8.5Nuclearradius Wecanusetwotechniquestofindtheradiusofanatomicnucleus. Closestapproachmethod IntheRutherfordscatteringexperiment,alphaparticlesarefiredatathingold foil.Someofthealphaparticlesaredetectedcomingstraightbackfromthegold foil.Thisindicatesthatthepositivelychargedalphaparticlesarebeingrepelled bythepositivelychargedgoldnucleus.Atthepointofclosestapproach.The initialkineticenergyofthealphaparticleπΈ" (whichisafixedvaluefora particularradioactivesource)istotallytransferredtoelectricalpotentialenergy πΈ# . gold alphaparticle r nucleus Wecanwrite: πΈ" = πΈ# 1 πβ ×π2 & β΄ ' πβ π£ ' = 4ππ/ π whereπβ = πππ π πππππβπππππ‘ππππ, πβ = πβππππππππππππππ‘πππ, π2 = πβππππππππππππ’ππππ’π , π = πππ π‘ππππππππππ ππ π‘πππππππβ. (1)!Rearrangetheequationtomakerthesubject. (2)!Takingπβ = 6.64×10H'I ππ, π£ = 2×10I ππ H& , πβ = 2×1.6×10H&L πΆ, π2 = 79×1.6×10H&L πΆ, π0 = 8.85×10H&' πΉπ H& ,workoutavalueforπ. (3)!Howdoesthevalueofrcalculatedcomparetothesizeofthenucleus? (4)!Theclosestapproachmethodproducesanupperlimitonthesizeofthe nucleus.Whyisthis? ©2016flippedaroundphysics.com Electrondiffraction Electronsacceleratedtoclosetothespeedoflight(c)havewavelikeproperties. Theycanbeusedtoinvestigatethesizeoftheatomicnucleus. Wavefrontspassing anobjectand diffracting. wavedirection ThedeBogliewavelengthcanbecalculated: β π= π whereβ = π‘βπππππππππππ π‘πππ‘ = 6.63×10HVW π½π , π = ππππππ‘π’ππππππππ‘πππ. [Astheelectronistravellingclosetothespeedoflight,relativisticcalculations needtobeusedtoworkoutthedeBrogliewavelength.Thisisbeyondthescope ofthiscourse.] Electronsneedtobeacceleratedinanelectricfieldacrossseveralhundred megavolts. (5)!Anelectronisacceleratedthrough400MV.Whatenergydoestheelectron gain? Thesehighenergyelectronsarepassedthroughathinsheetofasolidmaterial (inavacuum)andadetectorisusedrecordthearrivalofelectronsatdifferent angles(π)ontheotherside. detector high π energy electrons thinsheet material ©2016flippedaroundphysics.com ππ’πππππππππππ‘ππππ Agraphofthenumberofelectronsversusangleisshownbelow: π π& Therearetwoeffectstakingplace. 1) Electronsarebeingscattered.Thismeansthattheyarebeingdeflected whentheypassclosetothenucleus.Thisissimilartoalphaparticle scattering,exceptthat,inthiscasetheelectronsareattractedtothe nucleus.Scatteringproducestheoveralldecreaseinthenumberof electronsdetectedastheangleincreases. 2) Electronsarebeingdiffractedandproducinganinterferencepattern.This isresponsibleforthesmalldipatanangleπ& .Thisisafirstorderminima. Theangleatwhichthedipisdetectedcanbeusedtocalculatetheradiusrofthe nucleus. π sin π& = 0.61π whereπ = πππ΅πππππππ€ππ£ππππππ‘βπππ‘βππππππ‘ππππ . (6)!electronswithadeBrogliewavelengthπ = 3×10H&_ πarefiredatatarget containingoxygennuclei.Afirstorderminimaisdetectedatanangleof44°. Estimatetheradiusontheoxygennucleus. Nuclearradiifordifferentelements Electrondiffractionhasbeenusedtofindtheradiiofthenucleiofdifferent elementswithatomicnumberA.Itisfoundthatthereisasimplerelationship betweenatomicnumberAandradiusr. b π = π/ π΄c whereπ/ = 1.05×10H&_ π. (7)!Whatistheradiusofahydrogennucleus? (8)!Whatistheradiusofanitrogennucleus? ©2016flippedaroundphysics.com Nucleardensity Thenucleardensitycanbecalculated: ππ’ππππππππ π π π= = g c ππ’ππππππ£πππ’ππ chi (9)!Calculatethenucleardensityforhydrogen. b (10)!Substituteπ = π/ π΄c intheequationfordensity,above,andsimplify. (11)!Whatistherelationshipbetweenatomicnumberandnucleardensity? ©2016flippedaroundphysics.com
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