( x 0 U0 0 x>0 U wavenumbe k deceases by a facto fom left hand Hence to ight hand egion. Thus we should see wavelength of egion again E,U ositive but

Transcription

1 4 Sing 99 Poblem Set 3 Solutions Physics May 6, 999 Handout nal exam will given on Tuesday, May 8 fom 9:00 to :30 in Baton Hall. It will be witten as a 90 minute exam, but all am, will have full eiod fom 9:00 to :30 am to comlete it. students exam will cove mateial in lectues 9-8, oblem sets 0-3 otional oblems), comute lab 3 (schdg) and exeimental (including 6. exam will be closed-book, but with a fomula sheet ovided. lab Finite Ste Potential. ae given a -dimensional otential with U(x) =0fox0 and You nomalization condition, sketch eal at of a solution Ignoing vs x; to Schodinge equation fo a aticle with enegy E whee: (x);, m (E, U) = E,U0 E,U<0,q, U 0 E E, U<0 q tye of solution deends on sign of E, U. We will have Hence solutions fo E,U 0, and exonential solutions fo E,U < 0. oscillatoy Also we should note that both and d valued. case (a), E < U 0, E, U ositive fo x 0 but negative fo Fo Thus we have an oscillatoy solution in egion x 0. x>0. in ight hand egion, fo x>0, a decaying exonential, solution. Infomation on Final Exam time indeendent Schodinge equation in dimension given by Final Exam: d dx If we dene m (E, U) = ( k n Schodinge equation becomes CACUATORS WI NOT BE NEEDED OR AOWED = (, k d dx ae continuous, nite and single U(x) = U 0 fo x >0 whee U 0 a ositive constant. dx See gue below. which matches value and sloe of oscillating wave at x= 0. a) E<U 0.

2 ( x 0 U0 0 x>0 U wavenumbe k deceases by a facto fom left hand Hence to ight hand egion. Thus we should see wavelength of egion again E,U ositive but since E U 0 n dieence E,U Hee hadly changed fom left hand egion to ight hand egion. Thus wavenumbe k unchanged. Paticle in a Potential Well (Qualitative) 3. an electon of mass m taed in following otential well, Conside fo which V goes to innity fo x 0: b) E =U 0. Hee E, U ositive eveywhee and moe secically E, U = wave function incease. Th seen in following diagam. c) E U 0.

3 Sketch wave functions and fo two lowest bound standing a) wave states with enegies E and E. Do not woy about wave functions have a sinusoidal fom whee E, V (x) ositive and exonential whee it negative. wave function must to zeo fa fom well and be continuous, with continuous deivative, go x =. lowest state has zeo inteio nodes; next state has one at Does lowest enegy (\gound") state, E,havealowe o highe b) than lowest enegy state of an electon in a box, with innite enegy Each state has a lowe enegy in th otential than in a wavelength longe, as can be seen fom sketch in at a), box. Sketch eal at of wave function if electon has an enegy c) 4V 0 =3. Do not woy about nomalization condition in making you of Now E, V (x) ositive eveywhee, so wave function always oscillatoy, and has fom of a taveling wave. wavelength fo x, (E,V(x)),= Electon in a Box 4. electon of mass m conned in a otential which innite fo x<0, An fo x>. otential vaies linealy fom 0 to V 0 as x vaies fom and to(see gue below). 0 nomalization condition in making you sketch. sketch. 0 3=V ; ootional to (E, V (x)),=.fo x<,(e,v(x)),= = = 3=V 0. So wavelength longe by a facto of two fo x than fo x. inteio node. otential enegy on both sides, of same width? Why? so momentum, and efoe enegy, lowe. 3

4 Since V (x) vaies linealy fom 0 to V 0 wehavev(x)=v x 0. Using E = K(x)+V(x), we have, x K(x)=E,V(x)=E,V 0 In same ange, give an equation fo deboglie wavelength of b) electon, (x), in tems of E, V 0, x, m, h, and. deboglie wavelength given by = h. Using = wehave mk, h = (x) h x V0, m(e et total enegy of electon E be E ' V 0, and let deboglie c) of electon (0) =.Sketch a gah of a ossible standing wavelength Since E ' V 0, wavefunction will be sinusoidal thoughout otential well, with a wavelength equal to (x). At x =0, wavelength, coesonding to oscillations in well; but as x in- kinetic enegy deceases and wavelength inceases. ceases, must also go to zeo at x = 0 and x =, since otential wavefunction et total enegy of electon E be E<V 0.Sketch a gah of d) standing wave solution fo electon's wavefunction coesonding Since E<V 0, wavefunction will be sinusoidal though a otion of otential well, whee E > V(x). Fo egion only solution fo electon's wavefunction in th otential, in ange wave x. 0 innite e. Hence wavefunction looks like: In ange 0 x, give an equation fo electon's kinetic a) K(x), in tems of E, V 0, x and. enegy asxvaies fom 0 to to lowest enegy state in th otential, in ange 0 x. h = mk(x) (x)= ) 4

