THE WAVE NATURE OF PARTICLES

Transcription

1 THE WAVE NATURE OF PARTICLES IDENTIFY and SET UP: λ = = mv For an electron, 3 7 m = 9 kg For a roton, m = 67 kg EXECUTE: (a) 663 J s 3 6 (9 kg)(47 /s) λ = = = nm (b) λ is roortional to 3 m, so m e 9 kg 4 λ = λe = ( = m 67 kg c 39 IDENTIFY and SET UP: For a oton, E = For an electron or roton, = and E =, so E = λ λ m mλ 5 8 c (436 ev s)(3 /s) EXECUTE: (a) E = = = 6 kev 9 λ 663 J s 8 (b) E = = 63 J 38 ev 9 = = 3 mλ (9 kg) 3 m e 9 kg (c) E = Ee = (38 ev) = ev m 7 67 kg EVALUATE: For a given wavelengt a oton as muc more energy tan an electron, wic in turn as more energy tan a roton (663 J s) (a) λ = = = = 37 kg m s λ (8 4 (37 kg m s) 8 (b) K = = = 38 J = 93 ev 3 m (9 kg) 394 λ = = me (663 J s) = = (664 kg) (4 ev) (6 J ev) IDENTIFY and SET UP: Te de Broglie wavelengt is λ = = mv In te Bor model, mvr ( / ), n = n π so mv = n/( π r n ) Combine tese two exressions and obtain an equation for λ in terms of n Ten πrn πrn λ = = n n EXECUTE: (a) For n=, λ = πr wit r = a = 59, so λ = π(59 = 33 λ = πr ; te de Broglie wavelengt equals te circumference of te orbit (b) For n= 4, λ = πr4 / 4 rn = n a so r4 = 6 a 9 λ = π(6 a) / 4 = 4( πa) = 4(33 = 33 λ πr = 4 /4; te de Broglie wavelengt is n 4 = times te circumference of te orbit EVALUATE: As n increases te momentum of te electron increases and its de Broglie wavelengt decreases For any n, te circumference of te orbits equals an integer number of de Broglie wavelengts

2 39- Cater (a) For a nonrelativistic article, K =,so λ = = m Km 34 (b) (663 J s) -9-3 (8 ev)(6 J/eV)(9 kg) = IDENTIFY: A erson walking troug a door is like a article going troug a slit and ence sould exibit wave roerties SET UP: Te de Broglie wavelengt of te erson is λ = /mv EXECUTE: (a) Assume m = 75 kg and v = /s λ = /mv = ( J s)/[(75 kg)(/s)] = m EVALUATE: (b) A tyical doorway is about m wide, so te erson s de Broglie wavelengt is muc too small to sow wave beavior troug a slit tat is about 35 times as wide as te wavelengt Hence ordinary objects do not sow wave beavior in everyday life 398 Combining Equations 3738 and 3739 gives = mc γ (a) λ = = ( mc) γ = 443 (Te incorrect nonrelativistic calculation gives 55 ) 3 (b) ( mc) γ = IDENTIFY and SET UP: A oton as zero mass and its energy and wavelengt are related by Eq(38) An electron as mass Its energy is related to its momentum by E = / m and its wavelengt is related to its momentum by Eq(39) EXECUTE: (a) oton 8 c c (666 J s)(998 /s) E = so λ = = = 6 nm 9 λ E ( ev)(6 J/eV) electron E = /( m) so = me = (99 kg)( ev)(6 J/eV) = 46 kg m/s λ = / = 74 nm 9 (b) oton E = c/ λ = 7946 J = 496 ev 7 electron λ = / so = / λ = 65 kg m/s 4 5 E = /( m) = 3856 J = 4 ev (c) EVALUATE: You sould use a robe of wavelengt aroximately 5 nm An electron wit λ = 5 nm as muc less energy tan a oton wit λ = 5 nm, so is less likely to damage te molecule Note tat λ = / alies to all articles, tose wit mass and tose wit zero mass E = f = c/ λ alies only to otons and E = / m alies only to articles wit mass 39 IDENTIFY: Any moving article as a de Broglie wavelengt Te seed of a molecule, and ence its de Broglie wavelengt, deends on te temerature of te gas SET UP: Te average kinetic energy of te molecule is K av = 3/ kt, and te de Broglie wavelengt is λ = /mv = / EXECUTE: (a) Combining K av = 3/ kt and K = /m gives 3/ kt = av /m and av = 3mkT Te de Broglie wavelengt is λ = = 3mkT = 666 J s = kg 38 J/K 73 K ( )( )( ) (b) For an electron, λ = / = /mv gives 666 J s v = = = mλ ( 9 kg)( 8 6 m/s Tis is about % te seed of ligt, so we do not need to use relativity (c) For oton: E = c/λ = ( J s)(3 8 m/s)/(8 = 84 5 J For te H molecule: K av = (3/)kT = 3/ (38 3 J/K)(73 K) = 565 J For te electron: K = ½ mv = ½ (9 3 kg)(673 6 m/s) = 6 7 J EVALUATE: Te oton as about times more energy tan te electron and 3, times more energy tan te H molecule Tis sows tat otons of a given wavelengt will ave muc more energy tan articles of te same wavelengt

