Impurities in solids can be sometimes described by a particle-in-a-box model. Suppose He is substituted for Xe, and assume a particle-in-a-cubic-box model, the length of whose sides is equal to the atomic diameter of Xe (≈ 2.62 Å). Compute the lowest excitation energy for the He atom’s motion. (This is the energy difference between the ground state and the first excited state.)
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Impurities in solids can be sometimes described by a particle-in-a-box model. Suppose He is substituted for Xe, and assume a particle-in-a-cubic-box model, the length of whose sides is equal to the atomic diameter of Xe (≈ 2.62 Å). Compute the lowest excitation energy for the He atom’s motion. (This is the energy difference between the ground state and the first excited state.)
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- 3A simple harmonic oscillator is in the ground state with A-², where A=1/41/4 jm-1/2 and a = 1/2 m2, What is a (x²) and Ax? (Hint: [ [_xe-x²³ dx=0 and [_x²a-b²³dx==√m/6²³ ) Ax- μm m2 umA conduction electron is confined to a metal wire of length (1.46x10^1) cm. By treating the conduction electron as a particle confined to a one-dimensional box of the same length, find the energy spacing between the ground state and the first excited state. Give your answer in eV. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer
- A quantum system is described by a wave function (r) being a superposition of two states with different energies E1 and E2: (x) = c191(r)e iEit/h+ c292(x)e¯iE2t/h. where ci = 2icz and the real functions p1(x) and p2(r) have the following properties: vile)dz = ile)dz = 1, "0 = rp(x)T#(x)l& p1(x)92(x)dx% D0. Calculate: 1. Probabilities of measurement of energies E1 and E2 2. Expectation valuc of cnergy (E)I need the answer as soon as possible15 6x4 – 12Lx³ + 15 L²x? - 3,3 (- L²r? L3x 6r4 - N2 2me 12La3 dx = dx 2 2 2 Evaluate the integral in the numerator. p*H@ dx = N² 40m h² L³ e 40me The denominator of the E expression from Step 1 is 6 _ 3Lx + 13 ,2 4 -L²x 4 Lx³ + F8-3Lz' + 12rt - (%) Lz3 + +L교2 p dx = N² dx Evaluate the integral in the denominator. 1 1 φ φ αχx = N2 840 840 Step 3 of 6 Divide the numerator by the denominator (both from Step 2) and simplify. (Use the following as necessary: ħ, L, me, P, T, and x.) 21h? E L²m 21h? L²me e Step 4 of 6 Calculate the energy for an electron in a 0.43-nm box using the formula from Step 3. = 4.0 0.26 X 840 kJ•mol-1 φ Step 5 of 6 Calculate the exact energy for an electron in the first excited state in a 0.43-nm box. n2h2 Recall that for a particle in a one-dimensional box we can write En we can therefore calculate an exact solution. 8mL2' Eexact = 4.0 197 X kJ-mol-1 Submit
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