finalsol.nb In my frame, I am at rest. So the time it takes for the missile to reach me is just 8µ106 km

Transcription

1 finalsol.n Physis D, Winter 005 Final Exam Solutions Top gun a v enemy = 0.4 in my enemy's frame, v' missile = so, in my frame, v missile = Å º H0.4 H0.7 (it must e less than!) In my frame, I am at rest. So the time it takes for the missile to reah me is just 8µ06 km Å º.0 seonds Sustainale Fusion Will Solve Energy Crisis a The perentage of Deuterium whih eomes energy is Å H Å Å º 0.68 % H.06 The mass of the kg of Deuterium whih will eome energy then is kg. This means an energy of E = H kg º 5.74 µ 0 4 Joules will e released. We an get 0 9 Joules from 0 9 Å Å º 7, 400 kg of Deuterium 5.74µ0 4 Estimating The Plank Constant a We have two "experiments" here in whih energy is transfered from inoming photons, with wavelengths l = 00 nm and l = 405 nm, to eletrons in a lok of sodium, whih give up some of the energy in freeing themselves from the solid, and have the rest of their energy sapped while trying to go against the stopping potential. The equations for this proess is just a slightly ompliated form of energy onservation Putting together E g = f + K i, E g = h f = Å h, K i - K f = DU =» e V stop», and finally K f = 0 gives h Å l = f + e V stop, or e V stop = h f - f l Now, given two wavelengths, we get two suh equations and we an find the the slope h and the interept f for this linear relationship. H H -e V stop The slope, h, is given y e V stop Å ÅÅÅ = 9 ev nm ê º.97 µ 0-5 ev se º 6.6 µ 0-4 Joule se f - f The interept is found then as f = Å h H 9 ev nm l - e V stop = Å ÅÅÅ -.85 ev =. ev 00 nm 9 ev nm or y (as a hek) f = Å ÅÅÅ ev =. ev 405 nm Finally, the utoff wavelength is the one for whih ÅÅÅ h l ut = f, so we get 9 ev nm l ut = Å = 56 nm. ev

2 finalsol.n 4 Tiger Hunting in a Quantum Jungle a Dx = 4 meters. Dx Dp h ê 4 p. The minimum unertainty in momentum is Dp min = h ÅÅ 4 p Dx 50 Joule se = Å ÅÅÅ 4 p H4 m º kg meters ê se kg metersêse So the minimum unertainty in speed is Dv min = Å 00 Å kg ÅÅÅ = meters ê se Initially, the position unertainty is 4 meters. For eah of the 0 seonds with Dv min º 0.0 meters ê se, the unertainty in position inreases y 0.0 meters. Thus the final unertainty in position is 4. meters. 5 Triggering a Transition Between Quantum States a Normalization means Ÿ y y dx =, so we have = C Ÿ Hy + y Hy + y dx = C HŸ y y dx + Ÿ y y dx + Ÿ y y dx + Ÿ y y dx = C H The last line is so eause y and y are normalized and orthogonal. Hene we have C = ë è!!!. If we are measuring our ox's length in nm, C must have units of nm -ê, so with units, C = ÅÅÅ è!!! nm-ê To make this a funtion of time, we multiply eah part y its time-energy evolution exponential. YHx, t = Hx e - i E tê + y Hx e -i tê D where E n = ÅÅ to save time, I will refer to E n Å as w n. ÅÅ n p m Reall that the energy operator Y = i Y. My favorite way to hek whether a state is stationary is t to see if it is an eigenfuntion (a.k.a. eigenstate) of the energy operator. It must e for it to solve Shrodinger's Y = i t Y = i C H y Hx H-i w e -i w t + y Hx H-i w e -i w t = C H w y Hx e -i w t + w y Hx e -i w t This is not proportional to the wavefuntion Y eause w w, and so this state is not an energy eigenstate, nor is it a stationary state. Now for the average energy. Reall that all of the wavefuntions for the infinite square well are real, so y = y; however, the omplex onjugation will hange the sign of the exponents, i w n t. The average value of energy is < E >= C Ÿ Iy e +i ÅÅÅ E t + y e +i ÅÅÅ t M@i = C Ÿ Iy e +i ÅÅÅ E D Iy t e -i ÅÅÅ E t + y e -i ÅÅÅ t M dx t + y e +i ÅÅÅ t M I E y e -i ÅÅÅ E t + y e -i ÅÅÅ t M dx = C AE Ÿ y e +i E ÅÅÅ t y e -i E ÅÅÅ t dx + Ÿ y e +i ÅÅÅ t y e -i ÅÅÅ t dxe = H + H D = E +E ÅÅÅÅ

