Practice Problems for Midterm 2, 2017 - Lakehead University · Practice Problems for Midterm 2, 2017 Problem 1) Fermi Gas and MB gas comparison A) Gold is a dense metal with density

PracticeProblemsforMidterm2,2017Problem1)FermiGasandMBgascomparisonA)Goldisadensemetalwithdensity 1.93×10

4kg im−3 atT=300K.Calculateitsnumberdensity(n=N/V)inm−3 .Assumethatagoldatomcontributesoneconductingelectron,calculateitsFermiEnergy,EF ,inunitofeV.Repeatthecalculationbutwiththeassumptionthatonegoldatomcontributestwoconductingelectrons.Compareyouranswerswiththeexperimentalvaluesof5.53eV,andcomment.Molarmassofgoldis197g•mol-1Atomicmassofgold(Au)is197u.Weknow1u = 1.67 ×10−27 kg ,hencethemassofonegoldatomis mAu = 197u ×1.67 ×10

−27 kg iu−1 = 3.29 ×10−25 kg .Thenumber

densityis

NV

= ρmAu

= 1.93×104 kg im−3

3.29 ×10−25 kg= 5.87 ×1028m−3 .Usingthemassofthe

electron,me = 9.1×10−31kg ,andassumingoneelectronpergoldatom

NV

=5.87×1028m−3 .

EF = h2

8me3πNV

⎛⎝⎜

⎞⎠⎟2/3

=6.626 ×10−34 J i s( )28 × 9.1×10−31kg

3π× 5.87 ×1028m−3⎛

⎝⎜⎞⎠⎟2/3

= 8.83×10−19 J .

Using1eV = 1.6 ×10−19 J , EF = 8.83×10−19 J ÷1.6 ×10−19eV i J−1 = 5.52eV .(2points)

AssumingtwoelectronpergoldatomNV=2×5.87×1028m−3 =1.17×1029m−3 .

EF = h2

8me3πNV

⎛⎝⎜

⎞⎠⎟2/3

=6.626 ×10−34 J i s( )28 × 9.1×10−31kg

3π×1.17 ×1029m−3⎛

⎝⎜⎞⎠⎟2/3

= 1.4 ×10−18 J .

EF = 1.4 ×10−18 J ÷1.6 ×10−19eV i J−1 = 8.8eV .(2points)B) Use the answer of part A, to calculate themean Thermal energy,E =U /N ,whereUistheinternal(mean)energyofthesystem,asgivenbytheequationintheappendix.Foridenticalfermions,U = 3NεF / 5→ E =U / N = 3εF / 5 = 3.32eV (2points)C)Assumingthattheconductingelectronscanbetreatedaclassicalideal(MB)gas,findthemeanthermalenergyat300K.ExplainthediscrepancybetweenthevaluesofE obtainedinpartBandC.Usetheequipartitiontheoremthethermalenergyperparticleis

E = U

N= 32kBT = 3

21.381×10−23J iK−1( ) 300K( ) = 6.21×10−21J or0.038eV.Thisis

notsurprisingsinceclassicalgasareallowedtooccupythesamequantumstates,butidenticalfermions(suchaselectrons)cannotoccupythesamelow-energyquantumstatesmakingtheFermienergymuchgreaterthanthethermalenergy(3points)

Problem2)2DMaxwell-BoltzmanDistribution

a) In 2D the speed distribution becomesf2D v( ) = βmvexp −βmv

2

2⎛

⎝⎜⎞

⎠⎟with

v2 = vx

2 + v y2 ,material.Brieflyexplainthemathematicalmeaningoff2Ddv ,and

explain why the normalization off2D requires f2D dv =10

∞

∫ . Verify by direct

integrationthat f2D dv =10

∞

∫ .

f2Ddv istheprobabilitythattheparticlehasspeedintherangev tov +dv .Thetotalprobabilitymustequal1.Forcontinuousdistributionthenormalizationis

donebyintegration:

f2D dv = βmvexp −βmv2

2⎛

⎝⎜⎞

⎠⎟0

∞

∫ =10

∞

∫ .(1point)

