capítulo 41 (5th edition)

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Chapter 41 Solutions 41.1 (a) λ = = × × ( ) ( ) = h mv 6 626 10 0 400 34 . . Js 1.67 10 kg m/s -27 9 92 10 7 . × m (b) For destructive interference in a multiple-slit experiment, d m sinθ λ = + 1 2 with m = 0 for the first minimum. Then, θ λ = = ° sin . 1 2 0 0284 d y L = tan θ so y L = = ( ) ° ( ) = tan . tan . θ 10 0 0 0284 m 4 96 . mm (c) We cannot say the neutron passed through one slit. We can only say it passed through the slits. 41.2 Consider the first bright band away from the center: d m sinθ λ = 6 00 10 0 400 200 1 1 20 10 8 1 10 . sin tan . . × ( ) = = × m m λ λ = h mv e so mv h e = λ and K mv m v m h m e V e e e e = = = = ( ) 1 2 2 2 2 2 2 2 2 λ V h em e = = × ( ) × ( ) × ( ) × ( ) = 2 2 34 2 19 31 10 2 2 6 626 10 2 1 60 10 9 11 10 1 20 10 λ . . . . Js C kg m 105 V 41.3 (a) The wavelength of a non-relativistic particle of mass m is given by λ = = h p h mK / 2 where the kinetic energy K is in joules. If the neutron kinetic energy K n is given in electron volts, its kinetic energy in joules is K K n = × ( ) 1 60 10 19 . J/eV and the equation for the wavelength becomes λ = = × × ( ) × ( ) = h mK K n 2 6 626 10 2 1 67 10 1 60 10 34 27 19 . . . Js kg J/eV 2 87 10 11 . × K n m where K n is expressed in electron volts. (b) If K n = = 1 00 1000 . keV eV , then

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Chapter 41 Solutions41.1 (a) ( )( ) hmv6 626 100 40034.. J s1.67 10 kg m/s-279 92 107. m(b) For destructive interference in a multiple-slit experiment,d m sin +|.`,12with m 0 for the first minimum. Then, |.`, sin .120 0284dyL tan so y L ( ) ( ) tan . tan . 10 0 0 0284 m 4 96 . mm(c) We cannot say the neutron passed through one slit. We can only say it passed through the slits.41.2 Consider the first bright band away from the center: d m sin 6 00 100 4002001 1 20 108 1 10. sin tan.. ( )

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m m hm veso m v he andK m v m vmhm e Ve ee e ( )1222 2 222 2 V heme ( )( ) ( ) ( ) 2234219 31 10226 626 102 1 60 10 9 11 10 1 20 10.. . . J s C kg m105 V41.3 (a) The wavelength of a non-relativistic particle of mass m is given by h p h mK / 2 where thekinetic energy K is in joules. If the neutron kinetic energy Kn is given in electron volts, its kineticenergy in joules is K Kn ( )1 60 1019. J/ eV and the equation for the wavelength becomes ( ) ( ) hmKKn26 626 102 1 67 10 1 60 103427 19.. . J s kg J/ eV 2 87 1011. Kn mwhere Kn is expressed in electron volts.(b) If Kn 1 00 1000 . keV eV, thenChapter 41 Solutions 49541.4 hphmK 2, so K hm222 If the particles are electrons and ~ . 0 1 1010 nm m , the kinetic energy in electron volts isK ( )( )( ) |.

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6 626 102 9 11 10 10134231 102.. J s kg m eV1.602 10 J-19 ~ 102 eV41.5 hp p h 6 626 101 00 106 63 10341123...J s m kg m/ s(a) electrons: K pmee ( )( ) 23423126 63 102 9 11 10.. J s J 15 1 . keVThe relativistic answer is more precisely correct:K p c m c m ce e e +( ) 2 2 2 41 2214 9// . keV(b) photons: E pc ( ) ( ) 6 63 10 3 00 1023 8. . 124 keV41.6 The theoretical limit of the electron microscope is the wavelength of the electrons. IfKe 40 0 . keV, then E K m ce e + 2551 keV andpc E m ce ( ) ( )|.

