Problem Set 4 Solutions

Transcription

1 University of Alabama Department of Pysics and Astronomy PH 253 / LeClair Spring 2010 Problem Set 4 Solutions 1. Group velocity of a wave. For a free relativistic quantum particle moving wit speed v, te total energy is E =f = ω = p 2 c 2 + m 2 c 4 and te momentum is p=/λ= k =γmv. For te quantum wave representing te particle, te group speed is v g = dω/dk. Prove tat te group speed of te wave is te same as te speed of te particle. We can just brute-force tis one. Using te energy equation, we can write ω in terms of k: ω = 1 2 k 2 c 2 + m 2 c 4 (1) v g = dω dk = 1 ( kc 2) (2) 2 k 2 c 2 + m 2 c 4 kc 2 = 2 k 2 c 2 + m 2 c = pc 2 4 p 2 c 2 + m 2 c = p 2 c 4 4 p 2 c 2 + m 2 c 4 (3) In te last line, we substituted back in p= k. If we use p=γmv, we can reduce tis expression to te desired result: dω dk = p 2 c 4 p 2 c 2 + m 2 c 4 = c 2 v 2 = v 2 1 v 2 /c 2 + c2 γ 2 m 2 v 2 c 4 γ 2 m 2 v 2 c 2 + m 2 c 4 = 1 v 2 /c 2 1 v 2 /c ( 2 c 1 v 2 /c 2) = 2 v 2 γ 2 c 2 v 2 γ 2 v 2 + c 2 (4) v 2 + c 2 = ±v (5) v2 v g = v (6) 2. Uncertainty in everyday processes. A woman on a ladder drops small pellets toward a point target on te floor. (a) Sow tat, according to te uncertainty principle, te average miss distance must be at least x f = 2 m ( ) 2H 1/4 g (7)

2 were H is te initial eigt of eac pellet above te floor and m is te mass of eac pellet. Assume tat te spread in impact points is given by x f = x i +( v x ) t. (b) If H =2.00 m and m=0.500 g, wat is x f? Owing to te quantum nature of matter, te initial position of te pellet x i will ave some uncertainty x i. From te uncertainty principle, tis implies tere must also be an uncertainty in te initial momentum in te x direction p ix, and terefore te initial x component of te velocity as well, v ix. Tis random initial velocity will really ave bot a orizontal component v ix and a vertical component v iy. However, since te particle will quickly acquire a sizable vertical speed, te vertical uncertainty will be negligible, and it will not affect te orizontal position at wic te particle lands. Tus, we need only consider te initial random orizontal speed v ix wic will lead to a scatter in te orizontal position of te particle. Classically, te particle as a purely vertical velocity, so te total velocity in te orizontal direction is only tat wic comes from uncertainty. After a time t, tis will cause te particle to ave moved orizontally by v ix t compared to its expected position. Adding tis to te initial uncertainty in position, after a time t our total uncertainty is x(t) = x i + v ix t (8) If te particle starts from rest at eigt H, te time to reac te ground is just t= 2H/g. Tus, te final uncertainty in te orizontal position at wic te particle its te ground is x f = x i + v ix 2H g (9) From te uncertainty principle, we may relate te minimum uncertainty in initial velocity to te minimal uncertainty in initial position: x i p ix = m x i v ix = 2 = v ix = 2m x i (10) Using tis in our expression for te final spread in orizontal position, 2H x f = x i + 2m x i g (11) We can find te minimum spread in x f by minimizing wit respect to x i, and tis will tell us te degree of uncertainty in te final position required by te uncertainty principle it can be

3 more, but tis will give us te minimal value. d ( x f ) d ( x i ) = 1 2H 2m ( x i ) 2 g = x i = 2m ( ) 2H 1/4 (12) g Plugging tis in to our previous expression and simplifying, x f = 2m ( ) 2H 1/4 + 2m ( g ) 1/4 2H 2 g 2m 2H g = m ( ) 2H 1/ m (13) g 3. Significance of te Compton wavelengt. Sow tat te speed of a particle aving de Broglie wavelengt λ and a Compton wavelengt λ c =/mc is v = c 1 + (λ/λ c ) 2 (14) Tere are many ways to go about tis one. Here is a sort one. Since λ c = mc, ten λ cc= m. Te de Broglie wavelengt is tus λ = p = γmv = 1 m γv = λ cc γv (15) Squaring bot sides and simplifying, λ 2 = λ 2 c 2 ) ( ) c (1 v 2 v2 c c 2 = λ 2 2 c v 2 1 c 2 v 2 = 1 + λ2 λ 2 c v 2 c 2 = λ2 λ 2 c c v = 1 + (λ/λ c ) 2 (16) (17) (18) (19) 4. Zero point energy of a armonic oscillator. Te frequency f of a armonic oscillator of mass m and elasticity constant k is given by te equation f = 1 k 2π m (20)

