Name Solutions to Test 3 November 8, 2017


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1 Nme Solutions to Test 3 November 8, 07 This test consists of three prts. Plese note tht in prts II nd III, you cn skip one question of those offered. Some possibly useful formuls cn be found below. Brrier penetrtion: 34 h J s ev s J s 6.80 ev s T 6 9 V0 ev.600 J V0 Reflection off step: mv 0 V0 if V0 R V 0 n n if V0 n 0,,, Prt I: Multiple Choice [0 points] For ech question, choose the best nswer ( points ech) e L Hrmonic Osc. n Hydrogen 3.6 evz n D squre well: n n ml nx n x sin L L n,,3,. When prticle hits step boundry, under wht conditions is it t lest prtilly reflected? A) If nd only if the potentil is smller thn the energy B) If nd only if the potentil is greter thn the energy C) If the potentil step is positive, but not if it s negtive D) If the potentil step is positive or negtive (not zero) ) Never. If r is solution to the timeindependent Schrodinger eqution in three dimensions with energy, wht would be the corresponding solution to the timedependent eqution? r r C) re it it D) re ) None of these A) B) 3. The expecttion vlue x of the position opertor for wve function x tells you wht? A) The most likely plce to find the prticle B) The lest likely plce to find the prticle C) The position the prticle ctully is D) The vlue where the Hmiltonin must be evluted to get the energy ) The verge vlue of the position you would get if you mesured it multiple times
2 4. In quntum mechnics, in ddition to the wve function t t = 0, x,0 know to compute the wve function t lter times, x, t? A) The initil position x0, but not the initil momentum p0 B) The initil momentum x0, but not the initil position p0 C) The initil position x0 AND the initil momentum p0 x, t t D) The time derivtive of the wve function, ) None of these; the initil wve function x, t is sufficient ll by itself., wht else must we. An electron in 4p stte hs which of the two vlues? A) l 4, n B) n4, l C) m4, n D) n4, m ) l 4, m 6. When we hd wve impcting from the left on step function with V0 we found solutions on the right tht went like e x, but we threw out e x. Why? A) This wve function is not normlizble, since it blows up t infinity B) This solution would represent wve coming in from the right, which is not wht we hd C) This solution is redundnt, since it is proportionl to e x D) The wve function cnnot go t ll into the forbidden region, so it must be zero here ) It is only n pproximtion, since this wve quickly flls to zero on the left we felt sfe in ignoring it 7. Which of the following vlues is impossible for the ngulr momentum squred L? A) 0 B) C) D) 6 ) All of these re possible 8. Wht is the difference between bound stte nd n unbound stte? A) Bound sttes hve < V(0), unbound sttes hve > V(0) B) Bound sttes hve > V(0), unbound sttes hve < V(0) C) Bound sttes hve < V(), unbound sttes hve > V() D) Bound sttes hve > V(), unbound sttes hve < V() ) In quntum mechnics, becuse the position of prticle is uncertin, the distinction is meningless 9. Wht is the spin s of n electron, the quntum number determining its intrinsic ngulr momentum squred? A) ½ B) 0 C) D) Some other definite vlue ) Different for different electrons 0. Which subshell gets filled in immeditely fter filling in the 3p subshell in n tom? A) 3d B) d C) 4s D) 3s ) None of these
3 Prt II: Short nswer [0 points] Choose two of the following three questions nd give short nswer (4 sentences) (0 points ech).. lectrons in hydrogen re described by four numbers, n, l, m nd m s. Wht restrictions (if ny) re there on these four numbers? The principl quntum number n is positive integer, n,,3,. The integer l is nonnegtive integer smller thn l, so l 0,,,, n. The integer m is one of mgnitude no greter thn l, so ml, l, l,, l. Finlly, ms tkes on the vlues m. s. It is esy to show tht for the infinite squre well, the eqution sin n x nx L stisfies Schrodinger s eqution. Why then did we multiply it be L? In prticulr, give the relevnt eqution tht ws used to find this fctor (though you don t hve to do the mth). The wve function must not only stisfy Schrodinger s eqution, it must lso be x dx. If this is not true, sy normlized. This mens tht we need x dx A, we cn then define comes from. ˆ x x A. Tht s where the fctor of L 3. A single photon is sent into hlfsilvered mirror, nd then sent towrds two detectors, which we will ssume hve 00% detection efficiency. Wht is the probbility, for single photon, tht ech of the detectors see it? Wht is the probbility tht they both see it? Wht is the probbility tht neither sees it? The photon hs 0/0 chnce of going ech wy. It never goes both directions, nd it never goes both wys. B A