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1551 5113 00 101 60 101 001 10 102 2 42 281622keV keV m/ s J keV kg m/ s....The electron wavelength, and hence the theoretical limit of the microscope, is then hp6 626 106 03 103412..J s1.10 10 kg m/ s m226 03 . pm41.7 E K m ce + + 21 00 0 511 1 51 . . . MeV MeV MeVp E m ce2 2 2 4 2 21 51 0 511 c MeV MeV2 ( ) ( ) . . sop c 1 42 . MeV/ ( ) ( )( ) ( ) hphc1 426 626 10 3 00 101 42 10 1 60 108 74 1034 86 1913.. .. .. MeV J s m/ s J mSuppose the array is like a flat diffraction grating with openings 0 250 . nm apart:d m sin | `135 8 74 10Chapter 41 Solutions 49641.8 (a) p x m v x h/ 2 sov hm x ( )( ) 422 00 1 00 J s4 kg m . .0 250 . m/ s(b) The duck might move by 0 25 5 1 25 . . m/ s s m ( )( ) . With original position uncertainty of1 0 . 0 m, we can think of x growing to 1 00 1 25 . . m m + 2 25 . m41.9 For the electron, p m ve ( )( ) ( ) 9 11 10 500 1 00 10 4 56 1031 4 32. . . kg m/ s kg m/ sx hp ( ) 46 626 104 56 103432 .. J s4 kg m/ s1 16 . mmFor the bullet, p m v ( )( ) ( ) 0 0200 500 1 00 10 1 00 104 3. . . kg m/ s kg m/ sx hp 45 28 1032. mGoal SolutionAn electron ( me 9 11 1031. kg ) and a bullet ( . m 0 0200 kg) each have a speed of500 m/ s, accurate to within 0 0100 . %. Within what limits could we determine the position of theobjects?G: It seems reasonable that a tiny particle like an electron could be located within a more narrow regionthan a bigger object like a bullet, but we often find that the realm of the very small does not obeycommon sense.O: Heisenbergs uncertainty principle can be used to find the uncertainty in position from theuncertainty in the momentum.A: The uncertainty principle states: x px h/ 2 where p m vx and h h/ 2 .Both the electron and bullet have a velocity uncertainty, v ( )( ) 0 000100 500 0 0500 . . m/ s m/ sFor the electron, the minimum uncertainty in position isx hm v ( )( ) 46 63 109 11 10 0 05001 163431 .. ..J s4 kg m/ s mmFor the bullet,x hm v ( )( ) 46 63 100 0200 0 05005 28 103432 ..J s4 kg m/ s mChapter 41 Solutions 497Chapter 41 Solutions 49841.10 yxppyx and d p hy / 4 Eliminate py and solve for x .x p y dhx ( ) ( )( ) ( ) ( ) ( ) 4 4 1 00 10 100 1 00 102 00 106 626 103 2334 . ... kg m/ s m m J s3 79 1028. mThis is 190 times greater than the diameter of the Universe!41.11 p x h2so p m v hxe 4v hm xe ( ) ( ) 46 626 109 11 103431 .. J s4 kg 5.00 10 m111 16 106. m/ s41.12 With x 2 1015 m, the uncertainty principle requires pxx h22 6 1020. kg m/ sThe average momentum of the particle bound in a stationary nucleus is zero. The uncertainty inmomentum measures the root-mean-square momentum, so we take prms 3 1020 kg m/ s .For an electron, the non-relativistic approximation p m ve would predict v 3 1010 m/ s,while v cannot be greater than c.Thus, a better solution would be E m c pc m ce e ( ) +( )

]]] 2221 2256/ MeV 110112 2v c /sov c 0 99996 .For a proton, v p m / gives v 1 8 107. m / s, less than one-tenth the speed of light.41.13 (a) At the top of the ladder, the woman holds a pellet inside a small region xi. Thus, the uncertaintyprinciple requires her to release it with typical horizontal momentum p m v xx x i h/ 2 . Itfalls to the floor in time given by H gt + 0122 as t H g 2 , so the total width of the impactpoints is x x v t xm xHg x Axf i x iiii + ( ) + |.

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+h22, whereA H h 2Chapter 41 Solutions 499so x Ai , and the minimum width of the impact points is x x Ax Af iix Ai( ) +|.

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min 22 21 21 4hmHg|.`, |.

