedthattherewereAprotonsinthenucleusandA‐Zelectrons,sothenetposi<vechargewasZ.
Whatiswrongwiththismodel:GoodfinalquesBons?1)Uncertaintyprincipleforelectron.Whatistheenergyofanelectron
inanucleus?Youcandothiswithyour3Dinfinitewalledbox!2)Spin:Thespinoftwoprotonsandoneelectron(simplestcase)—1/2
or3/2.Measuredspinfordeuteriumis1!
WhiteDwarftoaNeutronStar—Degeneracypressure
Massesaremeasuredinatomicmassunitsu.Definedbytheequa<on
12C(6 protrons, 6 neutrons) has a mass ≡ 12u1u=1.660540x10−27kg or the rest energy m u( )c2 =931.494 MeV
Electron e +e 5.48x10−4u 0.511MeV 12
Z,AtomicnumberA,massnumberN,neutronnumber
ZA XN
Redundant
612C6 ⇒
12 C
Isotope:SameZbutdifferentA’s.ExampledifferentisotopesofH.
11H0 , 1
2H1, 13H2 ⇒
1 H , 2H , 3H : Hydrogen, Deutrium, Tritium
Example:Givethesymbolforthefollowing:(a)TheisotopeofHeliumwithmassnumber4.
(a)⇒ 24He2 ⇒ 4He
(b)TheisotopeofTinwith66neutrons. (b) Tin is Sn: Z=50
50116Sn66 ⇒
116 Sn
(c)TheisotopewitA=235,and143neutrons. (c) Z=A-N=92: U
92235U143 ⇒
235 U
RememberfromQuantumproper<esofelectronsitishardtodefinearadiusofanatom—fuzzyball.Samethingistrueforthenucleus.
Volume = 4πR3
3= A 4πR0
3
3R0 → radius of nucleon
R = R0A1/3
R0 1.2x10−15m = 1.2 fm
ρ(r) = ρ01+ e r−R( ) / a
ShapeofnuclearchargedensitydeterminedbyHofstadter(NoblePrize,1961)usinghighenergyelectronscaqering,seeFigure
Asimplees<mateoftheradius.ClosepackAnuclei.
1961NobelPrize,"forhispioneeringstudiesofelectronsca7eringinatomicnucleiandforhistherebyachieveddiscoveriesconcerningthestructureofthenucleons"
BornandraisedinNewYorkCity.MovedtoStanfordin1950
ElectronScaqeringfrom12C.Elas<cpeakat185MeV,andtheinelasBc‐scaGeringpeaksfromtheexcitedstatesof12C.Thepeaknear180.7MeVisassociatedwiththe4.43‐MeVlevel.
NuclearExcita<ons
1961NobelPrize,"forhispioneeringstudiesofelectronsca7eringinatomicnucleiandforhistherebyachieveddiscoveriesconcerningthestructureofthenucleons"
Thisfigureshowstheelas<candinelas<ccurvesforscaqeringof420‐MeVelectronsby12C.Thesolidcircles,represen;ngexperimentalpoints,showtheelas<c‐scaqeringbehaviorwhilethesolidsquaresshowtheinelas;c‐sca7eringcurveforthe4.43‐MeVlevelincarbon.
BornandraisedinNewYorkCity.WenttoStanfordin1950
Calculate the radius of 12C, 3270Ge, and 83
209Bi.
First determine the number of neutrons 12C→ 6
3270Ge→ 38 83
209Bi→126
12C→ R = R0A1/3 =(1.2fm)(12)1/3 = 2.7 fm
70Ge→ R = R0A1/3 =(1.2fm)(70)1/3 = 4.9 fm
209Bi→ R = R0A1/3 =(1.2fm)(209)1/3 = 7.1 fm
Problem:(a)FindanapproximateexpressionforthemassofanucleusofmassnumberA,(b) Anexpressionforthevolumeofthisnucleus
intermsofthemassnumber,and(c) Anumericalvalueforitsdensity.
(a) The mass of the proton is approximately equal to that of the neutron.M=Am0
(b) Assume that the nucleus is spherical and R=R0A1/3
V =4πr3
3=
4πR03
3A
(c) ρn =massVolume
=3Am0
4πR03A
=3m0
4πR03 =
3 1.67x10−27 kg( )4π 1.5x10−15m( )3 = 2.3x1017 kg / m3
WhatistheraBoofnucleardensitycomparedtoWater?
Rutherfordshothighenergyalphapar<cles(Heion)ataGoldFoil.Hemeasuredtheenergyoftherecoilingalphapar<clesthatwerescaqered180o.
The initial kinetic energy of the α particle 12mv2 to the total potential U=k
2e( ) Ze( )d
d =4ke2Zmv2 dwasobservedtobyafewfm.
TheRutherfordscaqeringequa<onforthenumberpar<clesscaqeredthroughananglethetawasbasedonpointpar<clesandcoulombWhentheexperimentdeviatedfromtheRutherfordFormula,thesizeofthenucleusisprobed.
27MeValphapar<clesonPb.
Modeledlikescaqeringoflightthroughaholeofradiusr.Butelectronsarerela<vis<c.
sinθ = 0.16λR
θC ρ(r) = ρ01+ e r−R( ) / a
Consideranisolatedelectronandproton.Rest energy m pc2 + mec
2
PuttheelectronandprotontogethertoformatomicH.
Rest energy mHc2
TheTotalEnergiesincluding13.6eVbindinghavetobeequal
m pc2 + mec
2 ≡ mHc2 +13.6eV
Binding Energy B=m pc2 + mec
2 − mHc2 = 13.6eV Thisisatomic,non‐p!
ConsiderDeuterium,p+n+e. Rest energy m pc2 + mNc
2 + mec2
Putthemtogethertoform2H=D.
Rest energy mDc2
B=m pc2 +m pc
2 + mec2 −mDc
2 Ignor me B=2.224 MeV c2 = 931.5MeV / u
B=u 1.008665 +1.00727 + 5.5x10−4 − 2.014102( )c2
Nuclear Binding Energy for ZAXN : B= Nmn + Zm
1H( ) − m ZA XN( ) c
2
Example:FindthetotalbindingenergyBandalsothebindingenergypernucleonB/Afor2H,56Feand238U.
We already did 2H : B=2.224 MeV: B/A=1.112 MeV
2656 F30 : B= 30x1.00865u + 26x1.007825u − 55.934939u[ ]931.5MeV / c2
B = 492.3MeV : B/A=8.791 MeV/nucleon
92238 U146 : B= 146x1.00865u + 92x1.007825u − 238.050785u[ ]931.5MeV / c2
B = 1802MeV : B/A=7.571 MeV/nucleon
92238 U146 : B/A=7.571 MeV/nucleon: Split nucleus into two parts gain energy: Fission
2H : B/A=1.112 MeV: Combine light elements gain energy: Fusion
TheShapeofthiscurvedeterminedbythreefactors
(1)Aconstanttermfromthefactthatnucleononlyinteractwithnearestneighbors.(2)AsharpdecreasewithsmallabecauseofsurfacetoVolumera<o.
(3)AgradualfalloffwithlargeAbecauseofcoulomb‐coulombrepulsion.
Morelater!
MostStable
Fusion
Fission