PracticeModernPhysicsII,W2018,Set1Question1EnergyLevelDiagramofBoronionB2+ ForneutralB,Z=5.(A)Drawthefine-structurediagramofB2+ thatincludesalln=3states.Labelthestatesinspectroscopicnotation.(B)Usingtheselectionrulesofequation8.8drawalltheallowedtransitionsofB2+ .SuchdiagramsarecalledGrotiandiagrams.Thegroundstate B2+ hasthreeelectrons,electronicconfigurationis1s2 2s1 .Asshowninclass,onlytheelectronintheunfilledsubshell 2s1contributestotheangularmomentumoftheatom.Weshowinclassthat

S = s1 =1/2,L= ℓ1 =0, J = L+ S = S =1/2 ,withtermsymbol22S1/2 .

Thefirstexcitedstateselectronicconfigurationis1s22p1 .Thereisasingleelectron

intheunfilled 2p1 ℓ1 =1( ) subshell.SoS =1/2 andL=1 .Thepossiblevaluesof

J = L+1/2,L−1/2 givestwostatesJ =3/2,1/2 .Thestatesare:1)S =1/2,L=1, J =1/2giving2

2P1/2 ;2)S =1/2,L=1, J =3/2giving22P3/2

Thenextexcitedstateelectronicconfigurationis1s2 3s1 .Itshouldbeclearthatthat

S = s1 =1/2,L= ℓ1 =0, J = L+ S = S =1/2 ,withtermsymbol32S1/2 .Thenextexcited

stateselectronicconfigurationis1s23p1 .The3p

1 electrongivesS =1/2andL=1 .ThepossiblevaluesofJ = L+1/2,L−1/2givestwostatesJ =3/2,1/2.Thestatesare:1)S =1/2,L=1, J =1/2giving3

2P1/2 ;2)S =1/2,L=1, J =3/2giving32P3/2 .

Thefinalexcitedstatesforthen=3shellhaselectronicconfigurationis1s23d1 .The

3d1 ℓ1 =2( ) electrongivesS =1/2 and L= ℓ1 =2 .Thepossiblevaluesof

J = L+1/2,L−1/2 givesJ =5/2,3/2 .Thestatesare:1)S=1/2,L=2,J=3/2giving

32D3/2 ;2)S =1/2,L=2, J =5/2giving3

2D5/2 .Theschematicenergydiagramisbelow:UsingHund’srule3,sincethe2p(3p)subshellislessthan

halffilled22P1/2 ( 3

2P1/2 )haslowerenergythan22P3/2 ( 3

2P1/2 ).

2s1 22S1/2

Inthediagram,athicklineindicatesastateofB2+,withthetermsymbol,n 2LJ ,ontherightoftheline.Theelectronicconfigurationsareontheleftofthelines.The

Justlikethepstates,Hund’srule3suggests haslowerenergythan

double-arrowlinesthatconnecttwostatesindicateallowedtransitionsthatobeythetransitionrulesofequation8.8.Question2AtomicPhysics(FineStructureandZeemanEffects)Excitedstatesofatoms,fine-structure,andZeemaneffects.Theground-stateconfigurationofSODIUM(Z=11)is1s2 2s2 2p6 3s1 ,intermsymbolis 32S1/2 .A)WriteisthefirstexcitedstateelectronicconfigurationofSodium.Findthespectroscopicnotation(termsymbol)ofthetwostatesassociatedwiththisexcitedstate.Hint:findallthetotalorbitalAngularMomentum(L)andtotalspin(S),andallthepossibletotalangularmomentum(J).Alsoseeequationsheetforenergyorder!!Theelectronicconfigurationofthefirstexcitedstateis1s2 2s2 2p6 3p1 .Forthisconfiguration,onlythe 3p1electroncontributetotheangularmomentum,andhenceS = 1 / 2 , L = 1 .Thetotalangularmomentumis J = L + S,... L − S .Therearetwostates:i) J = L + S = 3 / 2with S = 1 / 2 and L = 1 .Thetermsymbolis 32P3/2 .ii) J = L − S = 1 / 2withS = 1 / 2 and L = 1 .Thetermsymbolis 32P1/2 .B)WritethesecondexcitedstateelectronicconfigurationofSODIUM.Findthespectroscopicnotation(termsymbol)oftheonestateassociatedwiththisexcitedstate.Brieflyexplainwhythereisonlyonestate.Lookattheenergyorderlistinback!Theelectronicconfigurationofthesecondexcitedstateis1s2 2s2 2p6 4s1 .Forthisconfiguration,onlythe 4s1 electroncontributetotheangularmomentum,andhenceS = 1 / 2 , L = 0 .Thereisonlyvalueoftotalangularmomentumis J = S = 1 / 2 .For