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]]] min//....2 1 0546 105 00 102 2 009 803441 21 4 J s kg m m/ s25 19 1016. mChapter 41 Solutions 50041.14 Probability P x ax adx aaxaaaaaaa ( ) +( ) |.`,|.`, 22 211tanP ( )[ ] |.`,

]]] 11 114 41 1 tan tan 1 2 /41.15 (a) x A xA x ( ) |.`, ( )sin sin .25 00 1010so 25 00 1010 . m1 ( ) 25 00 1010.1 26 1010. m(b) p h 6 626 103410. J s1.26 10 m5 27 1024. kg m/ s(c) m 9 11 1031. kgK pm ( ) ( ) 22423117171925 27 102 9 11 101 52 101 52 101 602 10..... kg m/ s kg J J J/ eV95 5 . eV41.16 For an electron to fit into an infinitely deep potential well, anintegral number of half-wavelengths must equal the width of thewell.n21 00 109 . m so 2 00 109.nhp(a) Since K pmhmhmnne e e ( )( ) ( )22 22 29222 2 22 100 377/. eVFor K 6 eV , n 4(b) With n 4 , K 6 03 . eV41.17 (a) We can draw a diagram that parallels our treatment ofstanding mechanical waves. In each state, we measure thedistance d from one node to another (N to N ), and base oursolution upon that:Since dN to N 2and hChapter 41 Solutions 501Next,K pmhm d de e ( )( )

]]]]]2 22342312 816 626 108 9 11 10.. J s kgEvaluating, Kd 6 02 10382. J m2Kd 3 77 10192. eV m2In state 1, d 1 00 1010. mK137 7 . eVIn state 2, d 5 00 1011. mK2151 eVIn state 3, d 3 33 1011. mK3339 eVIn state 4, d 2 50 1011. mK4603 eV(b) When the electron falls from state 2 to state 1, it puts outenergyE hf hc 151 37 7 113 eV eV eV .i n t o e m i t t i n g a p h o to n o f w a v e l e n g t h ( ) ( )( ) ( ) hcE6 626 10 3 00 10113 1 60 1011 034 819. ... J s m/ s eV J/ eV nmThe wavelengths of the other spectral lines we find similarly:TransitionE eV ( ) nm ( )Chapter 41 Solutions 50241.18 E1192 00 3 20 10 . . eV JFor the ground-state, E hm Le1228(a) L hm Ee 84 34 10110. m 0 434 . nm(b) E E E hm Lhm Le e |.

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2 1222248 86 00 . eV41.19 E hc hm Lhm Le e |.

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[ ]222 22282 138L hm ce 387 93 1010. m 0 793 . nm41.20 E hc hm Lhm Le e |.

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[ ]222 22282 138so L 38hm ce41.21 E n hmLn 2 228so E E E hmLhcmc L ( )2 12222 23838and E hf hc Hence, ( ) ( ) ( )838 938 10 1 00 103 12402 26 52mc Lhc eV nm eV nm. 2 02 104. nm (gamma ray)E hc 6 15 . MeVFigureforGoalSolutiChapter 41 Solutions 503Goal SolutionThe nuclear potential energy that binds protons and neutrons in a nucleus is often approximated by asquare well. Imagine a proton confined in an infinitely high square well of width 10 0 . fm , a typicalnuclear diameter. Calculate the wavelength and energy associated with the photon emitted when theproton moves from the n 2 state to the ground state. In what region of the electromagneticspectrum does this wavelength belong?G: Nuclear radiation from nucleon transitions is usually in the form of high energy gamma rays withshort wavelengths.O: The energy of the particle can be obtained from the wavelengths of the standing wavescorresponding to each level. The transition between energy levels will result in the emission of aphoton with this energy difference.A: At level 1, the node-to-node distance of the standing wave is 1 00 1014. m, so the wavelength istwice this distance: h p / . 2 00 1014 m. The protons kinetic energy isK mv pmhm ( )( ) ( ) 1222 2234227 14213192 26 63 102 1 67 10 2 00 103 29 101 60 102 06.. .... J s kg m J J/ eV MeVIn the first excited state, level 2, the node-to-node distance is two times smaller than in state 1. Themomentum is two times larger and the energy is four times larger: K 8 23 . MeV.The proton has mass, has charge, moves slowly compared to light in a standing-wave state, and staysinside the nucleus. When it falls from level 2 to level 1, its energy change is2 06 8 23 6 17 . . . MeV MeV MeV Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at the speedof light, and that it has an energy of +6 17 . MeV.Its frequency is f Eh ( ) ( ) 6 17 1 60 106 63 101 49 10193421. ...10 eV J/ eV J s Hz6and its wavelength is cf3 00 101 49 102 02 10821 113...m/ s s mThis is a gamma ray, according to Figure 34.17.L:The radiated photons are energetic gamma rays as we expected for a nuclear transition. In the abovecalculations, we assumed that the proton was not relativistic ( v c < 0 1 . ), but we should check thisassumption for the highest energy state we examined ( n 2):v Km c ( ) ( ) 22 8 23 10 1 60 101 67 103 97 10 0 1336 19277. ... . eV J/ eV kg m/ sThis appears to be a borderline case where we should probably use relativistic equations, but ourclassical treatment should give reasonable results, within ( . ) % 0 133 12 accuracy.Chapter 41 Solutions 50441.22 2D for the lowest energy stateK pmhmhmD ( )[ ] ( ) 2 222 3427 142142 2 86 626 108 4 1 66 10 1 00 108 27 10( .. ..J s) kg m J20 517 . MeVp h hD ( ) 26 626 102 1 00 103414.. J s m3 31 1020. kg m/ s41.23 E hmL nn |.

