/
Answer Key HW  Chapter  assignment Conceptual Problem Answer Key HW  Chapter  assignment Conceptual Problem

Answer Key HW Chapter assignment Conceptual Problem - PDF document

olivia-moreira
olivia-moreira . @olivia-moreira
Follow
426 views
Uploaded On 2015-04-28

Answer Key HW Chapter assignment Conceptual Problem - PPT Presentation

Of the 10 electroweak particles Table 181 which ones travel at or near lightspeed To travel at or near lightspeed the mass needs to be zero or extremely close to zero The photon with zero mass will certainly travel at lightspeed The neutrinos with a ID: 56366

the electroweak

Share:

Link:

Embed:

Download Presentation from below link

Download Pdf The PPT/PDF document "Answer Key HW Chapter assignment Conce..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.


Presentation Transcript

canexistontheirownasasingleparticle,butquarksmustbecon ned ,appearingonlywithotherquarks.2Problems2.)\Aproton-antiprotonpair,atrest,annihilateandcreatetwophotons.Usingtheinfor-mationintheprecedingproblem,andthefactthataprotonis1,800timesmoremassivethananelectron, ndthefrequencyofeachphoton."Theinformationfromtheprecedingproblemstatesthatthefrequencyofphotonsemittedwhenanelectron-positronpairannihilateis1020Hz.Theprocessforthistohappenise�+e+!2 ,where standsforaphoton.Wewilluseenergyconservation,settingtheinitialenergyequaltothe nalenergy.Theinitialenergycomesjustfromrestmass:Ei=2mec2,wheremeisthemassofanelectron/positron(theyhavethesamemass),andthefactorof2comesfromthefactthattheelectronandpositionhavethesamemass.The nalenergyistheenergyofthetwophotons:Ef=2hfe,wherefe=1020Hzisthefrequencywearetoldfromproblem1.).EnergyconservationsaysEi=Ef!2mec2=2hfe.Simplifyingthisexpressionbydividingoutthecommonfactorof2:mec2=hfe.Howdowe gureoutthefrequencyforaprotonandantiproton?Theonlythingdif-ferentisthemass:themassoftheproton/antiprotonis1800timesthemassoftheelec-tron/position.Thus,wecanincreasethemassbyafactorof1800,andseewhathappenstothefrequency:(1800me)c2=h(1800fe).Thus,thefrequencyincreasesby1800aswell!Thenewfrequencyis 18001020Hz=1:81023Hz .4.)\Makingestimates.Alargeelectricpowerplantgenerates1000MWofelectricity.Iftheenergycamefrommatter-antimatterannihilation,aboutwhattotalmassofmatterandofantimatterwouldberequiredeachyear,assumingthattheelectricityisgeneratedatanenergyeciencyof50%?"Themainideaweneedisthis:(Energyneededfor1year)=(energyfrommatter-antimatterannihilation).Let'sworkontheleft-handside rst:togettheenergyneededfor1year,wewillmultiplythepowerby1year,inseconds(thisworksbecausepowerisenergydividedbytime-weneed1yearinsecondsbecauseaWatt=1Joule/1sec,andmultiplyingbysecondscancelsthoseunits,leavinguswithJoules).Energyneededfor1year=(Power)(1yearinsec)=1000MW(3:1107sec)=1000106W(3:1107sec)=3:11016J:(1)Nowletsworkontheright-handside:theenergyfrommatter-antimatterannihilationcomesmainlyfromthe(unknownamountof)restmassofthematterandantimatter(seeprob(2)above):2mc2.However,theplantoperatesat50%eciency,soweonlygethalfof2