J = 1 / 2 ,S = 1 / 2 and L = 0 .Thetermsymbolis 4 2S1/2 .C)ThefirstexcitedstateofLITHIUM(Z=3)haselectronicconfiguration1s2 2p1 ,andissplitintotwostates(doublet) 2 2P1/2 and 2

2P3/2 .ItisestimatedthattheinternalmagneticfieldforthisstateisBint = 0.36T .Findthefine-structureenergydifferencebetweenthe2 2P1/2 and2

2P3/2 states.Fine-structuresplittingisΔVfs = 2µBBint = 2 5.788 ×10

−5eV /T( ) 0.36T( ) = 4.17 ×10−5eV D)SupposetheLITHIUMatomisplacedinanexternalmagneticfieldofmagnitudeB = 0.5T .Howmanyenergylevelsdothe2 2P3/2 have?Drawaschematicdiagramtoillustratetheseenergylevels.Findtheenergyspacingbetweenadjacentenergylevels.HINT:ThisistheZeemanEffects!Landefactorisrelevant!TheZeemanenergyoftheselevelsaregivenbyVz = µBBextgmJ .For2

2P3/2 , J = 3 / 2

andmJ = −32,− 12, 12, 32givingfourdistinctenergylevels.

Theenergyseparationbetweenadjacentlevels(forexamplemj = 3 / 2 andmj = 1 / 2 ,ormj = 0 andmj = −1 / 2 )issimply,ΔVz = µBBextgΔmj = µBBextg ,where

Δmj = 1 foradjacentlevels.For2P3/2 state J =

32,L=1,andS=½

g = 1+ 1.5 1.5 +1( ) + 0.5 0.5 +1( )−1 1+1( )2 1.5( ) 1.5 +1( ) = 1.33

For Bext = 0.50T ,weuseµB = 5.788 ×10−5eV /T andg=1.33toobtainfortheenergy

differencebetweenadjacentZeeman 2P3/2 levels,ΔVz = µBgBext = 5.788 ×10

−5eV /T 0.5T( ) 1.33( ) = 3.84 ×10−5eV mJ = 3 / 2 mJ = 1 / 2 E 2P3/2

mJ = −1 / 2 mJ = −3 / 2 Question3MolecularSpectroscopy:ItisknownthatthehydrogenmoleculeH2

hasavibrationabsorption(emission)frequencyofν0 = 1.32 ×1014Hz .

A)ModeltheH2moleculeastwoHatomsconnectedbyaspring.Basedonthedatagivencalculatethespringconstantk.UsemH=1u.Reducedmassµ = mHmH

mH +mH

= 12mH = 8.33×10−28 kg.

ω = kµ→ν = 1

2πkµ→ k = 4π 2ν 2µ = 4π 2 1.32 ×1014 s−1( )2 8.33×10−28 kg( ) = 573Nm

B) Now consider a deuterium moleculeD2 , where D is a heavy hydrogen withnucleusofoneprotonandoneneutronwithamassmD=2amu.UsethedataofpartAtocalculatethevibrationfrequencyofthismolecule.

µ = mDmD

mD +mD

= 12mD = 1.67 ×10−27 kg.

ν = 12π

kµ= 12π

573N /m1.67 ×10−27 kg

= 9.3×1013Hz

C)AreH2andD2infraredactive?Brieflyexplainyouranswer.No.Onlyvibrationmodesthatinduceachangeinthedipolemomentofthemoleculeareinfraredactive.SincehomonucleardiatomicmoleculessuchasH2andD2donotevenhavedipolemoments,theycannotbeinfraredactive.