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2228E134227 142146 626 108 1 67 10 2 00 108 21 10 ( )( ) ( ) .. .. J s kg m JE1 0 513 . MeV E E2 14 2 05 . MeV E E3 19 4 62 . MeV41.24 (a) < > |.`, |.`, x xLxL dxL x xL dxL L2 2 2 1212420 0sin cos < > +

]]] xLxLL xLxLxLLL121164 4 420220 sin cos L/ 2(b) Probability |.`,

]]] 2 2 1 14420 4900 5100 4900 510LxL dxL xLL xLLLLLsin sin....Probability ( ) 0 20142 04 1 96 . sin . sin . 5 26 105. (c) Probability

]]] xLxL LL1440 2400 260sin..3 99 102. (d) In the n 2 graph in Figure 41.11 (b), it is more probable to find the particle either nearx L4or x L34than at the center, where the probability density is zero.Chapter 41 Solutions 505A n xL dx A LxL2 2 2021 sin |.`, |.`, or AL241.26 The desired probability is P dxLxL dxxx L L |.`, 2042042 2/ /sinwhere sincos21 22 Thus,P xLxLL |.`, +|.`, 144 140 0 004sin/0 250 .41.27 In 0 x L, the argument 2 x L / of the sine function ranges from 0 to 2 . The probabilitydensity 2 22/ sin / L x L ( ) ( ) reaches maxima at sin 1 and sin 1 at22 xL and2 32 xL The most probable positions of the particle are at x Lx L 434 and 41.28 (a) The probability is P dxLxL dxLxL dxL L L |.`, |.`, 203203032 2 12122/ / /sin cosP xLxLL |.`, |.`, |.

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122 131223133403sin sin/0 196 .(b) The probability density is symmetric about x L / 2. Thus,the probability of finding the particle between x L 2 3 / andx L is the same 0 196 . . Therefore, the probability of findingit in the range L x L / / 3 2 3 isP ( ) 1 00 2 0 196 0 609 . . . .(c) Classically, the electron moves back and forth with constantspeed between the walls, and the probability of finding theelectron is the same for all points between the walls. Thus, theclassical probability of finding the electron in any range equalto one-third of the available space is Pclassical 1 3 / .Chapter 41 Solutions 50641.29 The ground state energy of a particle (mass m) in a 1-dimensional box of width L is E hmL1228 .(a) For a proton m ( )1 67 1027. kg in a 0 200 . - nm wide box:E134227 102226 626 108 1 67 10 2 00 108 22 10 ( )( ) ( ) .. .. J s kg m J 5 13 103. eV(b) For an electron m ( )9 11 1031. kg in the same size box:E134231 102186 626 108 9 11 10 2 00 101 51 10 ( )( ) ( ) .. .. J s kg m J 9 41 . eV(c) The electron has a much higher energy because it is much less massive.41.30 (a) 12xLxL( ) |.`,cos P x xLxL1 1222( ) ( ) |.`, cos 22 2xLxL( ) |.`,sin P x xLxL2 2222 2( ) ( ) |.`, sin 32 3xLxL( ) |.`,cos P x xLxL3 32 2 3( ) ( ) |.`, cos41.31 We have ( )Aei kx tand 222x k Schrdingers equation: 22222x k mE U ( )hSince k p p22222222 2 ( ) ( )h hand E U p m ( ) 22 /Chapter 41 Solutions 507*41.32 x A kx B kx ( ) + cos sin x kA kx kB kx + sin cos 222 2x k A k x k B kx cos sin ( ) + ( )2 22 2mE U mEA kx B kxh h cos sinTherefore the Schrdinger equation is satisfied if 22 22xmE U |.`, ( )hor + ( ) |.`, + ( ) k A kx B kx mEA kx B kx222cos sin cos sinhThis is true as an identity (functional equality) for all x if E km h2 2241.33 Problem 45 in Ch. 16 helps students to understand how to draw conclusions from an identity.(a) x A xL( ) |.

