60. (a) We use the result of problem 3:Ephoton=hcλ=1240 nm·eV1.00 nm=1.24 keVand for the electronK=p22me=(h/λ)22me=(hc/λ)22mec2=12(0.511 MeV)µ1240 eV·nm1.00 nm¶2.50 eV.(b) In this case, we FndEphoton=1240 nm·eV1.00×10−6nm.24×109eV = 1.24 GeV,and for the electron (recognizing that 1240 eV·nm = 1240 MeV·fm)K=pp2c2+(mec2)2−mec2=p(hc/λ)2mec2)2−mec2=sµ1240 MeV·fm1.00 fm¶2+(0.511 MeV)2−0.511 MeV.24×103MeV = 1.
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.