3 Te Wave Nature of Particles IDENTIFY and SET UP: Use Eq(39) EXECUTE: 666 J s λ = = = = 39 3 mv (5 kg)(34 m/s) EVALUATE: Tis wavelengt is extremely sort; te bullet will not exibit wavelike roerties 39 (a) λ = mv v= mλ Energy conservation: eδ V = mv m 34 mv mλ (666 J s) Δ V = = = = = 669 V e e emλ (6 C) (9 kg) (5 8 c (666 J s) (3 s) 5 (b) Eoton = f = = = 33 J 9 λ 5 5 Eoton 33 J eδ V = K = E oton and Δ V = = = 83 V 9 e 6 C 393 (a) λ = nm 6 = = λ so = ( λ) = 73 s mv v m (b) E = mv = 5 ev (c) E = c / λ = KeV (d) Te electron is a better robe because for te same λ it as less energy and is less damaging to te structure being robed 394 IDENTIFY: Te electrons beave like waves and are diffracted by te slit SET UP: We use conservation of energy to find te seed of te electrons, and ten use tis seed to find teir de Broglie wavelengt, wic is λ = /mv Finally we know tat te first dark fringe for single-slit diffraction occurs wen a sin θ = λ EXECUTE: (a) Use energy conservation to find te seed of te electron: ½ mv = ev 9 ev ( 6 C )( V) v = = = m/s 3 m 9 kg wic is about % te seed of ligt, so we can ignore relativity (b) First find te de Broglie wavelengt: 666 J s λ = = = mv ( 9 kg)( 593 /s) = 3 nm For te first single slit dark fringe, we ave a sin θ = λ, wic gives λ 3 a = = = 66 = 66 nm sinθ sin(5 ) EVALUATE: Te slit widt is around 5 times te de Broglie wavelengt of te electron, and bot are muc smaller tan te wavelengt of visible ligt 395 For m =, λ = d sinθ = me (663 J s) E = = = 69 J = 43 ev 7 md sin θ (675 kg) (9 sin (86 ) m 396 Intensity maxima occur wen d sin θ = mλ λ = = so d sin θ = (Careful! Here, m is te order ME ME of te maxima, wereas M is te mass of te incoming article) m ()(663 J s) (a) d = = 3 9 ME sinθ (9 kg)(88 ev)(6 J/eV) sin(66 ) = 6 = 6 nm (b) m = also gives a maximum () (663 J s) θ = arcsin = (9 kg) (88 ev) (6 J ev) (6

4 39-4 Cater 39 Tis is te only oter one If we let m 3, ten tere are no more maxima m () (663 J s) (c) E = = 3 Md sin θ (9 kg) (6 sin (66 ) 8 = 749 J = 468 ev Using tis energy, if we let m =, ten sinθ > Tus, tere is no m = maximum in tis case m 397 Te condition for a maximum is dsin θ = mλ λ = =, so θ = arcsin (Careful! Here, m is te order of Mv dmv te maximum, wereas M is te incoming article mass) (a) m = θ = arcsin dmv 663 J s = arcsin = (6 (9 kg) (6 s) () (663 J s) m = θ = arcsin = (6 (9 kg) (6 s) π radians (b) For small angles (in radians!) y Dθ,so y (5 cm) (7 ) = 8cm, 8 π radians y (5 cm) (44 ) = 36 cm and y y = 36 cm 8cm = 8 cm IDENTIFY: Since we know only tat te mosquito is somewere in te room, tere is an uncertainty in its osition Te Heisenberg uncertainty rincile tells us tat tere is an uncertainty in its momentum SET UP: Te uncertainty rincile is ΔxΔx EXECUTE: (a) You know te mosquito is somewere in te room, so te maximum uncertainty in its orizontal osition is Δ x = 5 m (b) Te uncertainty rincile gives ΔxΔx, and Δ x = mδv x since we know te mosquito s mass Tis gives ΔxmΔv x, wic we can solve for Δv x to get te minimum uncertainty in v x 55 J s Δ v x = = = 4 9 m/s -6 mδ x (5 kg)(5 m) wic is ardly a serious imediment! EVALUATE: For someting as large as a mosquito, te uncertainty rincile laces a negligible limitation on our ability to measure its seed 399 (a) IDENTIFY and SET UP: Use ΔxΔx / π to calculate Δ x and obtain Δ vx from tis 666 J s 8 EXECUTE: Δx = = 55 kg m/s 6 πδ x π( 8 Δ x 55 kg m/s 3 Δ vx = = = 879 /s m kg (b) EVALUATE: Even for tis very small Δ x te minimum Δ vx required by te Heisenberg uncertainty rincile is very small Te uncertainty rincile does not imose any ractical limit on te simultaneous measurements of te ositions and velocities of ordinary objects 39 IDENTIFY: Since we know tat te marble is somewere on te table, tere is an uncertainty in its osition Te Heisenberg uncertainty rincile tells us tat tere is terefore an uncertainty in its momentum SET UP: Te uncertainty rincile is ΔxΔx EXECUTE: (a) Since te marble is somewere on te table, te maximum uncertainty in its orizontal osition is Δ x = 75 m (b) Following te same rocedure as in art (b) of roblem 398, te minimum uncertainty in te orizontal velocity of te marble is 55 J s Δ v x = = = m/s mδx kg (75 m) ( ) (c) Te uncertainty rincile tells us tat we cannot know tat te marble s orizontal velocity is exactly zero, so te smallest we could measure it to be is m/s, from art (b) Te longest time it could remain on te