4 Te energy of te oscillator is given by E = p2 2m kx2 (21) were p is te system s linear momentum and x is te displacement from its equilibrium position. Use te uncertainty principle, x p /2, to express te oscillator s energy E in terms of x and sow, by taking te derivative of tis function and setting de/dx=0, tat te minimum energy of te oscillator (its ground state energy) is E min =f/2. Te minimum uncertainty in momentum p, given an uncertainty x in position is given by te uncertainty principle: x p = 2 = p = 2 x (22) Te minimum uncertainty is also ten te minimum average value we can expect eiter variable to take on: p min = p p, x min = x x. Te energy equation may te be rewritten in terms of te minimal x and p: E = p2 2m kx2 = 2 8mx kx2 = 2 8mx mω2 x 2 (23) In te last line, we used ω = k/m, so k =mω 2. Minimizing te energy wit respect to x, de dx = 2 2 4mx 3 + mω2 x = 0 = x 2 = 2mω (24) Plugging tis back in to te energy equation, we ave te minimum energy: E min = 2 2mω + 1 8m 2 mω2 2mω = 1 4 ω ω = 1 ω (25) 2 5. Deduction of atomic energy levels. Te following lines appear in te emission spectrum of an element: λ 1 =122 nm λ 2 =102.5 nm λ 3 =97 nm λ 4 =95 nm λ 5 =656 nm λ 6 =486 nm λ 7 =434 nm λ 8 =1880 nm λ 9 =1280 nm λ 10 =4050 nm (a) Does te spectrum exibit te Ritz combination rule? (b) Calculate te energy of te potons associated wit eac spectral line. (c) Sketc an energy level diagram tat is consistent wit te element s spectral lines.

5 In fact, te wavelengts correspond to te ydrogen spectrum. Armed wit tat knowledge, te problem is muc simpler. Eac given wavelengt corresponds to te energy difference between two levels of te ydrogen atom (say, n = 3 and n = 1). Besides te direct transition from level 3 to level 1, te electron could also go from 3 to 2 and ten 2 to 1, meaning tere must be two oter poton energies tat sum to tat of te first poton: E 3 1 =E 3 2 +E 2 1. Using te energy level diagram below i, it is a simple matter to write down all possible energy level transitions and teir energies, and matc tem to te list. Figure 1: In te Bor model of te ydrogen atom, te emission lines correspond to an electron jump from a iger energy level n to a lower n wit te emission of a poton corresponding to te energy difference between te two levels. For te visible Balmer series, n =2 and n runs from 3 upward. From ttp:// yperpysics.py-astr.gsu.edu/. Te table below lists te energies corresponding to te given wavelengts, teir identification, and teir decomposition into sums of oter given energies. If no decomposition is listed, it means tat tis wavelengt corresponds to te first in a series, and cannot be decomposed into smaller individual transitions. If you did t realize te given wavelengts were from te ydrogen spectra, one could also brute-force i From ttp://yperpysics.py-astr.gsu.edu/base/yde.tml.

6 level λ (nm) f (PHz) E (ev) transition Decompositions E Lyman 5 1 E 1 + E 9 ; E 2 + E 7 ; E 3 + E 4 E 2 +E 8 +E 9 ; E 3 + E 5 + E 9 ; E 3 +E 6 +E 7 E 3 +E 6 +E 8 +E 9 E Lyman 4 1 E 8 +E 6 +E 3 ; E 5 +E 3 ; E 8 +E 2 E Lyman 3 1 E 6 +E 3 E Lyman 2 1 E Balmer 5 2 E 9 +E 8 +E 6 ; E 9 +E 5 ; E 7 +E 6 E Balmer 4 2 E 8 +E 6 E Balmer 3 2 E Pascen 5 3 E 9 +E 8 E Pascen 4 3 E Brackett 5 4 te problem. I wrote a small C program to searc for combinations of te energies corresponding to te given wavelengts. Here are te results, wic matc te predictions of te Bor model nicely. I ordered te wavelengts given from lowest to igest, meaning te corresponding energies E 0 to E 9 are ordered from igest to lowest. Te brute-force searc was carried only as far as four terms. E 0 = 13.1 E 1 + E 9 = 13.1 E 0 = 13.1 E 2 + E 7 = 13.1 E 0 = 13.1 E 3 + E 4 = 13.0 E 1 = 12.8 E 2 + E 8 = 12.8 E 1 = 12.8 E 3 + E 5 = 12.7 E 2 = 12.1 E 3 + E 6 = 12.1 E 4 = 2.86 E 5 + E 9 = 2.86 E 4 = 2.86 E 6 + E 7 = 2.86 E 5 = 2.55 E 6 + E 8 = 2.55 E 7 = E 8 + E 9 = E 0 = 13.1 E 2 + E 8 + E 9 = 13.1 E 0 = 13.1 E 3 + E 5 + E 9 = 13.0 E 0 = 13.1 E 3 + E 6 + E 7 = 13.0 E 1 = 12.8 E 3 + E 6 + E 8 = 12.7 E 4 = 2.86 E 6 + E 8 + E 9 = 2.86 E 0 = 12.8 E 3 + E 6 + E 8 + E 9 = 13.0

