8.5 Nuclear radius - Flipped Around Physics

8.5Nuclearradius
Wecanusetwotechniquestofindtheradiusofanatomicnucleus.
Closestapproachmethod
IntheRutherfordscatteringexperiment,alphaparticlesarefiredatathingold
foil.Someofthealphaparticlesaredetectedcomingstraightbackfromthegold
foil.Thisindicatesthatthepositivelychargedalphaparticlesarebeingrepelled
bythepositivelychargedgoldnucleus.Atthepointofclosestapproach.The
initialkineticenergyofthealphaparticle𝐸" (whichisafixedvaluefora
particularradioactivesource)istotallytransferredtoelectricalpotentialenergy
𝐸# .
gold
alphaparticle
r
nucleus
Wecanwrite:
𝐸" = 𝐸# 1 π‘„βˆ ×𝑄2
&
∴ ' π‘šβˆ 𝑣 ' =
4πœ‹πœ€/ π‘Ÿ
whereπ‘šβˆ = π‘šπ‘Žπ‘ π‘ π‘œπ‘“π‘Žπ‘™π‘β„Žπ‘Žπ‘π‘Žπ‘Ÿπ‘‘π‘–π‘π‘™π‘’, π‘„βˆ = π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’π‘œπ‘“π‘Žπ‘™π‘π‘Žπ‘π‘Žπ‘Ÿπ‘‘π‘π‘™π‘’, 𝑄2 =
π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’π‘œπ‘“π‘”π‘œπ‘™π‘‘π‘›π‘’π‘π‘™π‘’π‘’π‘ , π‘Ÿ = π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’π‘œπ‘“π‘π‘™π‘œπ‘ π‘’π‘ π‘‘π‘Žπ‘π‘π‘Ÿπ‘œπ‘Žπ‘β„Ž.
(1)!Rearrangetheequationtomakerthesubject.
(2)!Takingπ‘šβˆ = 6.64×10H'I π‘˜π‘”, 𝑣 = 2×10I π‘šπ‘  H& , π‘„βˆ = 2×1.6×10H&L 𝐢, 𝑄2 =
79×1.6×10H&L 𝐢, πœ€0 = 8.85×10H&' πΉπ‘š H& ,workoutavalueforπ‘Ÿ.
(3)!Howdoesthevalueofrcalculatedcomparetothesizeofthenucleus?
(4)!Theclosestapproachmethodproducesanupperlimitonthesizeofthe
nucleus.Whyisthis?
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Electrondiffraction
Electronsacceleratedtoclosetothespeedoflight(c)havewavelikeproperties.
Theycanbeusedtoinvestigatethesizeoftheatomicnucleus.
Wavefrontspassing
anobjectand
diffracting.
wavedirection
ThedeBogliewavelengthcanbecalculated:
β„Ž
πœ†= 𝑝
whereβ„Ž = π‘‘β„Žπ‘’π‘ƒπ‘™π‘Žπ‘›π‘π‘˜π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ = 6.63×10HVW 𝐽𝑠, 𝑝 = π‘šπ‘œπ‘šπ‘’π‘›π‘‘π‘’π‘šπ‘œπ‘“π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›.
[Astheelectronistravellingclosetothespeedoflight,relativisticcalculations
needtobeusedtoworkoutthedeBrogliewavelength.Thisisbeyondthescope
ofthiscourse.]
Electronsneedtobeacceleratedinanelectricfieldacrossseveralhundred
megavolts.
(5)!Anelectronisacceleratedthrough400MV.Whatenergydoestheelectron
gain?
Thesehighenergyelectronsarepassedthroughathinsheetofasolidmaterial
(inavacuum)andadetectorisusedrecordthearrivalofelectronsatdifferent
angles(πœƒ)ontheotherside.
detector
high
πœƒ
energy
electrons
thinsheet
material
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π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘œπ‘“π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘ 
Agraphofthenumberofelectronsversusangleisshownbelow:
πœƒ
πœƒ& Therearetwoeffectstakingplace.
1) Electronsarebeingscattered.Thismeansthattheyarebeingdeflected
whentheypassclosetothenucleus.Thisissimilartoalphaparticle
scattering,exceptthat,inthiscasetheelectronsareattractedtothe
nucleus.Scatteringproducestheoveralldecreaseinthenumberof
electronsdetectedastheangleincreases.
2) Electronsarebeingdiffractedandproducinganinterferencepattern.This
isresponsibleforthesmalldipatanangleπœƒ& .Thisisafirstorderminima.
Theangleatwhichthedipisdetectedcanbeusedtocalculatetheradiusrofthe
nucleus.
π‘Ÿ sin πœƒ& = 0.61πœ†
whereπœ† = π‘‘π‘’π΅π‘Ÿπ‘œπ‘”π‘™π‘–π‘’π‘€π‘Žπ‘£π‘’π‘™π‘’π‘›π‘”π‘‘β„Žπ‘œπ‘“π‘‘β„Žπ‘’π‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›π‘ .
(6)!electronswithadeBrogliewavelengthπœ† = 3×10H&_ π‘šarefiredatatarget
containingoxygennuclei.Afirstorderminimaisdetectedatanangleof44°.
Estimatetheradiusontheoxygennucleus.
Nuclearradiifordifferentelements
Electrondiffractionhasbeenusedtofindtheradiiofthenucleiofdifferent
elementswithatomicnumberA.Itisfoundthatthereisasimplerelationship
betweenatomicnumberAandradiusr.
b
π‘Ÿ = π‘Ÿ/ 𝐴c whereπ‘Ÿ/ = 1.05×10H&_ π‘š.
(7)!Whatistheradiusofahydrogennucleus?
(8)!Whatistheradiusofanitrogennucleus?
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Nucleardensity
Thenucleardensitycanbecalculated:
π‘›π‘’π‘π‘™π‘’π‘Žπ‘Ÿπ‘šπ‘Žπ‘ π‘ 
𝑀
𝜌=
= g c
π‘›π‘’π‘π‘™π‘’π‘Žπ‘Ÿπ‘£π‘œπ‘™π‘’π‘šπ‘’ chi
(9)!Calculatethenucleardensityforhydrogen.
b
(10)!Substituteπ‘Ÿ = π‘Ÿ/ 𝐴c intheequationfordensity,above,andsimplify.
(11)!Whatistherelationshipbetweenatomicnumberandnucleardensity?
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