5 Te Wave Nature of Particles 39-5 table is te time to travel te full widt of te table (75 m), so t = x/v x = (75 m)/(63 33 m/s) = 9 3 s = 9 4 years Since te universe is about 4 9 years old, tis time is about 4 9 yr 4 6 times te age of te universe! Don t old your breat! 9 4 yr EVALUATE: For ouseold objects, te uncertainty rincile laces a negligible limitation on our ability to measure teir seed 39 Heisenberg s Uncertainty Princiles tells us tat ΔxΔx We can treat te standard deviation as a direct π 5 35 measure of uncertainty Here ΔxΔ x = ( (3 kg m s) = 36 J s but = 5 J s π Terefore ΔxΔ x < so te claim is not valid π 39 (a) ( Δx)( mδvx ) π, and setting Δ vx = () vxand te roduct of te uncertainties equal to / π (for te minimum uncertainty) gives v = ( πm() Δ x) = 579 m s x (b) Reeating wit te roton mass gives 36 mm s (663 J s) Δ E > = = 3 J = 7 ev 3 πδ t π(5 s) 394 IDENTIFY and SET UP: Te Heisenberg Uncertainty Princile says ΔxΔx Te minimum allowed π ΔxΔ x is / π Δ x = mδ vx 663 J s 4 EXECUTE: (a) mδxδ vx = Δ vx = = = 3 /s 7 π πmδ x π(67 kg)( 663 J s 4 (b) Δ x = = = 46 3 πmδ vx π(9 kg)(5 m/s) (663 J s) ΔEΔ t = Δ E = = = 39 J = 869 ev = 869 MeV π πδ t π(76 s) Δ E 869 MeV c = 5 8 = E 397 MeV c ΔEΔ t = Δ E =Δmc Δ m= 6 ev c = 33 J c π 663 J s 5 Δ t = = = 3 s πδ mc π(33 J) 5 c 99 J m 397 IDENTIFY and SET UP: For a oton E = = For an electron Ee = = = λ λ m m λ mλ 5 99 J m 7 EXECUTE: (a) oton 9 99 E = = J (663 J s) electron Ee = = 4 J 3 9 (9 kg)( 7 E 99 J 3 = = 86 Ee 4 J (b) Te electron as muc less energy so would be less damaging EVALUATE: For a article wit mass, suc as an electron, E ~ λ For a massless oton E ~ λ ( λ) ( λ) 398 (a) ev = K = =, sov = = 49V m m me 3 9 kg (b) Te voltage is reduced by te ratio of te article masses, (49 V) = 9 V 7 67 kg 399 IDENTIFY and SET UP: ψ ( x) = Asin kx Te osition robability density is given by ψ ( x) = A sin kx EXECUTE: (a) Te robability is igest were sin kx = so kx = πx / λ = nπ/, n =, 3, 5, x= nλ/ 4, n=, 3, 5, so x = λ/ 4, 3 λ/ 4, 5 λ/4,