5 well whee E < V(x), wavefunction will exhibit exonential of Fo lowest enegy state, e will be no inteio nodes. decay. must also go to zeo at x = 0 and x =, since otential wavefunction innite e. Hence wavefunction will look like: Electon Tunnelling 5. a baie of height V 0 =6 ev, and of thickness =700 m. ( Conside, m). Calculate enegy of an incident electon such that its m=0 of tansmsion in 000. obability tansmsion obability given by, E) m(v0 T ' e,k (V0, E) mc ln T =,K =, mc mc can convet c fom J-m to ev-m using ev = :6 0,9 J and We,5 m: m=0 = 6:63 0,34 J, s m=s c ev J m, 97; 500 ev =, m m 0,9 :60 using V 0=6 ev, =700 m, mc =5,000 ev, and ln 0,3 = , n, have we, m6:907 ev m ) 400 Angula Momentum 6. an electon in a hydogen atom in a state with l = 5, what If lagest ossible value of z fo l =5 z=m l=5. of = l(l +)= 30. angle between and its magnitude on z-ax, z,given by ojection 5 cos = 5 =0:93 = In th oblem, we know T and we want to nd E. So (V0, E) c Solving fo E gives = V 0, (,c ln T ) E =3:6 0,6 J, m E=6eV, ( =5:07 ev 5; 000 ev smallest ossible angle between and z-ax? = z cos = z 30 Th angle smallest when z takes on its lagest value, 5. n which gives =4:. whee K = c = 5

6 Quantum States of Hydogen Atom 7. atom state has a maxiumum m l value of 4. What can you Ahydogen maximum value of m l equals value of l fo state, so l =4. l = 4, n since l anges u to n,, we must have n 5. quantum If m s = always. numbe You ae doing mables, of mass 30 g, fom oof of a building, a) onto a small taget 0 m below. You ae tying to be as accuate ossible in hitting taget, and ae concened that Heenbeg as Pincile may limit you accuacy. Using uncetainty e- Uncetainty lation in fom x x gue below shows aangement. oof, you do best job you can to line u mable with At taget, achieving a localization of mable equal to x i. Because uncetainty incile, th localization causes an uncetainty in of of momentum given by x-comonent x : i x which m mass of mable. time which it takes mable in fall fom oof to taget given by to H t ; = g Roof x i say about est of its quantum numbes? H FOR EXTRA CREDIT: Heenbeg Uncetainty Pincile 8., estimate dtance by which you will Taget x f ms taget due to th eect. x total Th, in tun, means that x-velocity of mable has an uncetainty x x v ; = i mx m 6

7 H height of oof and g acceleation of gavity. Because whee of x-velocity vaiation v x caused by uncetainty inicle, H xt=v x x H f =v g : i mx g H g : i mx says that as we decease x i, second tem, due to x, Th lage. e some otimum value of x i which coesonds to gets of x i. value of x i that minimizes x total solution function equation to total dx =0: i dx m m 4 H : g 4 H : g dtance a few ecent of diamete of nucleus of a uanium Th So e no cause fo concen about eos due to uncetainty atom. You decide to change fom doing mables to doing heium atoms b) 0,7 kg). What you estimated ms dtance in th (mass=6:64 m =6:64 0,7 kg, we have Using 4 =0:5 mm: e will be a vaiation of whee mable hits gound of x total;minimum s,34, 6:63 J 0,7 0 s m 9:8 m=s kg 0 6:64 4 Th now a signicant eect. total eo in whee mable hits gound n given by x total =x i+x f x i+ a minimum value of x total. We nd th by minimizing x total as a Caying out indicated dieentiation, we get x i;minimum = n, lugging th value in to equation fo x total gives x total;minimum Evaluating th with given numbes yields s 6:63 0,34 J, s total;minimum x 4 0:03 kg m 4 0,6 =:0 m=s 9:8 m: incile in th case. case? 7