Since

vexp −av2( )dv = − 12aexp −av2( )∫

dvf2D0

∞

∫ = exp −βmv2

2⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥0

∞

= −exp −βm∞2

2⎛

⎝⎜⎞

⎠⎟+exp −βm0

2

2⎛

⎝⎜⎞

⎠⎟=1

(3points)

b) Find the average speed in 2D,v of a neutron at T = 300K. Data: Mass of

neutroninequationsheet.HINT: x2exp −ax2( )dx0

∞

∫ = π( )1/2 / 4a3/2( ) .

v = dvvf2D0

∞

∫ = βmv2exp −βmv2

2⎛

⎝⎜⎞

⎠⎟0

∞

∫ = βm π 1/2

4 βm/2( )1/2=

πkBT2m

⎛

⎝⎜⎞

⎠⎟

1/2

m=1.67×10−27kg , v = π ×1.381×10−23 J iK −1 ×300K

2×1.67×10−27kg⎛

⎝⎜⎞

⎠⎟

1/2

=1974ms(3points)

c)Usingdf2Ddv

=0 find themostprobablespeedofaneutronatT=300K,andcomparewithyouranswerinpartb.

df2Ddv

= βmexp −βmv2

2⎛

⎝⎜⎞

⎠⎟− β 2m2v2exp −βmv

2

2⎛

⎝⎜⎞

⎠⎟=0→ v* =

kBTm

⎛

⎝⎜⎞

⎠⎟

1/2

v* = 1.381×10−23 J iK −1 ×300K

1.67×10−27kg⎛

⎝⎜⎞

⎠⎟

1/2

=1575ms(3points)

Problem3)VibrationalEnergyLevelofHClandHBrmolecule.A)AssumethattheHClmoleculebehaveslikeaharmonicoscillatorwithaforceconstantof481N/m.Findtheenergy(ineV)ofitsground(n=0)andfirstexcited(n=1)vibrationalstates.DATA:For1H mH = 1u ;35Cl mCl = 35u ,and1 u = 1.66 ×10−27 kg .Notethedifferencebetweenangularfrequencyω (in s−1 )andν =ω / 2π (unitHz).

Use En = n +1/ 2( )!ω , n = 0,1,2... ,whereω = kµ, andµ =

m1m2

m1 + m2

.

µ = mHmCl

mH +mCl

= 1u × 35u1u + 35u

= 0.97u ×1.66 ×10−27 kg iu−1 = 1.614 ×10−27 kg (1point)

ωHCl =kµ= 481N /m

1.614 ×10−27 kg= 5.46 ×1014 s−1 (1point)

Ground(n=0)

E0HCl = 0.5!ω = 0.5 ×1.055 ×10−34 J • s × 5.46 ×1014 s−1 = 2.88 ×10−20 J = 0.18eV (1

point)FirstExcited(n=1)

E1HCl = 1.5!ω = 1.5 ×1.055 ×10−34 J • s × 5.46 ×1014 s−1 = 8.64 ×10−20 J = 0.54eV (1

point)B)Bromide(80Br mBr = 80u )andCholrine(35Cl mCl = 35u )aregroup17intheperiodictable.Inoneortwosentences,usetheprevioussentencetojustifywhyHBRmoleculeshouldhavesimilarforceconstant,k,asHCl.Thensssumekarethesame,andcalculatethegroundstatevibrationalenergyofHBr.

HBr µ = mHmBr

mH +mBr

= 1u × 80u1u + 80u

= 0.988u ×1.66 ×10−27 kg iu−1 = 1.64 ×10−27 kg (1point)

IfweassumethatHBrhasthesameforceconstant(sincebondingisachemicalpropertythatdependsonelectronclouddistributions,andisindependentofnuclearmass)asHBrthen

ωHBr =kµ= 481N /m

1.64 ×10−27 kg= 5.42 ×1014 s−1 (1point)

Ground(n=0)

E0HBr = 0.5!ω = 0.5 ×1.055 ×10−34 J • s × 5.42 ×1014 s−1 = 2.86 ×10−20 J = 0.178eV ,which

isslightlysmallerthanthevalueforHCl.(1point)

C)SpectroscopicdatafoundthatthetransitionvibrationalfrequencyofHClandHBrare vHCl = 8.66 ×10

13Hz and vHBr = 7.68 ×1013Hz .Isthisdataconsistentwiththe

findingofpartB?Iftheanswerisno,discusswhyinnomorethanfoursentences.FrompartA, vHCl

Theory =ωHCl / 2π = 5.46 ×1014 s−1 / 2π = 8.69 ×1013Hz FrompartB, vHBr

Theory =ωHBr / 2π = 5.42 ×1014 s−1 / 2π = 8.63×1013Hz (1point)Hence vHCl

Theory ≈ vHBrTheory .However,theobservedvaluesare vHCl

Theory > vHBrTheory .Thismeans

thattheassumptionthattheforceconstantofHClandHBrarethesameisnotcompletelyvalid.(2points)Problem4)RotationalspectroscopyTherotationalenergyofadiatomicmolecule

is Erot =

ℓ ℓ+1( )"22I

.