3 finalsol.n d The average position at time t is < x > = C Ÿ Iy e +i ÅÅÅ E t + y e +i ÅÅÅ t M x Iy e -i ÅÅÅ E t + y e -i ÅÅÅ t M dx = = Å nm A Ÿ y x y dx + Ÿ y x y dx + Ÿ y x y Ie i ÅÅÅ E t-i ÅÅÅ t + e -i ÅÅÅ t+i ÅÅÅ t M dxe Å E -E Ÿ y x y dx + Ÿ y x y dx + Ÿ y x y H osh t dxd Now, just use the given definitions < x > = x 0 + A osh E -E ÅÅÅÅ t = x 0 + A oshw t e x 0 will e the middle of the well of ourse. The amplitude will e, if we take our ox to have its left wall at the origin and its right wall at x = nm, A = Ÿ x sinhk x sinhk x dx where k n = n p ê Now use the formula given, sinhq sinhq = oshq - q - oshq + q D. Notie that k - k = k and k + k = k. Then A = Ÿ x oshk x - oshk x D dx ÿ H nm - Now we have to i..p. (integrate y parts). It is quiker to do it like this: Hene, Ÿ 0 x oshkn x dx = x k n sinhk n x 0 -Ÿ 0 k n sinhk n x dx = Å k oshk n x n 0 = H ÅÅ n p HosHn p - A = nm A H ÅÅÅÅ p Hos p - - H ÅÅ p HosH p - E = Å nm A = - Å 8 Å = - ÅÅÅ 8 nm º nm 9 nm p 9 p - ÅÅÅÅ p 9 H-D = + ÅÅÅÅ p A negative amplitude! That is a surprise! But it does make sense if you graph the funtion that you are integrating. There is learly more area under the x axis than aove. See the figure elow, in units where =. This just means that the eletron starts out traveling to the left instead of to the right at t = 0. 9 D

4 finalsol.n 4 f (for finally) The angular frequeny is W = -E = ÅÅÅÅ ÅÅ So the period is = ÅÅ m h p ÅÅ p m e Å p I4 Å - p Å M T = ÅÅ p W = ÅÅÅÅ H p me p Å ÅÅ h ÿ H nm = ÅÅÅÅ ev ÅÅ Å Å H 40 ev nm ÿ µ0 7 nmêse nm º.66 µ 0-5 seonds 6 Rapping Aout a -D Harmoni Osillator a TISE: - ÅÅ m ÅÅÅ yhx + ÅÅÅÅ x m Hw x + w y yhx = E yhx with w < w We must separate variales like so, y = y x Hx ÿ y y Hy. Then we must use the fat that the first exited state will have an exitation in the diretion with the smaller exitation energy, w. I.e., sine w < w, y Hx, y ~ y x, Hx ÿ y y,0 Hy and you are given that y x, Hx ~ "######### m w Å x e - m w ÅÅÅ ÅÅÅ x Å y y,0 Hy ~ e - m w ÅÅÅ ÅÅÅ y so you should get that y Hx, y = A "######### m w Å x e - m w x ÅÅÅ ÅÅÅ Å e - m w ÅÅÅ ÅÅÅ y Sine "######### m w Å is a onstant, it is also okay to work with y Hx, y = A x e - ÅÅÅ m w ÅÅÅ x Å - ÅÅÅ m w ÅÅÅ y Å Å = A "######### m w Å x e - m w x ÅÅÅ ÅÅÅ Å - m w ÅÅÅ ÅÅÅ y Å Normalization: = A x e - ÅÅÅ m w ÅÅÅ x Å dx = A A- ÅÅ w ÅÅ m e - ÅÅÅ m w ÅÅÅ y Å dy e - ÅÅÅ m w ÅÅÅ x dxe A "######### p Å E p m w = A A- ÅÅ w ÅÅ "######### Å m m w E A "######### p Å m w E = A A- ÅÅ "####### p w ÅÅ w H- ÅÅÅÅ m-ê E "######### p Å m w = A A ÅÅÅÅ Å p w ê w ê E m A = "################ w ê ######### w ê m ÅÅ p

5 finalsol.n 5 d This is a stationary state. In the x diretion there is ÅÅÅÅ w worth of energy, and in the y diretion, there is ÅÅÅÅ w worth of energy. The total energy is E = ÅÅÅÅ w + ÅÅÅÅ w e This is not a degenerate state, sine w w. The two losest states are n = n = 0 and n =, n = 0. Both have energies different from this one. f The average potential energy of this state turns out to e ÅÅÅÅ of the total energy, ÅÅÅÅ w 4 + ÅÅÅÅ w 4. 7 An Exited Hydrogen Atom a This is H, so Z =. The s state orresponds to R,0. The proaility as a funtion of r alone is PHr = Hr R,0, so we have R,0 Hr ~ I - PHr ~I r - r Å M e -rê r M e -rê Now, we an find maxima in PHr y finding extreme points for its square root, whih is easier in most ases. r Hr R,0 ~ I r - r M H- ê + H - r ê = -r ê + r ê a r ê = - r ê + r ê a 0 Of ourse, we have extrema for r Hr R,0 = 0, i.e., r ê = è!!! 5, just like in homework. You an see from plotting or from evaluating PHr at the two points that r ê = + è!!! 5 º 5.4. This is a it more than the seond Bohr radius r Bohr = 4 The p state orresponds to R, R, Hr ~ r e-rê a0 è!!!!!!!! PHr ~r e -rê r Hr e -rê a0 ~ r - r ê r ê = 4 This is preisely the seond Bohr radius. Three heers for Bohr!