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