Midterm #1 - TakeHome - AnswersEngel 5.1 (p. 77-8) describes how to obtain total energy eigenfunctions and eigenvalues for an electron in a finite depth box of width 1.00 x 10-9 m with Vo = 1.2 x 10-18 J. Engel also lists the five lowest eigenvalues for this box.1. What are the five lowest eigenvalues (in J) for an electron in an infinite depth box of the same width?Solution: En= h2 n2ê8 ma2h = 6.626 10−34Jsme = 9.109 10−31kga = 10−9 mn = 81, 2, 3, 4, 5<6.626 × 10−34kg m2s9.109 × 10−31kgm100000000081, 2, 3, 4, 5<En =h2 n28mea296.02479× 10−20kg m2s2,2.40992× 10−19kg m2s2,5.42231× 10−19kg m2s2,9.63967× 10−19kg m2s2,1.5062 × 10−18kg m2s2=2. Which system has lower eigenvalues, the finite depth box or the infinite depth box? Why?Solution: Engel p. 78 gives the five lowest energies of the finite depth box as:4.61 μ 10-20J,1.84 μ 10-19J,4.09 μ 10-19J,7.13 μ 10-19 J,1.07 μ 10-18JEach energy state of the finite depth box is lower than the corresponding state of the infinite depth box. This can be explained by the boundary condition for the infinite depth box that requires the wave function to vanish at x = a. This forces the infinite depth box to curve faster over the interval 0 § x § a, and larger curvature corresponds to larger energy.3. Use the formulas provided in Engel, and any graphing tools at your disposal, to determine the lowest eigenvalue (1% accuracy in E is adequate) for an electron confined to a finite depth box of the same width, but with Vo = 2.4 x 10-18 J.Solution: Engel, p. 77 shows two equations that must be satisfied by the energy of a stationary state (the energy must satisfy one of the equations). The top equation is used to generate the lowest energy state (see figure on p. 78), and it is written below in two pieces (LHStop = left-hand side, RHS = right-hand side).2Exam1takehome_answers.nbClear@EnergyDV0 = 2.4 × 10−18me = 9.109 × 10−31a = 10−9hbar = 1.055 × 10−34LHStop =$%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%me HV0 − EnergyLa22 hbar2RHStop =$%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%me Energy a22 hbar2 TanA$%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%me Energy a22hbar2E2.4 × 10−189.109 × 10−31110000000001.055 × 10−346.39688× 109è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!2.4 × 10−18− Energy6.39688× 109è!!!!!!!!!!!!!!!!Energy TanA6.39688 × 109è!!!!!!!!!!!!!!!!Energy EAny E that satisfies this equation, i.e., makes LHStop = RHStop, is acceptable, but we want the lowest energy possible. Unfortunately, the equations cannot be solved algebraically, so we must 1) plot the functions on the left-hand side for a range of possible energies, and 2) plot the functions on the right-hand side for the same energy range. The state's energy is given by the crossing point of these two graphs.Note #1: We already know from the previous problems that the ground state energy of the finite depth box must be less than the ground state energy of the infinite depth box (6.02 × 10-20J). Since Vo for this problem is larger than Vo for the previous problem, we also know that the answer must be greater than ground state energy given in the previous problem (4.61 × 10-20J). So we need to graph the energy range 4.61-6.02 × 10-20 J. The following graph was arrived at by using trial-and-error to adjust the horizontal (Energy) and vertical ranges. The crossing point occurs at about 4.97 × 10-20J.Exam1takehome_answers.nb3Plot@8LHStop, RHStop<, 8Energy, 4.96 10−20, 5.0 10−20<, PlotRange → 89.6, 10<D4.96×10-204.97×10-204.98×10-204.99×10-209.69.79.89.9 Graphics 4Exam1takehome_answers.nbFindRoot@LHStop == RHStop, 8Energy, 5 ∗ 10−20<D8Energy → 4.97192 × 10−20<Referring to Engel’s original finite depth box (Vo = 1.2 x 10-18 J, a = 1 x 10-9 m) and the general formulas that he provides for the eigenfunctions,4a. What are the normalized eigenfunctions for E = 4.61 x 10-20 J and 4.09 x 10-19 J?Solution: The general formulas for the wave functions in regions I-III are given on Engel p. 77. Engel states that A' = B' = 0, which greatly simplifies the formulas for region I and region III. To take the next step, it is helpful to note the symmetry of the potential function. The box has a point of symmetry at x = 0, and all wave functions must have either even or odd symmetry with respect to this point. The ground state should be described by an even function (no nodes). The first and second excited states should be described by odd (one node) and even (two nodes) functions, respectively.The symmetry requirements introduce additional restrictions on A, B, C, and D. An even wave function has the same value at x = a/2 and -a/2. This means C = 0 and A = B. As it happens, E = 4.61 x 10-20 J is the ground state, and E = 4.09 x 10-19 J is the second excited state, so both energies correspond to states described by even wave functions.Using the formulas in Engel, p. 77, and letting A = B, we can obtain the relative values of A, B, and D:2A e-kHaê2L = 2D cosIk aÅÅÅÅ2MAÅÅÅÅÅD= ekHaê2L cosIk