Question4VibrationalEnergyLevelofoxygenmoleculeO2A) Assume that the O2 molecule behaves like a harmonic oscillator with a forceconstantof210N/m.Findtheenergy(ineV)ofitsground(n=0)andfirstexcited(n=1)vibrationalstates.For16O,mO =16u ,where1u =1.66×10

−27kg .For16O,mO =16u ,where1 u = 1.66 ×10−27 kg .First mO = 1 6u × 1.66 ×10-27kg /u( ) = 2.66 ×10−26 kg

Reducedmassµ = mOmO

mO +mO

= mO

2=2.66 ×10−26 kg( )

2= 1.33×10−26 kg

Thefundamentalvibrationalfrequencyisω = kµ,wherek=210N/m

ω = 210N /m1.33×10−26 kg

= 1.26 ×1014Hz , Evibr = n +1/ 2( )!ω , n = 0,1,2...

Groundstatev=0, E0 =

!ω2

=1.055 ×10−34 J • s( ) 1.26 ×1014Hz( )

2= 6.63×10−21J

IneV,ΔE0 =6.63×10−21 J × 1

1.6×10−19eVJ

⎛⎝⎜

⎞⎠⎟=0.0415eV (0.5point)

Firstexcitedstatev=1,

E1 = 3hν0

2=

3 1.055 ×10−34 J • s( ) 1.26 ×1013Hz( )2

= 1.99 ×10−20 J (0.5point)

IneV,ΔE1 = 1.99 ×10−21J × 1

1.6 ×10−19eVJ

⎛⎝⎜

⎞⎠⎟ = 0.124eV (0.5point)

B)Findthevibrationquantumnumberthatapproximatelycorrespondstoits1.5-eVdissociationenergy.Hint:seedissociationequationontheequationsheet.

Evibr = n +1/ 2( )!ω −U0 ,whereU0isthedissociationenergy.ThedissociationenergyisfoundwhenEvibr = 0 ,whichgives

0 = n +1/ 2( )!ω −U0 ⇒ n = U0

!ω− 1

2

UsingU0 =1.5eV ,ω = 1.26 ×1014Hz ,andh= 4.136×10−5eV • s .Thevibrationalquantumnumberthatapproximatelycorrespondstoits1.5-eVdissociationenergy

is v = U0

!ω− 12= 1.5eV6.582 ×10−16eV • s( ) 1.26 ×1014Hz( ) −

12= 17.6

Afterroundingoffv=18.C)IsO2infraredactive?Brieflyexplainyouranswer.Foravibrationalmodetobeactive,thevibrationmustchangethedipolemoment.SinceO2isahomonulceardiatomicmolecule,itpossessesnodipolemoment,anditsonestretchvibrationmodedoesnotchangethedipole.Henceitisinfraredinactive.

Question5MicrowaveSpectroscopy:Therotationaltransitionfromthe ℓ = 2 tothe ℓ = 1 stateinCOisaccompaniedbytheemissionofa9.55 ×10−4 eV photonA)UsethisinformationtofindtherotationalinertiaoftheCOmolecule.

Use Eℓ = ℓ ℓ+1( ) "

2

2I,andthetransition.Forthe ℓ→ ℓ−1 transition,thechangeof

energyis ΔE = Eℓ − Eℓ−1 =

"2ℓI,whichequalsthephotonenergyis

ΔE = 9.55 ×10−4 eV ×1.6 ×10−19 J i eV −1 = 1.53×10−22 J == !

2ℓI

→ I = !2ℓ

ΔE.Use

h = 6.626 ×10−34 J i s or ! =

h2π

= 1.054 ×10−34 J is ,and ℓ = 2 ,

I = !

2ℓΔE

= 1.46 ×10−46 kg im2 .

B) What is the bond length between the C and O atoms. Data: Mass of carbonmC = 12u ;MassofoxygenmO = 16u ;1u =1.66×10

−27kg .MassofcarbonmC = 1 2u × 1.66 ×10-27kg /u( ) = 1.99 ×10−26 kg

MassofoxygenmO = 1 6u × 1.66 ×10-27kg /u( ) = 2.66 ×10−26 kg

ReducedMassµ = mCmO

mC +mO

=1.99 ×10−26 kg( ) 2.66 ×10−26 kg( )1.99 ×10−26 kg + 2.66 ×10−26 kg

= 1.14 ×10−26 kg

I = µR2 → R = I / µ = 1.46 ×10−46 kg•m2 /1.14 ×10−26 kg = 1.13×10−10m