{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1895_001

1895_001 - 8mm “We_C tsetse;ng 6.58 X1046€V°S 5~29...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 -- ' ...
View Full Document

{[ snackBarMessage ]}