7 Tus, te Ritz combination rule is obeyed, even if one didn t realize te spectrum was for ydrogen. 6. Macroscopic quantum matter. Under usual conditions of temperature and density, te beavior of a gas can be interpreted in terms of classical pysics, for example, by Maxwell s kinetic teory (treating te gas as a collection of point particles). However, wen te de Broglie wavelengt of te atoms is of te order of te mean distance d between tem, te classical approac fails. (a) Sow tat te classical approac fails wen d 2 mk b T (26) (b) Assuming tat te mean distance between te atoms is 1 nm, at wat temperature will quantum effects begin to appear in te gas elium? Te average speed of a particle in an ideal gas is v = 3kB T m (27) wic lets us write te average de Broglie wavelengt as λ = p = m v = 3kB T m (28) Te classical approac is expected to fail wen te spacing between atoms approaces te de Broglie wavelengt, or d 3kB T m kb T m (29) At room temperature, tis is about 0.07 nm; at 4.2 K (te boiling point of elium), it is about 0.6 nm. Tis already ints tat quantum effects sould arise in elium near its boiling point, and certainly in te liquid. If we assume a mean spacing of 1 nm for te elium atoms, te temperature at wic te classical approac breaks down is T = 2 3k λ 2 m 2 K (30) 7. How big are atoms? A simple but sopisticated argument olds tat te ydrogen atom as its observed size because tis size minimizes te total energy of te system. Te argument rests on te assumption tat te lowest-energy state corresponds to a pysical size comparable to a de Broglie wavelengt of te electron. Larger size means larger de Broglie wavelengt, ence smaller

8 momentum and kinetic energy. In contrast smaller size means lower potential energy, since te potential well is deepest near te proton. Te observed size is a compromise between kinetic and potential energies tat minimizes te total energy of te system. Develop te argument explicitly, for example as follows: (a) Write down te classical expression for te total energy of te ydrogen atom wit an electron of momentum p in a circular orbit of radius r. Keep kinetic and potential energies separate. (b) Failure of classical energy minimization. Use te force law to obtain te total energy as a function of radius. Wat radius corresponds to te lowest possible energy? (c) For te lowest-energy state, demand tat te orbit circumference be one de Broglie wavelengt. Obtain an expression for te total energy as a function of radius. Note ow a larger radius decreases te kinetic and increases te potential energy, wereas a smaller radius increases te kinetic and decreases te potential energy. (d) Take te derivative of te energy versus radius function and find te radius tat minimizes te total energy. How large is tat radius for a ydrogen atom? For a He + ion wit one electron? Te total energy of te classical system is: E = p2 2m k ee 2 r (31) Using te force equation, we can express te kinetic energy in terms of radius, noting tat te electric force must provide te centripetal force: mv 2 = k ee 2 r r mv2 = k ee 2 2r (32) = p2 2m = E = k ee 2 2r = 1 2 U (33) Te total energy is alf te potential energy, as it must be for a stable bound obit in a 1/r 2 central force. By inspection, we can see tat te minimum energy is wen r 0, corresponding E, i.e., te electron crases into te proton. Bad! If we demand tat te circumference of te orbit, 2πr, equal te de Broglie wavelengt, 2πr = λ = mv = mvr = 2π = = mv = p = r (34)

9 Substituting tis into te energy equation, E = p2 2m k ee 2 r = 2 2mr 2 k ee 2 r (35) Minimizing tis wit respect to r, de dr = 2 2 2mr 2 + k ee 2 r 2 = 0 2 mr = k ee 2 2 r = e 2 mk e (36) (37) (38) For ydrogen, we find r m. For He +, we ave two protons in te nucleus (carge 2e), and te electrical potential energy is modified to 2k e e 2 /r. Tis gives r m. 8. (a) Early neutron model. Te neutron is an electrically neutral particle wit a mass approximately equal to te proton mass. An early model considered te neutron to be an object were te electron is confined inside te proton. Assuming tat te proton radius is r m, estimate te electron s kinetic energy due to Heisenberg uncertainty, and compare it to te neutron rest mass. (b) Energy spread of electron beam. A monocromatic beam of electrons of energy E = 1 kev is incident on a sutter tat opens for t = 1 ns. Wat is te fractional energy spread v/v of te electron velocity v after te sutter? (Decide first weter to perform a relativistic or a non-relativistic calculation.) (a) Confinement on a scale r means tat te uncertainty in position is x r, and uncertainty dictates p x r p = p r (39) Te energy of te electron is E 2 = m 2 c 4 + p 2 c 2. Te uncertainty in energy comes only from te second term, since te electron rest mass is fixed, and tus E c p c 2r 0.1 GeV (40) Tis is about 10% of te neutron s rest mass, a sizable fraction. (b) Since te energy scale is far less tan te electron s rest energy (E mc 2 ), we may neglect relativistic effects. If te pulse is t in lengt, it as a spatial extent of x=v t. Tis implies an

10 uncertainty in momentum: x p = v t p 2 = p = m v 2v t (41) Te uncertainty in velocity can be related to te uncertainty in momentum: v v = m v mv = 2v t 1 mv = 2mv 2 t (42) Te kinetic energy of an electron in te beam is 1 2 mv2, and tus we can relate te uncertainty in velocity to te time uncertainty and te electron energy: v v = 4E t (43)