1895_001 - 8mm “We _C; tsetse;ng ' 6.58 X1046€V°S...

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Unformatted text preview: 8mm “We _C; tsetse;ng ' 6.58 X1046€V°S 5~29. AEAtzh —> Men/At: 21.99x10'21eV 3.823d(8.64x10‘s/d) The energy uncertainty of the excited state is AE, so the (1 energy can be no sharper than m Ag. . 5-30. AxApzh ——> Mpzh —+ Apzh/it. Because-1:111:17,p=hll;thus,Apnp. 5—3 1 . For the cheetah p = mv : 30kg(40ml s) : 1200kg-mls. Because Ap = p * (see Problem 5-30), Ax z h/Ap = SOJ-sllzookg-mls z 4.2 x10'2m = 4.2cm m__ 532. Because c =f2, for photon, it=clf=hclhf zlzclE, so e— h 24 - . .. fl E:~—c:= 1 082m =2.48><1052V —-——~~ _ l 5.0 x10 Hm I ' g h 2.48x105eV - and p: =8.3x10‘7eV-s/m —-e—_ c _ 3x10811115 For electron: h 4.14x10-*5eV.S _ fi___-_m_ “-«—=—-—v«-——__= p4 ’ prm V S'Oxwum 8.3x10 eV S/m “9 Notice that Ap for the electron is 1000 times larger than p for the photon. (a) n(;1/2)=L —>fl=2L/n. Becausel=hlp=:hI\/2mE,then 2 2 2 2 _ 2 2 E: h 2 2.1—2: h "2. IfE, =h2/8mL2, then En = h "2 =n2Ei. 2m/1 2m(L/n) SmL 8mL 2 he 2 1240eV-n 2 (b) For L=0.1nm, El = h 2 =( 22 =__(__,__;”_1.)_{ 8mL 8mc L 8(0.511x106eV)(0.1nm) __ E‘ =37.6eV and E2 = 37.6n2eV 1000 800 600 400 “MM—m mmmm w 200 __ 0 he (0) f=AElh —> clit=AElh -—> 1=W AE For n = 2 -> n :21 transition, AE =112.8eV and 11 = W =11.0nm 112.8eV (d) For n = 3 —> n = 2 trtmsition, AE =188eV and H = 6.6mm 188eV (e) For n = 5 —~—> n ml transition, AB = 903eV and A = M =1.4nm 903eV h 2 5-45. (a) For proton: EI = ( c1 2 from Problem 5—44. 8mpc L ,7,_Wm fi__Wm___ 1240M V- 2 1 = _(.__e_£n..l.i. = 205'MeV and En = ZOSnZMeV 8(933MeV)(1fm) E2 = 820MeV and E3 = 1840MeV (b) Forn = 2 -—>n =1 transition, l———- —————=2.02fin (C) For n = 3 —> n = 2 transition, 1:3 22% 1020MeV AE _ 1240MeV-fm ((1) Porn : 3 —> n =1 transition, 1:11: .— 1635MeV AE =1.22fin = 0.76 fm 5-46. (a) E 2 a2 Izmir} (Equation 5-28) and E = h212m2 (b) For electron with A ‘-= 10—mm : E: (he)2 _ (197.3eronmf =3.816V 2 chzAz 2(0.511x106eV)(1.0'1nm) For electron with A 2 Ian orA 210'2m : E =liner/(10“):1(107nm)2 =3.81x10“‘5eV . 2' 1.055 10-34 - 2 __r(c) E=fi _ ( ><_ J3} 2122152 E 2(100x10'3g ><1(>‘31ciz:/g)(2>404)2 =1.39x20““J = 8.7x10-"43eV -.= #khl’. Also, ES?- = —iaJ‘P. fl mm 2 For the Schrodinger equation: %=ik‘l’ and d T Substituting these into the Schrodinger equation yields: - 7?th I 2m + W? = hw‘I’, which is true, provided ha) = iisz [211: + V, i.e., if E = Ex + V. For the classical wave equation: (from Equation 6-1) (13‘? (12‘? From above: 2 = 4:"? and also dt 2 ~—' 2‘11. Substituting into Equation 6-1 fl k (with T replacing and v replacing c) —k2‘P = (I I v2 )(éwz‘l’), which is true for v = a) I k. ’ do! d 21/! )c x 1 x2 1 WW ——-»~fi 6‘3- (a) a; = "(x’Lzlv’ and a2 = F” 7*" w k Substituting into the time-independent Schrodinger equation, iile . 1‘12 7:2 — + + V x =1? = ' fin [ 2mL4 2an2 W ' ( ) 9” 2minZ W . if hzxz a2 fizxz 1 ”—“—“7i—' Solvmg for V(x), V(x) = ZmLZ -[- 2mL4 + 2mg ] = 2mg =§kx1' where k = h2 [11:12". This is the equation of a parabola centered at x = O. O 7 x _ (b) The classical svstem with thiq damn/1mg“; :a +1“, gamma "Maine... 6—8. I w * wdx = A2 j alt—“’0 x e‘lh‘wldx = 1 rd —a :3 = A2 (2a)_=1 m- = Aszxz Azx I (2a)“2 '.A== Normalization between *00 and +00 is not possible because the value of the integral is infinite. 6—9. (a) The ground State of an infinite well is E1 2 I12 I 8mL2 ; (he)2 [81120213 2 For m : mp. L = 0.1nm: E1 = (1240M6V.fin) 2 = 0.0218V J 8(938.3x105eV)(0.lnm) - (1240MeV-fin)? —0—_ F = ,L=l : E: -- :205MV (b) or m m" . fin l 8(938.3x106eV)(1fm)2 e 6-10. The ground state wave function is (n = 1) 0/1 (x) = 2/ L sin (27x! L) (Equation 6432) _ w The probability of finding the particle in A25 is approximately: =—2.Sin2 fiEé‘ES'i-nz E - r W , L L L L ‘ L 2 0.00215 7 @‘“’** (a) For x = 3 and Ax = 0.002L. P(x)Ax == «LT-15in2 = 0.004 3012 12?— = 0004 2L 2L I 2 0.002L , — (b) For x = ? and P (X)Ax = —(————ls:n2 = 0.004 sin2 3375 = 0.0030 m (c) For x=L and P(x)Ax=0.004sin22r =0 - 22 h2 6-16. E =”" = _E = n2+2n+l _ .1 8m]? and ma. ,, W4 ) h2 he or, AEH =(2n+1)8mL2 w]: —_M “2 NZ [I2 . . MM 30 L: M = M“: = W :0_459nm ’ 8mx 8m:2 8(0.511x1066V) 15-19. (A) y/5(x) = (2/L)”2 sin(2£le)dx —_mfi 0.4L P: I (2iL)sin2(57rle)dx 0.2L Letting 571'le = u, then 57rdx/L =' du andx 2 0.2L —> u = 7: andx = 0.4L ——>.u = 222-, so 2:: ~—V~-Sin2x mfigm P: i Isinzuduz _2_ .1; .2____ =_1_ L 5:: I L 5:: 4 5 7_____._u_—_m__(b) _ .1? = (21L).s.in2 53%, 2)V(0.01L) : 0.02 where 0.9m: Aux - - v -- ' ...
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This note was uploaded on 07/21/2011 for the course PHYS 2213 taught by Professor Erbill during the Summer '10 term at Georgia Institute of Technology.

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1895_001 - 8mm “We _C; tsetse;ng ' 6.58 X1046€V°S...

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