(A)Inaphotonabsorptionexperimentadiatomicmoleculeabsorbsaphoton,andmakesatransitionfromthe ℓ rotationalstatetothe ℓ +1 rotationalstate.Show

explicitlythattheenergyofthephotonis ΔEphoton =

ℓ+1( )"2I

.

For ℓth rotationalstate Erot =

ℓ ℓ+1( )"22I

and ℓ+1th rotationalstate

Erot =

ℓ+1( ) ℓ+ 2( )"22I

=ℓ2 + 3ℓ+ 2( )"2

2I,

whichgives ΔEphoton =

ℓ2 + 3ℓ+ 2( )"22I

−ℓ ℓ+1( )"2

2I=2ℓ+ 2( )"22I

=ℓ+1( )"2I

.(2points)

(B)Aphotonoffrequencyν = 5.1×1011Hz isabsorbedbythediatomicmoleculeHBrresultinginatransitionfromthe ℓ = 0 tothe ℓ = 1 rotationalstate.Usethisinformationtodeterminetheaveragebondlengthofthemolecule.For1H,

mH =1amu ;80Br,mBr =80amu ;where1amu =1.66×10−27kg .80amu =Brm ;where kg271066.1amu 1 −×= .

mH = 1 amu × 1.66 ×10-27kg / amu( ) = 1.66 ×10−27 kg mBr = 80amu × 1.66 ×10-27kg / amu( ) = 1.32 ×10−25 kg (0.5point)

Reducedmassµ =mHmBr

mH + mBr

=1.66 ×10−27 kg( ) 1.32 ×10−25 kg( )1.66 ×10−27 kg +1.32 ×10−25 kg

= 1.64 ×10−27 kg.

(0.5point)Considertheenergyofphotonsproducedby ℓ→ ℓ +1transition:

hν = ΔEphoton =

ℓ +1( )"2I

→ν =ℓ +1( )h2π( )2 I

,set ℓ = 0 anduse I = µR2 toobtain

v = h2π( )2 µR2

,whichgives(3points)

R =h

2π( )2 µv=

6.626 ×10−34 J • s2π( )2 ×1.64 ×10−27 kg × 5.1×1011Hz

= 1.42 ×10−10m (2points)

(C)Microwavecommunicationsystems(suchaswirelessconnectionsignals)operateoverlongdistancesintheatmosphere.Molecularrotationalspectraareinthemicrowaveregion.Brieflyexplainwhyatmosphericgasesdonotabsorbmicrowavestoanygreatextent.Theearth’satmosphereisfilledmostlywithsymmetricmoleculessuchasO2andN2,whichdonotpossesssymmetricdipolemoments,andconsequentlycannotinteractwithmicrowaveradiations.Thisallowsmicrowavetopropagateunhinderedintheatmosphere.(2points)Problem5SuperconductivityInthesuperfluidstate,liquidhelium-4atomsformBose-Einstein(BE)condensatethatformscirculatingvorticesthatdonotdissipate(i.e.donotslowdown).Inthesuperconductingstate,electronsinmetalsformBEcondensatethatconductwithnodissipation(theresistanceofasuperconductoriszero).(A)Notethatelectronsarefermions,butliquidhelium-4atomsarebosons.ExplainhowelectronsinsuperconductingmetalsformBEcondensate.Yourexplanation(nomorethan4sentences)shouldincludetheenergygapEg .Inthesuperconductingstate,twoelectronsareattractivetoeachotherduetophonon-mediatedinteractions.Thetwoelectronsformsacooperpairwithtotalspinzero,andhencearebosonsthatcanformBEcondensate.(1.5points)ThebondingenergyofacooperpairiscalledthegapenergyEg .Thisistheenergyneededtobreakthebondbetweenacooperpair.(1.5points)(B)ThecriticaltemperatureofthesuperconductingstateofVanadiumisTC = 5.4KanditszerotemperatureenergygapisEg 0( ) = 15.8 ×10−4 eV .WhatfractionofelectronsinVanadiumisinthesuperconductingstateattemperatureT = 3.0K ?WhatistheenergygapatT = 3.0K ?Note:assumesuperconductorsobeythesamecondensatefractionequationassuperfluidhelium–seeequationssheet!!