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122ddxAxL 22 22 22xAL Schrdingers equation 22 22xmE U ( )hbecomes |.

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+( ) |.

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( )2 21212 222 22 2222 2 2ALmEA xLmx A xLmL L x h hh + 12 222 224LmE mExLxL h hThis will be true for all x if both12 2LmEhand mEL L h2 2 410 Both of these conditions are satisfied for a particle of energy EL m h22.(b) For normalization,1 1 12222222244 |.

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A xL dx A xLxL dxLLLL123 523 523 51615232542 2 +

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]]] A x xLxL A L L LL L LA LLLAL1516(c)Chapter 41 Solutions 508P 47810 580 .Chapter 41 Solutions 50941.34 (a) Setting the total energy E equal to zero and rearranging the Schrdinger equation to isolate thepotential energy function givesU xmddx( ) |.

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h2 2221If x Axe x L( ) 2 2/ Then ddx Ax AxL eLx L 223 244 62 2 ( ) /orddxx LL x222 244 6 ( )( )andU xmLxL( ) |.

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h2222246S ee figure to the right .41.35 (a) S ee figure to the right .(b) The wavelength of the transmitted wavetraveling to the left is the same as the originalwavelength, which equals 2L.441.37 T e CL 2 (Use Equation 41.17)22 2 9 11 10 8 00 101 055 102 00 10 4 5831 193410CL ( ) ( ) ( ) . .. . FiChapter 41 Solutions 510orGoalSolutionChapter 41 Solutions 511Goal SolutionAn electron with kinetic energy E 5 00 . eV is incident on a barrier with thickness L 0 100 . nmand height U 10 0 . eV (Fig. P41.37). What is the probability that the electron (a) will tunnelthrough the barrier and (b) will be reflected?G: Since the barrier energy is higher than the kinetic energy of the electron, transmission is not likely,but should be possible since the barrier is not infinitely high or thick.O: The probability of transmission is found from the transmission coefficient equation 41.18.A: The transmission coefficient isC m U E ( )( ) ( ) ( ) 22 9 11 10 10 0 5 00 1 60 106 63 101 14 1031 193410h. . . ... kg eV eV J/ eV J s/ 2 m-1(a) The probability of transmission isT e e eCL ( ) ( ) 22 1 14 10 2 00 104 5810 1 100 0103. ...m m(b) If the electron does not tunnel, it is reflected, with probability 1 0 0103 0 990 . .L:Our expectation was correct: there is only a 1% chance that the electron will penetrate the barrier. Thistunneling probability would be greater if the barrier were thinner, shorter, or if the kinetic energy ofthe electron were greater.41.38 C ( ) ( ) ( ) 2 9 11 10 5 00 4 50 1 60 1031 19. . . . kg m/ s1.055 10 J s-34T e CL ( ) ( )[ ] ( ) 2 9 1 122 3 62 10 950 10 6 88 exp . exp . m mT 1 03 103. 41.39 From problem 38, C 3 62 109 1. m10 2 3 62 106 9 1 ( )[ ]exp . m LTaking logarithms, ( )13 816 2 3 62 109 1. . m LNew L 1 91 . nmChapter 41 Solutions 512*41.40 With the wave function proportional to e CL , the transmission coefficient and the tunneling currentare proportional to 2, to e CL .Then, IIeee0 5000 5152 10 0 0 5002 10 0 0 51520 0 0 015... .. .. .nm nm / nm nm / nm nm( )( ) ( )( ) ( )( ) ( )1 35 .*41.41 With transmission coefficient e CL , the fractional change in transmission ise eeeL LL ( ) ( ) + ( ) ( ) ( ) 2 10 0 2 10 0 0 002002 10 029 0 0 002001 0 0392. . ... .. / nm / nm nm / nm3 92 . %41.42 ( )Be m x / 22hso ddxmx |.`,hand ddxmx m2222 |.`, + |.`,h hSubstituting into Equation 41.19 givesmx m mE mx h h h h|.`, + |.`, |.`, + |.`,222222which is satisfied provided that E h2.41.43 Problem 45 in Chapter 16 helps students to understand how to draw conclusions from an identity. Axe bx2so ddx Ae bx Aebx bx 2 222andddx bxAe bxAe b x Ae b b xbx bx bx222 3 2 22 4 4 6 42 2 2 + + Substituting into Equation 41.19, + |.`, + |.`,6 422 222b b x mE mx h hFor this to be true as an identity, it must be true for all values of x .So we must have both 622b mEhand 422b m |.`,h(a) Therefore b m 2h(b) and E bm 32h 32h(c) The wave function is that of the f irst excited state .Chapter 41 Solutions 513*41.44 The longest wavelength corresponds to minimum photon energy, which must be equal to the spacingbetween energy levels of the oscillator:hc km h h so ( ) |.

