CH5&6

CH5&6 - 5-25 To find the energy width of the...

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Unformatted text preview: 5-25 To find the energy width of the "-ray use "E"t#h2or "E#h2"t#6.58$10%16eV&s2( )0.10$10%9s( )#3.29$10%6eV. As the intrinsic energy width of ~3"10#6eVis so much less than the experimental resolution of 5 eV, the intrinsic width cant be measured using this method. 5-26 The full width at half-maximum (FWHM) is 110 MeV. So "E=55 MeVand using "Emin"tmin=h2, "tmin=h2"E=6.58#10$16eV%s2 55#106eV( )&6.0#10$24s=lifetime ~2"tmin=1.2#10$23s5-27 For a single slit with width a, minima are given by sin"=n#awhere n=1, 2, 3,Kand sin"#tan"=xL, x1L="aand x2L=2"a#x2$x1L="aor "=a#xL=5 $2.1 cm20 cm=0.525 E=p22m=h22m"2=hc( )22mc2"2=1.24$104eV%( )22 5.11$105eV( )0.525 ( )2=546 eV5-29 With oneslit open P1= "12or P2= "22. With bothslits open, P="1+"22. At a maximum, the wavefunctions are in phase so Pmax="1+"2( )2. At a minimum, the wavefunctions are out of phase and Pmin=1 2( )2. Now P1P2="12"22=25or "1"2=5, and PmaxPmin="1+"2( )2"1# "2( )2=5"2+"2( )25"1# "2( )2=6242=3616=2.25. 5-32 (a) f=Eh=1.8( )1.6"10#19J( )6.63"10#34J$s=4.34"1014Hz(b) "=cf=691 nm(c) "E#h"t=6.63$10%34J&s22$10%6s( )"E#5.276$10%29J=3.30$10%10eV6-2 (a) Normalization requires 1="2#$$%dx=A2cos2#L4L4%2&xL()*+,dx=A22()*+,1+cos4&xL()*+,()*+,#L4L4%dxso A=2L. (b) P="2L8#dx=A2cos2L8#2$xL%&()*dx=4L%&()*12%&()*1+cos4$xL%&()*dx%&()*L8#=2L%&()*L8%&()*+2L%&()*L4$%&()*sin4$xL%&()*L8=14+12$=0.4096-3 (a) Asin2"x#$%&()=Asin 5*1010x( )so 2"#$%&()=5*1010m+1, "=2#5$1010=1.26$10%10m. (b) p=h"=6.626#10$34Js1.26#10$10m=5.26#10$24kg m s(c) K=p22mm=9.11"10#31kgK=5.26"10#24kg m s( )22"9.11"10#31kg( )=1.52"10#17JK=1.52"10#17J1.6"10#19J eV=95 eV6-6 "x( ) =Acoskx+Bsinkx#"#x=$kAsinkx+kBcoskx#2"#x2=$k2Acoskx$k2Bsinkx$2mh2%&()*E$U( )"=$2mEh2%&()*Acoskx+Bsinkx( )The Schrdinger equation is satisfied if "2#"x2=$2mh2%&()*E$U( )#or "k2Acoskx+Bsinkx( ) ="2mEh2#$%&(Acoskx+Bsinkx( )....
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CH5&6 - 5-25 To find the energy width of the...

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