VanadiumisinthesuperconductingstateattemperatureT = 3.0K ?WhatistheenergygapatT = 3.0K ?Note:assumesuperconductorsobeythesamecondensatefractionequationassuperfluidhelium–seeequationssheet!!

UseF = 1− T /Tc( )3/2 = 1− 3.0K5.4K

⎛⎝⎜

⎞⎠⎟3/2

= 0.58 (Thisequationisnotpartof2018W

midterm2)so0.58offreeelectronsareinsuperconductingstate.Eg T = 3( ) = 1.74Eg 0( ) 1− T /Tc( )( )1/2 = 1.74 ×15.8 ×10−4 eV 1− 3K / 5.4K( )( )1/2

= 1.83×10−3eV (2points)(C)TheisotopeeffectequationstatesthatM 0.5TC ≡ constant ,whereMistheatomicmassoftheelement.Innomorethanfoursentencesexplaintheunderlyingphysicalbasisofthisequation.Twonegativelychargeelectronsareelectricallyrepulsive,sotheremustbeanattractiveforcethatmediatesthe“bonding”oftwoelectronstoformaCooperpair.Asdiscussedthisforceisduetolatticevibrationthatcreatesquantummodescalledphonons.Thelatticevibrationeffectisgreaterforlighteratoms,whichhassmalleratomicmass,M.Hencethecriticaltemperature,TC ,ishigher(superconductingstateoccursathighertemperature)forlighterelements,andTC ∝M −0.5 .(2points)UsingM 0.5TC ≡ constant → MV

0.5TC ,V = MHg0.5TC ,Hg ,whereV ≡ Vadanium,Hg ≡Mercury.

TC ,Hg = MV /MHg( )0.5TC ,V = 50u / 201u( )1/2 5.4K = 2.7K (2points)Thecalculatedvalueissignificantlysmallerthanactualexperimentalvalue.HoweveritisstillconsistentwiththeisotopeeffectthatpredictsalowerTC forheavierelements(1point)(D)VanadiumhasatomicmassmV = 50u ,andMercury(Hg)hasatomicmassmHg = 201u .Usethistoestimatethecriticaltemperatureofmercury.TheactualexperimentalonHgisTC = 4.2K .Innomorethanthreesentences,discusswhetheryourcalculationsvalidatetheisotopeeffectofsuperconductivity.UsingM 0.5TC ≡ constant → MV

0.5TC ,V = MHg0.5TC ,Hg ,whereV ≡ Vadanium,Hg ≡Mercury.

TC ,Hg = MV /MHg( )0.5TC ,V = 50u / 201u( )1/2 5.4K = 2.7K (2points)Thecalculatedvalueissignificantlysmallerthanactualexperimentalvalue.HoweveritisstillconsistentwiththeisotopeeffectthatpredictsalowerTC forheavierelements(1point)

Problem6Ro-vibrationalspectrumofCOisshownbelow:UsethedataabovetocalculatethebondlengthandbondspringconstantkofCO.Experimentaldata:R=1.13×10−10m,k=1860N/m.Usethesamemethodasforquestion5ofassignment5

UsefulEquationsCanonicalEnsemble:ConstantVolume,V,particlenumber,N,andtemperature,T.ProbabilityofsystemoccupyingastatewithenergyEisproportionaltotheBoltzmanfactor∝exp −βE( ) ,Maxwell-Boltzman3Dspeeddistribution.FMB v( )dv = 4πN m / 2πkBT( )( )3/2 v2 exp −mv2 / 2kBT( )( )dv ;root-mean-squarespeed

vrms = v2 = 3kBTm

;meanspeedv =

8kBTπm

2Dspeeddistributionf2D v( ) = βmvexp −βmv

2

2⎛

⎝⎜⎞

⎠⎟withv

2 = vx2 + v y

2 ,v = dvvf2D v( )0

∞

∫ ;

3DspeeddistributionIdealgasU =f2kBT fisquadraticdegreeoffreedom;

PV = NKBT = 2U / 3 ;U = 3NkBT / 2 .