4 Prt III: Clcultion: [60 points] Choose three of the following four questions nd perform the indicted clcultions (0 points ech). 4. An electron is plced in potentil of one of three types: hydrogenlike tom, infinite squre well, or hrmonic oscilltor. It is found tht it hs evenly spced energies, so to go from level n to level n+ lwys tkes 40.8 ev of energy. () Which type of potentil is it: hydrogenlike tom, infinite squre well, or hrmonic oscilltor? The hydrogen tom hs energies proportionl to energies proportionl to leds to uniform spcing of the levels, since n, nd the infinite squre well hs n, which n. The hrmonic oscilltor hs energies given by n n n n n n. Hence it must be hrmonic oscilltor. (b) For the pproprite potentil found in prt (), find the corresponding unknown: Z for hydrogenlike, L for infinite squre well, or for the hrmonic oscilltor. We hve n n, so we hve 40.8 ev s ev s (c) Wht is the ground stte energy for this system? The ground stte energy is ev 0.4 ev. 0 (d) If you wnt to go from level n = 7 to n = 4, wht would you emit photon or bsorb one, nd wht would be the energy of tht photon? Since we re going down in energy, energy is lost, so we would emit photon. The energy would be ev.4 ev. 7 4
5 . A set electrons ( m9.0 kg. 0 ev/ c ) hve energy = 7.0 ev nd impct brrier of height V ev nd thickness L = 0.0 nm. () How mny electrons re reflected from the brrier? This is pretty strightforwrd. We first clculte the dmping coefficient, which is.0 ev73.0 ev 7.0 ev mv0 mc V0 c m ev s.9980 m/s We then substitute it into the formul for the penetrtion probbility, to get 7.0 ev 7.0 ev T L e 6 exp m.0 0 m V0 V ev 73.0 ev e We then multiply this by the number of toms to get NT NT I intended to sk for the number trnsmitted, but I sked for the number reflected, so we just subtrct from the totl to get N N N T T (b) The brrier is now incresed until it is the right thickness so exctly one electron is expected to mke it through the brrier. Wht is the thickness L of the brrier now? The number trnsmitted is now NT NT, so we hve T N.3 0. Using exctly the sme formuls s before, we now hve L 7.0 ev 7.0 ev L L N T 6 e 6 e 3.79 e, V0 V ev 73.0 ev L 0 e 3.79N , L ln.6 0.8,.8.8 L m m 0.37 nm.
6 6. A prticle of mss m lies in the hrmonic potentil, V x kx This wve function hs x nd 30 xx 0 x, 0 otherwise. x x 7. The wve function is. It hs lredy been properly normlized. () Find the expecttion vlue of the momentum p nd momentum squred Since the wve function is rel, the expecttion vlue of the momentum vnishes, p 0. For the momentum squred, we hve * * * d op p p dx dx dx i x dx 30 d 30 d x x x x dx x x xdx 0 dx 0 dx x x 60 0 x x dx. 3 3 (b) Find the uncertinties 0 0 x nd p. p. We hve x x x 0 0 p p p 0., 8 Though we weren t sked, we cn check the uncertinty principle, x p. 4 (c) If you were to mesure the energy of the prticle, wht would be the verge vlue you would obtin? The verge energy is the expecttion vlue of the Hmiltonin, so we hve p 0 k k H V p k x m m m 7 m 7.
7 7. An electron is in the stte 3,,0 of hydrogen. Some hydrogen wve functions re written below. () Write explicitly the wve function 3,,0 s function of r,, nd. The wve function is given by 4 r r r 3 3 r r r 3 3,,0 R3,Y,0 e cos e cos (b) Find the probbility density for this wve function. * The probbility density is just, which since it is rel is just the sme thing s, so we hve * r r r 3 8r r r 3 3,,0 3,,0 3,,0 e cos e cos (c) Find ll the plces tht the electron definitely is not. This is ny plce tht the wve function vnishes. So we just look t ech fctor, nd sk where it vnishes. The first fctor vnishes t r = 0, which is the origin. The second fctor, r 6, vnishes t r = 6. And the finl fctor vnishes when cos 0, which hppens only t (keeping mind tht we lwys choose 0 ). This corresponds to the xyplne. So in summry, it vnishes in the xyplne nd on sphere of rdius 6 (note tht the former criterion includes the origin, so it is unnecessry to list r = 0 seprtely). r r r r R e R e R e r r3 4 r3,0, 3 3,0, 3 3, 3 7 Y,0 cos, Y3, sin cos e, Y3,0 cos 3cos i 3
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