6 39-6 Cater 39 (b) Te robability of finding te article is zero were ψ =, wic occurs were sin kx = and kx = πx / λ = nπ, n =,,, x= nλ/, n=,,, so x=, λ/, λ, 3 λ/, EVALUATE: Te situation is analogous to a standing wave, wit te robability analogous to te square of te amlitude of te standing wave * * 393 Ψ = ψ sin ωt, so Ψ = Ψ Ψ = ψψsin ωt = ψ sin ωt Ψ is not time-indeendent, so Ψ is not te wavefunction for a stationary state 393 IDENTIFY: To describe a real situation, a wave function must be normalizable SET UP: ψ dv is te robability tat te article is found in volume dv Since te article must be somewere, ψ must ave te roerty tat ψ dv = wen te integral is taken over all sace EXECUTE: (a) In one dimension, as we ave ere, te integral discussed above is of te form ψ ( x) dx= (b) Using te result from art (a), we ave ( e ) dx= e dx= = Hence tis wave function cannot ax ax ax e a be normalized and terefore cannot be a valid wave function (c) We only need to integrate tis wave function of to because it is zero for x < For normalization we ave bx bx bx Ae A = ψ dx = ( Ae ) dx = A e dx = b = A, wic gives b b =, so A= b EVALUATE: If b were ositive, te given wave function could not be normalized, so it would not be allowable ψ x, and is inversely roortional to 393 (a) Te uncertainty in te article osition is roortional to te widt of ( ) 3933 α Tis can be seen by eiter lotting te function for different values of α, finding te exectation value x = ψ xdxfor te normalized wave function or by finding te full widt at alf-maximum Te article s uncertainty in osition decreases wit increasing α Te deendence of te exectation value x on α may be found by considering = x xe α x e α x dx α x u = ln e dx α = ln e du =, dx α α 4α were te substitution u = α xas been made (b) Since te uncertainty in osition decreases, te uncertainty in momentum must increase x iy * x+ iy * x iy x+ iy f( x, y) = and f ( x, y) = f = f f = = x+ iy xiy x+ iy xiy 3934 Te same * ψ( xyz,, ) = ψ ( xyzψ,, ) ( xyz,, ) i φ * i φ + i φ ψ( xyze,, ) = ( ψ ( xyze,, ) )( ψ( xyze,, ) ) = ψ * ( xyzψ,, ) ( xyz,, ) iφ iφ Te comlex conjugate means convert all i s to i s and vice-versa e e = 3935 IDENTIFY: To describe a real situation, a wave function must be normalizable SET UP: ψ dv is te robability tat te article is found in volume dv Since te article must be somewere, ψ must ave te roerty tat ψ dv = wen te integral is taken over all sace EXECUTE: (a) For normalization of te one-dimensional wave function, we ave bx bx bx bx ψ dx Ae dx + Ae dx = A e dx + A e dx = = ( ) ( ) bx bx e e A = A + = b b b, wic gives A= b = m = 4 m / (b) Te gra of te wavefunction versus x is given in Figure m (c) (i) P + ψ dx + 5 m = = bx Ae dx 5 m, were we ave used te fact tat te wave function is an even function of x Evaluating te integral gives A (5 m) ( m ) P = ( e b ) ( e = ) = 865 b m Tere is a little more tan an 86% robability tat te article will be found witin 5 cm of te origin

7 Te Wave Nature of Particles 39-7 bx bx A m (ii) P = ( Ae ) dx = A e dx = b = ( m ) = = 5 Tere is a 5-5 cance tat te article will be found to te left of te origin, wic agrees wit te fact tat te wave function is symmetric about te y-axis (iii) P = EVALUATE: m 5 m Ae bx dx A e - e - e e = = 585 b Tere is little cance of finding te article in regions were te wave function is small ( m )( m) ( m )(5 m) 4 = ( ) ( ) Figure 3935 d ψ 3936 Eq (398): + Uψ = Eψ Let ψ = Aψ+ Bψ m dx d ( Aψ ) ( ) ( ) + Bψ + U Aψ+ Bψ = EAψ+ Bψ mdx d ψ d ψ A + Uψ Eψ + B + Uψ Eψ = mdx mdx But eac of ψ and ψ satisfy Scrödinger s equation searately so te equation still olds true, for any A or B d ψ Uψ = BE ψ + CE ψ If ψ were a solution wit energy E, ten BEψ + CEψ = BEψ+ CEψ or mdx BE ( Eψ ) = CE ( E) ψ Tis would mean tat ψ is a constant multile of ψ, andψ andψ would be wave functions wit te same energy However, E E, so tis is not ossible, and ψ cannot be a solution to Eq (398) (663 J s) 3938 (a) λ = = = mk (9 kg)(4 ev)(6 J ev) 3 R R (5 m)(9 kg) 7 (b) = = = 667 s v 9 Em (4 ev)(6 J ev) λ (c) Te widt wis w= R ' and w=δ vyt =Δ yt m, were t is te time found in art (b) and a is te slit widt a mλr 8 Combining te exressions for w, Δ y = = 65 kg m s at (d) Δ y = = 4 μm, wic is te same order of magnitude πδ y 3939 (a) E = c λ = ev 6 7 (b) Find E for an electron wit λ = λ = so = λ = 666 kg m s E m 4 = ( ) = 5 ev E = qδv Δ V = 4 so 5 V v = m= = (666 kg m s) (99 kg) 73 s (c) Same λ so same E = /( m) but now m = 673 kg so E = 8 ev and Δ V = 8 V v = m = = 7 7 (666 kg m s) (673 kg) 4 m s (a) Single slit diffraction: asinθ = mλ λ = asin θ = (5 sin = J s 4 λ = mv v= mλ v = = 4 s 3 8 (9 kg)(53 8 λ 53 (b) asinθ = λ sinθ =± =± =± θ =± 43 a 5