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2 2 3 00 109 11 108 998311 2c mk.../ m/ s kg N/ m600 nm*41.45 (a) With ( )Be m x / 22h, the normalization condition 21 dx allbecomes 1 2 2122 2 2 2 2 2022 2 ( ) ( ) B e dx B e dx Bmm x m x / //h hhwhere Table B.6 in Appendix B was used to evaluate the integral.Thus, 12 Bmhand B m |.

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h1 4 /(b) For small , the probability of finding the particle in the range / / 2 2 x is 22222 00 ( ) //dx B e mh|.

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1 2 /41.46 (a) With < > x 0 and < > px0, the average value of x2 is x ( )2 and the average value of px2 ispx( )2. Then x px h/ 2 requiresE pmkpxx + 2 222 2 4h pmkpxx2 222 8+ h(b) To minimize this as a function of px2, we require dEdp mkpx x224012 811 + ( )hThen kp mxh24812p mk mkx221 228 2 |.

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h h/and E mkmkmkkmkm( ) + +h hhh h2 238 4 42.Chapter 41 Solutions 514*41.47 Suppose the marble has mass 20 g. Suppose the wall of the box is 12 cm high and 2 mm thick.While it is inside the wall,U mgy ( )( )( ) 0 02 9 8 0 12 0 0235 . . . . kg m/ s m J2and E K mv ( )( ) 1221220 02 0 8 0 0064 . . . kg m/ s JThen C m U E ( ) ( )( ) 2 2 0 02 0 01711 055 102 5 103432 1h. ...kg J J s mand the transmission coefficient ise e e eCL ( ) ( ) ( ) 22 2 5 10 2 1010 102 30 4 3 104 3 1032 329292910. . ..~ 101030*41.48 (a) 2L 2 00 1010. m(b) p h 6 626 103410. J s2.00 10 m3 31 1024. kg m/ s(c) E pm 220 172 . eV41.49 (a) S ee the first figure to the right .(b) S ee the second figure to the right .(c) is continuous and 0 as x t(d) Since is symmetric, 2 202 1 dx dx or22212 2020A e dx Ae ex |.

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( ) This gives A Chapter 41 Solutions 515(a) Use Schrdingers equation 22 22xmE U ( )hwith solutions 11 1 + Ae Beik x ik x[region I]22 Ceik x[region II]Where k mE12handk m E U22 ( )hThen, matching functions and derivatives at x 0: 1020( ) ( ) + A B Candddxddx k A B k C 10201 2|.`, |.`, ( ) Then B k kk k A +112 12 1Ck k A +212 1/Incident wave Aeikx reflects Be ikx , with probability R BAk kk k ( )+ ( ) 222 122 1211//k kk k1 221 22 ( )+ ( )(b) With E 7 00 . eV and U 5 00 . eV, kkE UE212 007 000 535 ...The reflection probability is R ( )+ ( ) 1 0 5351 0 53522..0 0920 .The probability of transmission is T R 1 0 908 .41.51 R k kk kk kk k ( )+ ( ) ( )+ ( )1 221 222 122 1211//h2 22km E U for constant Uh km E2122 since U 0 (1)Chapter 41 Solutions 516Dividing (2) by (1), kkUE22121 11212 sokk2112and therefore,R ( )+( ) ( )+( ) 1 1 21 1 22 12 12222//0 0294 .Chapter 41 Solutions 51741.52 (a) The wave functions and probability densities are the same as those shown in the two lower curves inFigure 41.11 of the textbook.(b)P dx xdx x x1 120 15020 1500 1500 35021 00 1 002 0021 00421 00 |.`,|.`, |.`,

]]] .sin.. .sin.. ... nm nm nm nm nm nm0.350 nm nm0.350 nm nm nmIn the above result we used sin / / sin22 1 4 2 axdx x a ax ( ) ( ) ( )P x x10 1500 3501 00 1 00221 00 |.`,|.