RotationalLevels: Eℓ =

L2

2I=ℓ ℓ +1( )2I

"2 ,ℓ = 0,1,2...,where I = µR2 andµ =m1m2

m1 + m2

.

Emission/absorption of photons: i) selection rule Δℓ = ±1; ii) hν = ΔEℓ↔ℓ−1.

ΔEℓ↔ℓ−1 = Eℓ − Eℓ−1 = ℓ"

2

I.Separationbetweenadjacentlines

!

2

I.VibrationalLevel

Evibr = n +1 / 2( )!ω , n = 0,1,2... ,where ω =kµ, and µ =

m1m2

m1 + m2

.

ΔEn+1↔n = En+1 − En = !ω . Absorption and Emission of photons: i) selection ruleΔn = ±1; ii)hν = ΔEn+1↔n . DissociationEnergy( 0U ): Evibr = n +1 / 2( )!ω −U0 .

Ro-vibrational energy is En ,ℓ = "ω n+ 12

⎛⎝⎜

⎞⎠⎟+ "

2

2I ℓ ℓ+1( ) , with the first termassociatedwithvibrationwithquantumnumber,n=0,1,2,…,andthesecondwiththe rotational modes wit quantum numbers ℓ =0,1,2.. Selection Rules

Δn= nf −n0 = ±1 and Δℓ = ℓ f − ℓ0 =0,±1 ,Q-branch Δℓ =0 ;R-branch Δℓ = +1 ;P-branch Δℓ = −1 . Wavenumber ν =1/λ→ν = Eph /hc , Eph is photon energy,

�

E = hν and

�

λ = c /ν ,where

�

c = 2.998 ×108m /s.Fermi-Dirac(FD) fFD = 1/ e ε−µ( )/kBT +1( ) . Bose-Einstein(BE) fBE = 1/ e

ε−µ( )/kBT −1( ) .Fermi Energy εF = h2 / 2m( ) 3N / 8πV( )( )2/3 ; TF = εF / kB ; uF = 2εF /m ;U = 3NεF / 5 ;PV = 2U / 3 Usefulconstants:Atomicmassunit(u)1u

�

=1.66 ×10−27kg;electronrestmass

�

me = 9.11×10−31kg;protonmass

�

mp =1.66 ×10−27kg;

�

e =1.6 ×10−19C;

�

1 eV = 1.6 ×10−19J;

�

h = 6.62 ×10−34 J • s = 4.136 ×10−15eV • s;

�

! =1.055 ×10−34 J • s = 6.582 ×10−16eV • s;Bohr

magneton µB =

e!2m

= 9.274 ×10−24 J /T ; kB = 1.381×10−23J /K = 8.617 ×10−5eV /K .

ε0 =8.854×10−12C2 /N •m2

SuperconductorM 0.5Tc = constant,Mistheatomicweightofthesubstance,whileTcisthecriticalortransitiontemperaturefromthenormaltothesuperconductingstate.EnergyGapEg 0( ) = 3.54kBTC ,Eg T( ) = 1.74Eg 0( ) 1− T /Tc( )( )1/2 ;criticalmagneticfield BC T( ) = BC 0( ) 1− T /Tc( )2( ) .Ferromagnetism:assumeoneunpairedelectroncancontributemagneticmomentof µB (up)or−µB (down),with µB = .MagnetizationM = N+ − N−( )µB ,N+ ≡ numberdensity of up spin,N− ≡ number density of down spin. Maximum magnetizationMmax = NµB ,Nisnumberdensityofunpairedspin.

UsefulIntegrals: exp −ax2( )dx0

∞

∫ = π( )1/2 / 2a1/2( ) ; x2exp −ax2( )dx0

∞

∫ = π( )1/2 / 4a3/2( ) ;

x4 exp −ax2( )dx0

∞

∫ =3 π( )1/2 / 8a3/2( ) ; xexp −ax2( )dx0

∞

∫ = 12a ;

x3exp −ax2( )dx0

∞

∫ = 12a2 ;

exp −ax( ) = 1a0

∞

∫ ;xexp −ax( ) = 1

a20

∞

∫ ;x3exp −ax( ) = 6

a40

∞

∫ .