8 39-8 Cater IDENTIFY: Te electrons beave like waves and roduce a double-slit interference attern after assing troug te slits SET UP: Te first angle at wic destructive interference occurs is given by d sin θ = λ/ Te de Broglie wavelengt of eac of te electrons is λ = /mv EXECUTE: (a) First find te wavelengt of te electrons For te first dark fringe, we ave d sin θ = λ/, wic gives (5 nm)(sin 8 ) = λ/, and λ = 775 nm Now solve te de Broglie wavelengt equation for te seed of te electron: 666 J s v = = 3 9 mλ (9 kg)(775 = 94 5 m/s wic is about 3% te seed of ligt, so tey are nonrelativistic (b) Energy conservation gives ev = ½ mv and V = mv /e = (9 3 kg)(94 5 m) /[(6 9 C)] = 5 V EVALUATE: Te ole must be muc smaller tan te wavelengt of visible ligt for te electrons to sow diffraction 394 IDENTIFY: Te ala articles and rotons beave as waves and exibit circular-aerture diffraction after assing troug te ole SET UP: For a round ole, te first dark ring occurs at te angle θ for wic sinθ = λ /D, were D is te diameter of te ole Te de Broglie wavelengt for a article is λ = / = /mv EXECUTE: Taking te ratio of te sines for te ala article and roton gives sinθα λα λα = = sinθ λ λ Te de Broglie wavelengt gives λ = / and λ α = / α, so sin θ / sin / α α θ = = Using K = /m, we ave α = mk Since te ala article as twice te carge of te roton and bot are accelerated troug te same otential difference, K α = K Terefore = mk and α = mαkα = mα( K) = 4mαK Substituting tese quantities into te ratio of te sines gives sinθ mk m sin m α θ = = α 4mK = α 7 67 kg Solving for sin θ α gives sin θ α = sin5 and θ 7 α = 53 (664 kg) Since sin θ is inversely roortional to te mass of te article, te larger-mass ala articles form EVALUATE: teir first dark ring at a smaller angle tan te ring for te ligter rotons 3943 IDENTIFY: Bot te electrons and otons beave like waves and exibit single-slit diffraction after assing troug teir resective slits SET UP: Te energy of te oton is E = c/λ and te de Broglie wavelengt of te electron is λ = /mv = / Destructive interference for a single slit first occurs wen a sin θ = λ EXECUTE: (a) For te oton: λ = c/e and a sinθ = λ Since te a and θ are te same for te otons and electrons, tey must bot ave te same wavelengt Equating tese two exressions for λ gives a sin θ = c/e For te electron, λ = / = mk and a sin θ = λ Equating tese two exressions for λ gives a sin θ = mk Equating te two exressions for asinθ gives c/e = mk, wic gives 7 / E = c mk = (45 J ) K E c mk mc (b) = = Since v << c, mc > K, so te square root is > Terefore E/K >, meaning tat te K K K oton as more energy tan te electron EVALUATE: As we ave seen in Problem 39, wen a oton and a article ave te same wavelengt, te oton as more energy tan te article 6 d sin θ (4 sin(3 rad) 3944 According to Eq(354) λ = = = 6 nm Te velocity of an electron wit m tis wavelengt is given by Eq(39) (663 J s) 3 v = = = = s 3 9 m mλ (9 kg)(6 α

9 Since tis velocity is muc smaller tan c we can calculate te energy of te electron classically (9 3 kg)( 3 m s) 67 5 J 49 ev K = mv = = = μ 3945 Te de Broglie wavelengt of te blood cell is (663 J s) 7 λ = = = mv ( kg)(4 s) We need not be concerned about wave beavior v c 3946 (a) λ = = v v v λ mv = = λ mv + = mv c c c c c v = = v = λ m c λ m + + c mcλ + c mcλ mcλ (b) v = c ( ) c = Δ Δ= λ + ( mc) 5 (c) λ = << mc (9 kg) (3 s) ( 8 Δ= = 85 (663 J s) 8 v= ( ) c= ( 85 ) c 3947 (a) Recall λ = = = So for an electron: me mqδv 663 J s λ = λ = 3 9 (9 kg)(6 C)(5 V) 663 J s 3 (b) For an ala article: λ = = (664 kg)(6 C)(5 V) Te Wave Nature of Particles IDENTIFY and SET UP: Te minimum uncertainty roduct is ΔxΔ x = Δ x = r, were r is te radius of te π n = Bor orbit In te n = Bor orbit, mv r = and = mv = π π r 663 J s 4 EXECUTE: Δ x = = = = kg m/s Tis is te same as te magnitude of πδ x πr π(59 te momentum of te electron in te n = Bor orbit EVALUATE: Since te momentum is te same order of magnitude as te uncertainty in te momentum, te uncertainty rincile lays a large role in te structure of atoms c 3949 IDENTIFY and SET UP: Combining te two equations in te int gives PC = K( K + mc ) and λ = K( K + mc ) EXECUTE: (a) Wit (b) (i) K = 3mc tis becomes c λ = = 3 mc (3mc + mc ) K = = = = 5mc 3mc 3 8 3(99 kg)(998 /s) J 53 MeV 666 J s λ = = = 5mc 5(99 kg)(998 /s) (ii) K is roortional to m, so for a roton K = ( m/ me)(53 MeV) = 836(53 MeV) = 8 MeV λ is roortional to /m, so for a roton λ = ( m / m )(66 = (/836)(666 = 34 e EVALUATE: Te roton as a larger rest mass energy so its kinetic energy is larger wen K = 3 mc Te roton also as larger momentum so as a smaller λ