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. .sin...nm nm nm nm nmP11 000 350 0 11 0020 700 0 300 ( ) ( )[ ]'' .. ..sin . sin .nm nm 50 nm nm 0 200 .(c) P xdx x x220 1500 3500 1500 35021 0021 002 0021 00841 00 |.`, |.`,

]]] .sin...sin..... P x x20 1500 3501 001 00441 001 00 0 350 0 1501 0041 40 0 600 |.`,

]]] ( ) ( ) ( )[ ]'' ..sin.. . ..sin . sin ... 0 351 .(d) Using E n hmLn 2 228, we find that E1 0 377 . eV and E2 1 51 . eV41.53 (a) mgy mvi f122v gyf i ( )( ) 2 2 9 80 50 0 31 3 . . . m/ s m m/ s2 ( )( ) hmv6 626 1031 334.. J s75.0 kg m/ s2 82 1037. m (not observable)(b) E t h/ 2 so E ( ) 6 626 105 00 10343.. J s4 s 1 06 1032. J(c) EE ( )( )( ) 1 06 1075 0 9 80 50 032. Jkg m/ s m22 87 1035. % Chapter 41 Solutions 51841.54 From the uncertainty principle E t h/ 2 or mc t22( ) h/ . Therefore, mmhc t mht ER( ) ( ) ( )( ) |.

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4 46 626 104 8 70 10 135123417 .. J s s MeV MeV1.60 10 J-132 81 108. Chapter 41 Solutions 51941.55 (a) f Eh |.

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180 1 60 101 0019 eV6.626 10 J s J eV34..4 34 1014. Hz(b) cf3 00 104 34 106 91 108147...m/ s Hz m 691 nm(c) E t h2so Etht ( ) h2 46 626 102 00 102 64 1034829 ...J s4 s J 1 65 1010. eV41.56 (a) f Eh(b) cfhcE(c) E t h2so Et h2hT 441.57 x x dx2 22 For a one-dimensional box of width L, nLn xL |.`,2sinThus, xL x n xL dxL2 2 202 |.`, sin L Ln2 22 23 2(from integral tables)41.58 (a) 21 dx becomesA xL dx A L xLxL A LLLLL2 2442442221442 21 cos sin//// |.`, |.`, + |.`,

]]] |.`,|.`, or AL24 and AL2(b) The probability of finding the particle between 0 and L/ 8 is 0822 2082 1412L Ldx A xL dx/ /cos |.`, + 0 409 .Chapter 41 Solutions 52041.59 For a particle with wave function xae x a( ) 2/ for x > 0 and 0 for x < 0(a) x ( ) 20, x < 0 and 2 22xae x a( ) /, x > 0(b) Prob x x dx dx < ( ) ( ) ( ) 0 020 0 0(c) Normalization x dx dx dx ( ) + 2 202010 2 0 1 102 20 0dx a e dx e ex a x a + ( ) ( ) // /Prob 0 2 12020202< < ( ) ( ) x a dx a e dx e eax aax a a // /0 865 .41.60 (a) +( ) ( )hphcE m chcm c K m c ee e2 2 42222 ( ) ( )( ) ( ) |.

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6 626 10 3 00 10576 511134 82 2. . J s m/ s keV keV keV1.60 10 J164 68 1012. m(b) 50 0 . 2 34 1010. m41.61 (a) x p h 2 so if x r , p h 2r(b) Choosing pr h, K pmpme e ( )222 2 h222m reU k ere 2, so E K U + h2222m rk eree(c) To minimize E,dE k eeh2h2Chapter 41 Solutions 521Then, Emm k ek e m k e m k eee ee e e e e |.

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hh h h2 2222222 422 2 13 6 . eV41.62 (a) The requirement that nL2 so p h nhL 2 is still valid.E pc mc E nhcL mcn ( ) +( ) |.`, +( )2222222 K E mcn n 2 nhcL mc mc22222 |.`, +( ) (b) Taking L 1 00 1012. m , m 9 11 1031. kg , and n 1, we find K1 4 69 1014. JNonrelativistic, E hmL1223431 1221486 626 108 9 11 10 1 00 106 02 10 ( )( ) ( ) .. .. J s kg m J2Comparing this to K1, we see that this value is too large by 28 6 . % .41.63 (a) U eded + + +|.`, + ( )

]]] ( )202041121311217 34 e e/732k ede(b) From Equation 41.9, K E hm de ( ) 228 9122hm de2236(c) E U K + and dEdd 0 for a minimum: 73 1802223k edhm dee d hk e mhm k ee e e e( )( ) ( )( ) ( ) ( ) ( ) 37 18 426 626 1042 9 11 10 8 99 10 1 602 10222234231 9 192.. . . C0 0499 . nm(d) Since the lithium spacing is a, where Na V3 , and the density is Nm V / , where m is the mass ofone atom, we get:a VmNmm |.`, |.

