Midterm1Solutions - ‘3 v ‘ 1 Up to this point the only...

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Unformatted text preview: ‘3 v; ‘ 1) Up to this point, the only model systems we have discussed were the free particle and the particle-in-a-box. One we have not discussed yet is the so-called particle—on—a-ring, which is our first model to consider rotational kinetic energy. In this model, the particle is restricted to follow a perfectly circular path with constant radius R. It can crudely model the motion of electrons in the conjugated 7T- orbital system of benzene and other aromatic molecules. The natural coordinates for circular motion are polar coordinates. In this case, the Schrodinger equation can be written as: h2 62 — —— 6 = E 6 This is a simple linear differential equation with constant coefficients. The boundary conditions for this equation must be 111(6) = 1/)(6 + 211), which results in the following solutions for the energy eigenfunctions and eigenvalues: _ _1_ 'm9 1MB) — met} h2 m2 = szz m = O,'_|'1,-_l’2, Em 3. Describe the degeneracy of the energy levels of a particle—on—a-ring. Tim 45me SH; W87 lave/K (fly/‘30» l5 §ivgl7 ASMWJLC. A” am“ [Ci/J5 [Vi/l #0) Ma clawing Atdwm‘zae‘ - ; Whit" Wall _ v E ' 2mg“ 27” ii. b. What is the average angular position, (6), of the particle? <97 : Sawwetiig ” ‘Zfil +£M6 | 'LMQ — m 597,76 lo 0 ._l lit LN? 4W? i ZN, ‘_ —,~ 9 Zita 96 ‘5 QH’LT‘I) a} : l f 62 1:? erL 7' c. What' IS the most probable angle to find the electron? W93— W‘Me lfléfi mam“ 4W; .L. 02% 6/ Zii WW” W‘Hc WW1 Walls WKWEQX‘ Thwacm ale [WAG-WM 415$ ' W) o Tit”: i5 “0 We??? fWDLr/Hc M‘cJflr [lax/3‘97”!” me/éy d. What quantum mechanical justification do we have to require that boundary conditions for the Schrodinger equation be 111(6) = 1M6 + 2n)? 6%)4T ICQWMSfaw-«Qs to wig“ L9,) Mowfi “HAL #9ng HM Alf/[6?) Wit/15+ é: fngVkMt/kavtgz 30 SL musf meef a? War“ 3154.; (Alum Limit (emf; .. (We): ’Z/Jfiadn) 2) Quantum corrals are a class of nanoscale structures composed of a collection of individual atoms arranged on the surface of a metal in the form of a polygon. Outside the walls ofthis corral, electrons are free to move along the surface of the atom. Electrons within the confines of the corral are restricted to stay within the corral. STM Image of a Quantum Corral. M.F. Crommie, C.P. Lutz, D.M. Eigler, E.J. Heller. Waves on a metal surface and quantum corrals. Surface Review and Letters 2 ( 1), 127—137 (1995). Consider an electron existing inside a square-shaped quantum corral of iron atoms on the surface of a copper crystal. lfthe longest wavelength of light that this electron can absorb is 7.50 x 10'6 m, what is . the length of the side of the corral? ”(roll/l awl Vt; .c Yaw‘lfilc m Z”? lama 3) All of the free particles we have discussed up until this point have been described by a single quantum state and wave vector k. When we considered such a free particle, we asserted that we could not determine the particle’s position at all (0x = 00) because the momentum of the particle was known exactly (0px = 0). Now, consider a free particle in a superposition of 2 momentum states, such that its wave function is: 9’06) = ie—iklx + ie—ikzx {2' \/§ In this case, there is Uncertainty in the momentum because we will have a finite probability of observing the particle with 2 different momenta. We call such a superposition a “wave packet”, and it allows us to more precisely predict the position of free particles. For this problem, let k1 = 4.31 x 1010 In‘1 and let k2 = 4.74 x 1010 m”1 3. Determine the average momentum of this particle. 4W: isfigak‘xfaflzctzmw .me )le 4k»: Jlf;ixtk>/KX I ($7335 (0‘760H£+ fiefihx)(fi:‘0 ~ka \ b: wizlagifl c. How precisely can we determine the position of this particle? I. QUE?— 1 JF/_. -lO ()7: Z M?) ”1001M? M ...
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