10 39- Cater (a) (b) (666 J s) 5 = π(5 kg m s K = c + mc mc = = 3 ( ) ( ) 3 J 8 MeV 6 (c) Te result of art (b), about MeV = ev, is many orders of magnitude larger tan te otential energy of an electron in a ydrogen atom 395 (a) IDENTIFY and SET UP: ΔxΔx / π 5 Estimate Δx as Δx J s EXECUTE: Ten te minimum allowed Δ x is Δx = = kg m/s 5 πδ x π(5 (b) IDENTIFY and SET UP: EXECUTE: Assume E = ( mc ) + ( c) mc = (99 kg)(998 /s) = 887 J c = = 8 ( kg m/s)(998 /s) 696 J E = + = 4 (887 J) (696 J) 697 J kg m/s Use Eq(3739) to calculate E, and ten K = E mc 4 9 K = E mc = 697 J 887 J = 65 J( ev/6 J) = 39 MeV (c) IDENTIFY and SET UP: Te Coulomb otential energy for a air of oint carges is given by Eq(39) Te roton as carge +e and te electron as carge e 9 9 ke (8988 N m /C )(6 C) 4 EXECUTE: U = = = 46 J =9 MeV 5 r 5 EVALUATE: Te kinetic energy of te electron required by te uncertainty rincile would be muc larger tan te magnitude of te negative Coulomb otential energy Te total energy of te electron would be large and ositive and te electron could not be bound witin te nucleus 395 (a) Take te direction of te electron beam to be te x-direction and te direction of motion erendicular to te Δ y 666 J s beam to be te y-direction Δ vy = = = = 3 m/s 3 3 m πmδ y π(9 kg)(5 (b) Te uncertainty Δ r in te osition of te oint were te electrons strike te screen is Δy x x Δ r =Δ vyt = = = 956, m v πmδy Km (c) Tis is far too small to affect te clarity of te icture 3953 IDENTIFY and SET UP: ΔEΔt Take te minimum uncertainty roduct, so Δ E = π 7 ΔE Δ t = 84 s m= 64me Δ m = c 663 J s 7 x 8 6 J 8 π Δ t, wit 8 35 EXECUTE: Δ E = = 6 J Δ m = = 4 kg π (84 s) (3 /s) 35 Δ m 4 kg 8 = = 58 3 m (64)(9 kg) 3954 IDENTIFY: Te insect beaves like a wave as it asses troug te ole in te screen SET UP: (a) For wave beavior to sow u, te wavelengt of te insect must be of te order of te diameter of te ole Te de Broglie wavelengt is λ = /mv EXECUTE: Te de Broglie wavelengt of te insect must be of te order of te diameter of te ole in te screen, so λ 5 mm Te de Broglie wavelengt gives 666 J s v = = = 33 5 m/s 6 mλ 5 kg 4 m ( )( ) (b) t = x/v = (5 m)/(33 5 m/s) = 377 s = 4 yr Te universe is about 4 billion years old (4 yr), so tis time would be about 85, times te age of te universe EVALUATE: Don t exect to see a diffracting insect! Wave beavior of articles occurs only at te very small scale 3955 IDENTIFY and SET UP: Use Eq(39) to relate your wavelengt and seed 666 J s 35 EXECUTE: (a) λ =, so v = = = /s mv mλ (6 kg)()