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1 31 3271 3101 66 105302 80 10// /..density kg 7 kg m m 0 280 . nm(5.62 times larger than c).Chapter 41 Solutions 522*41.64 (a) ( )Bxe m x / 22hddx Be Bx mxe Be B mx em x m x m x m x + |.`, |.`,( ) ( ) ( ) ( ) / / 2 2 2 2 22 2 2 222h h h hh hddx Bx mxe B mxe B mx mxem x m x m x222 2 2 22 2 22 |.`, |.`, |.`,|.`,( ) ( ) ( )h h h hh h h / /ddx B mxe B mx em x m x222 3 232 2 |.`, + |.`,( ) ( )h hh h / /Substituting into the Schrdinger Equation (41.19), we have |.`, + |.`, + |.`,( ) ( ) ( ) ( )322 3 22222 22 2 2 2B mxe B mx e mEBxe mx Bxem x m x m x m x h h h hh h h h / / / /This is true if 32 Eh; it is true if E 32h(b) We never find the particle at x 0 because 0 there.(c) is maximized if ddx x m |.`,0 12h, which is true at xm t h(d) We require 21 dx :1 2 214 22 2 2 20232 1 2 3 23 22 2 ( ) ( )( ) ( ) B x e dx B x e dx BmBmm x m x / // ///h hhhThen B m |.`, 21 21 43 4 ///h43 331 4m h|.

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/(e) At x m 2 h/ , the potential energy is 122 21224 2 m x m m ( ) h h / . This is largerthan the total energy 3 2 h/ , so there is z ero classical probability of finding the particle here.(f) Probability |. `, ( ) ( ) 2222 22 2dx Bxe B x em x m x / / h hProbability |.`,|.`, ( ) ( ) 2 41 23 24/// /mm e m mhh h h81 24 meh|.`,/Chapter 41 Solutions 52341.65 (a) 201 dxL: A xLxLxLxL dxL2 2 20162821 sin sin sin sin |.`, + |.`, + |.`,|.`,

]]] A L L xLxL dxL202162821|.`, + |.`, + |.`,|.`,

]]] sin sin A L xLxL dx A L L xLLxx L2 202 30172161721631 + |.`,|.`,

]]] + |.`,

]]]] sin cos sin AL2217 , so the normalization constant is A L 2 17 /(b) 21 dxaa:A xa B xa A B xaxa dxaa22222221 cos sin cos sin |.`, + |.`, + |.`,|.`,

]]] The first two terms are A a2 and B a2. The third term is:2222 242 283 202 3A B xaxaxa dx A B xaxa dx a A B xaaaaaaacos sin cos cos sin cos |.`,|.`,|.`,

]]] |.`,|.`, |.`, so that a A B2 21 +( ) , giving A B a2 21 + / .*41.66 With one slit open P1 12 orP2 22 With both slits open, P + 1 22At a maximum, the wave functions are in phase Pmax +( ) 1 22At a minimum, the wave functions are out of phase Pmin ( ) 1 22Now PP1 12225 0 . , so 15 00 .Chapter 41 Solutions 524*41.67 (a) The light is unpolarized. It contains both horizontal and vertical field oscillations.(b) The interference pattern appears, but with diminished overall intensity.(c) The results are the same in each case.(d) The interference pattern appears and disappears as the polarizer turns, with alternately increasingand decreasing contrast between the bright and dark fringes. The intensity on the screen is preciselyzero at the center of a dark fringe four times in each revolution, when the filter axis has turned by 45,135, 225, and 315 from the vertical.(e) Looking at the overall light energy arriving at the screen, we see a low-contrast interference pattern.After we sort out the individual photon runs into those for trial 1, those for trial 2, and those for trial3, we have the original results replicated: The runs for trials 1 and 2 form the two blue graphs inFigure 41.3, and the runs for trial 3 build up the red graph.