11 Te Wave Nature of Particles 39- distance 8 m (b) t = = = 73 s( y/356 s) = 3 y 35 velocity /s Since you walk troug doorways muc more quickly tan tis, you will not exerience diffraction effects EVALUATE: A kg object moving at m/s as a de Broglie wavelengt λ = 66, wic is exceedingly small An object like you as a very, very small λ at ordinary seeds and does not exibit wavelike roerties 9 c (a) E = 58 ev = 43 J, wit a wavelengt of λ = = 48 = 48 nm E (663 J s) 8 9 (b) Δ E = = = 643 J = 4 ev 7 πδ t π(64 s) (c) λe = c, so ( Δ λ ) E+ λδ E =, and Δ E E = Δ λ λ, so J 6 7 Δ λ = λ Δ EE= ( nm 9 = = 43 J 3957 IDENTIFY: Te electrons beave as waves wose wavelengt is equal to te de Broglie wavelengt SET UP: Te de Broglie wavelengt is λ = /mv, and te energy of a oton is E = f = c/λ EXECUTE: (a) Use te de Broglie wavelengt to find te seed of te electron 666 J s v = = = mλ ( 9 kg)( 5 m/s wic is muc less tan te seed of ligt, so it is nonrelativistic (b) Energy conservation gives ev = ½ mv V = mv /e = (9 3 kg)(77 5 m/s) /[(6 9 C)] = 5 V (c) K = ev = e(5 V) = 5 ev, wic is about ¼ te otential energy of te NaCl crystal, so te electron would not be too damaging (d) E = c/λ = (436 5 ev s)(3 8 m/s)/( 9 m) = 4 ev wic would certainly destroy te molecules under study EVALUATE: As we ave seen in Problems 39 and 3943, wen a article and a oton ave te same wavelengt, te oton as muc more energy λ 3958 sin θ = sin θ, and λ = ( ) = ( me ), and so θ = arcsin sinθ λ λ me (663 J s)sin 358 θ = arcsin = (3 (9 kg)(45 )(6 J ev) 3959 (a) Te maxima occur wen d sinθ = mλ as described in Section 387 (b) λ = = me (663 J s) λ = = 46 = 46 nm 37 9 (9 kg)(7 ev) 6 J/eV ( ) mλ θ = sin (Note: Tis m is te order of te maximum, not te mass) d ()(46 sin 533 = (9 (c) Te work function of te metal acts like an attractive otential increasing te kinetic energy of incoming electrons by eφ An increase in kinetic energy is an increase in momentum tat leads to a smaller wavelengt A smaller wavelengt gives a smaller angle θ (see art (b)) 396 (a) Using te given aroximation, ( ( ) ),( ) 3 E = x m + kx de dx = kx ( mx ), and te minimum energy 3 occurs wen kx = ( mx ), or x = Te minimum energy is ten k m mk (b) Tey are te same 396 (a) IDENTIFY and SET UP: U = A x Eq(77) relates force and otential Te sloe of te function Ax is not continuous at x = so we must consider te regions x > and x < searately d( Ax) EXECUTE: For x>, x = x so U = Ax and F = = A For x<, x = x so U = Ax and dx d( Ax) F = =+ A We can write tis result as F = A x / x, valid for all x excet for x = dx

12 39- Cater 39 (b) IDENTIFY and SET UP: Use te uncertainty rincile, exressed as ΔΔx, and as in Problem 395 estimate Δ by and Δ x by x Use tis to write te energy E of te article as a function of x Find te value of x tat gives te minimum E and ten find te minimum E EXECUTE: E = K + U = + A x m x, so / x Ten E + A x mx For x>, E = + Ax mx de To find te value of x tat gives minimum E set dx = = + A 3 mx /3 3 x = and x= ma ma Wit tis x te minimum E is /3 /3 ma /3 /3 /3 /3 /3 /3 E = A m A m A + = + m ma /3 3 A E = m EVALUATE: Te otential well is saed like a V Te larger A is te steeer te sloe of U and te smaller te region to wic te article is confined and te greater is its energy Note tat for te x tat minimizes E, K = U iωt iωt 396 For tis wave function, Ψ = ψ e + ψ e,so iωt iωt iωt iωt * i( ωω) t i( ωω) t Ψ =Ψ Ψ = ( ψe + ψe )( ψe + ψe ) = ψψ + ψψ + ψψe + ψψe Te frequencies ω and ω are given as not being te same, so Ψ is not time-indeendent, and Ψ is not te wave function for a stationary state 3963 Te time-deendent equation, wit te searated form for Ψ ( xt, ) as given becomes d ψ i ψ( iω) = + U( x) ψ mdx Since ψ is a solution of te time-indeendent solution wit energy E, te term in arentesis is Eψ, and so ω = E,and ω = ( E ) π E E π π ( k) k 3964 (a) ω= π f = = k = = = ω = E = K = = ω = λ m m m ψ( xt, ) (b) From Problem 3963 te time-deendent Scrödinger s equation is + m x ψ( xt, ) ψ( xt, ) mi ψ( xt, ) U( x) ψ( xt, ) = i Ux ( ) = for a free article, so = t x t Try ψ( xt, ) = cos( kx ωt) : ψ ( xt, ) = Aω sin( kx ωt ) t ψ( xt, ) ψ =Ak sin( kx ωt) and = Ak cos( kxωt) x x mi Putting tis into te Scrödinger s equation, Ak cos( kx ωt) = Aω sin( kx ωt) Tis is not generally true for all xand t so is not a solution

13 Te Wave Nature of Particles 39-3 (c) Try ψ( xt, ) = Asin( kx ωt) : ψ( xt, ) =Aωcos( kx ωt) t ψ( xt, ) ψ( xt, ) = Ak cos( kx ωt) and =Ak sin( kxωt) x x mi Again, Ak sin( kx ωt) = Aω cos( kx ωt) is not generally true for all x and t so is not a solution (d) Try ψ( xt, ) = Acos( kx ωt) + Bsin( kx ωt) : ψ( xt, ) =+ Aω sin( kx ωt) Bω cos( kx ωt) t ψ( xt, ) ψ( xt, ) =Ak sin( kx ωt) + Bk cos( kx ωt) and =Ak cos(( kxωt) Bk sin( kxωt) x x Putting tis into te Scrödinger s equation, mi Ak cos( kx ωt) Bk sin( kx ωt) = ( + Aω sin( kx ωt) Bω cos( kx ωt)) k Recall tat ω = Collect sin and cos terms m ( A+ ib) k cos( kx ωt) + ( iab) k sin( kx ωt ) = Tis is only true if B = ia 3965 (a) IDENTIFY and SET UP: Let te y-direction be from te trower to te catcer, and let te x-direction be orizontal and erendicular to te y-direction A cube wit volume V = 5 cm = 5 as side lengt l V (5 ) 5 m /3 3 3 /3 = = = Tus estimate x as x 5 m Δ Δ Use te uncertainty rincile to estimate Δ x 663 J s EXECUTE: ΔxΔx / π ten gives Δx = = kg m/s πδ x π(5 m) (Te value of in tis oter universe as been used) (b) IDENTIFY and SET UP: Δ x= ( Δ vx) t is te uncertainty in te x-coordinate of te ball wen it reaces te catcer, were t is te time it takes te ball to reac te second student Obtain Δ vx from Δ x Δx kg m/s EXECUTE: Te uncertainty in te ball s orizontal velocity is Δ vx = = = 84 m/s m 5 kg m Te time it takes te ball to travel to te second student is t = = s Te uncertainty in te x-coordinate 6 m/s of te ball wen it reaces te second student tat is introduced by Δvx is Δ x= ( Δ vx) t = (84 m/s)( s) = 7 m Te ball could miss te second student by about 7 m EVALUATE: A game of catc would be very different in tis universe We don t notice te effects of te uncertainty rincile in everyday life because is so small ( αx βy γz 3966 (a) ψ = Axe + + ) To save some algebra, let u x αu =, so tat ψ = ue f ( y, z) ψ = ( αu) ψ ; te maximum occurs at u =, x =± u α α (b) ψ vanises at x =, so te robability of finding te article in te x = lane is zero Te wave function vanises for x =± 3967 (a) IDENTIFY and SET UP: Te robability is EXECUTE: αr αr ψ = Ae so P= 4πAre dr dp (b) IDENTIFY and SET UP: P is maximum were dr = d EXECUTE: ( re α r ) = dr P= ψ dv wit dv = 4πr dr αr 3 αr 3 re 4α r e = and tis reduces to r 4α r = r = is a solution of te equation but corresonds to a minimum not a maximum Seek r not equal to so divide by r and get 4αr =

14 39-4 Cater (a) Tis gives r = (We took te ositive square root since r must be ositive) α EVALUATE: Tis is different from te value of r, r =, were ψ is a maximum At r =, ψ as a maximum but te volume element dv = 4π r dr is zero ere so P does not ave a maximum at r = Bk e B B α k ( ) = () = max = Bk ( ) e ln( ) k α k = = = α k = ln() = ω k α k π x /4 (b) Using integral tables: ψ( x) = α α e cos kxdk = ( e ) ψ( x ) is a maximum wen x = α π x /4α x (c) ψ( x ) = wene = = ln(/) x = α ln = ωx 4α 4α (d) ω k ln ωω x = ωx = ln ( α ln ) = (ln) = π π α π π k sin kx sin kx 3969 (a) ψ( x) = B( k)coskxdk = coskxdk = = k kx kx k π (b) ψ( x) as a maximum value at te origin x = ψ( x) = wen kx = π so x = Tus te widt of tis k π π function wx = x = If k =, wx = L Bk ( ) versus k is graed in Figure 3969a Te gra of ψ( x) versus k L x is in Figure 3969b π (c) If k = wx = L L wk π wk k (d) ww x = = = = Te uncertainty rincile states tat ww x For us, no matter wat π k k k π is,, k wwx = wic is greater tan π Figure 3969 ( λ) n 397 (a) For a standing wave, nλ = L, and En = = = m m 8mL 7 (b) Wit L= a = 59, E = 5 J = 34 ev

15 Te Wave Nature of Particles Time of fligt of te marble, from a free-fall kinematic equation is just t = y (5m) = = 6 s g 98 m s Δx t Δ xf =Δ xi + ( Δ vx) t =Δ xi + t = +Δxi m πδ xm i d( Δxf ) t To minimize Δx f wit resect to Δ xi, = = + d( Δx) πm( Δx) t Δ xi (min) = πm i t t t (663 J s)(6 s) 6 7 f (min) 8 8 nm Δ x = + = = = = πm πm πm π( kg) i

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