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Physics 249 Homework 5 Due Oct 19th 1) Consider two potential steps. One from V(x<0)=0 to V(x>0)=V0 and the other from V(x<0)=0 to V(x>0)=-V0. Consider a particle incident on this step potential from βx with total energy E=2V0. Note this is the total energy including the potential energy. For each calculate. a) The wave numbers in regions two (x>0) in terms of the wave number in region 1 (x<0) and V0. b) The reflection coefficients. c) The transmission coefficients. Note since the potential energy is zero in region one the total energy is the same as the kinetic energy. a) The wave number for both ases i (V0) and ii (-V0) in region one is.: π1 = β2ππΈ β4πV0 = β β in region two the wave number are π2π = β2π(2V0 β V0 ) β2πV0 = β β β2π(2V0 β (βV0 )) π2π = β = β6πV0 β b) The reflection coefficients. 2 π1 β π2 2 2 β β2 π π = ( ) =( ) = 0.0294 π1 + π2 2 + β2 2 π1 β π2 2 2 β β6 π ππ = ( ) =( ) = 0.0102 π1 + π2 2 + β6 c) The transmission coefficients. ππ = 4π1 π2 8β2 = 2 = 0.9706 2 (π1 + π2 ) (2 + β2) πππ = 4π1 π2 8β6 = 2 = 0.9898 2 (π1 + π2 ) (2 + β6) Note that T+R=1 in both cases! 2) Consider potential barrier 0.5nm wide and 10eV high. Electrons accelerated through 5V approach the barrier. What percentage of electrons will tunnel through the barrier? β1 π = π π ( |πΉ|2 |π΄| 2 ) = [1 + 2 π ππβ πΌπ ] πΈ πΈ 4 π (1 β π ) 0 0 Note that there was an error in the formula in the book and in my notes. The sinh should have been squared. The formula has been corrected in my notes from lecture 16. Solutions with either sinh2 or sinh were accepted. β1 β2π(π0 β πΈ) π ππβ ( )π β π = 1+ πΈ πΈ 4 π (1 β π ) 0 0 [ ] 2 β2ππ 2 (π0 β πΈ) π ππβ ( )π βπ = 1+ πΈ πΈ 4 π (1 β π ) 0 0 [ ] β1 2 β1 β2 β 0.511π₯106 (10 β 5) π ππβ ( ) 0.5 197.3ππππ 2 π = 1+ [ = 4.23π₯10β5 = 4.23π₯10β3 % 4 β 0.5(1 β 0.5) ] 3) Consider a infinite 3D box potential with L2=2L1 and L3=4L1. What are the quantum numbers of the lowest degenerate energy levels? List, ordered by energy, the quantum numbers and energies of all the levels up two the lowest energy set of degenerate energy levels. πΈ= β2 2 β2 π 2 π12 π22 π32 (π1 + π22 + π32 ) = ( + + ) 2π 2ππΏ2 1 4 16 1, 1, 4 and 1, 2, 2 Systematically you increment up the third eigenstate comparing it to incrementing up the second (or first) eigenstate at each step. 1, 2, 1 is smaller then 1, 1, 4 and 1, 2, 2 is the same as 1, 1, 4. Both give fractions of ¼ and 1. Define β2 π 2 πΈ0 = 2ππΏ2 for the constant term in front of the terms 1, 1, 1: πΈ = 1.313πΈ0 1, 1, 2 πΈ = 1.5πΈ0 1, 1, 3 πΈ = 1.813πΈ0 1, 2, 1 πΈ = 2.063πΈ0 1, 1, 4 πΈ = 2.25πΈ0 1, 2, 2 πΈ = 2.25πΈ0 4) The radial probability distribution function for the hydrogen atom ground state can be writer as π(π) = πΆπ 2 π β2π/π0 , where C is a normalization constant. a) Why is the factor of π 2 included? b) Show that P(r) has itβs maximum value at r=a0. c) If you considered equal infinitesimal volumes, what at what radius is the maximum probability to find an electron? d) Calculate the expectation value of r. a) The probability at a give radius is integrated over a spherical shell of area 4ππ 2. b) ππ(π) 2 β2π β2π/π0 2 = 2πΆππ β πΆπ π π0 = 0 ππ π0 1 π β π π2 = 0 0 π = π0 β2π π 2 π(π) 2 β2π 2 β2π 2 2 β2π 2 π π π = 2πΆπ 0 β πΆ2π π 0 β πΆ2π π 0 + πΆπ ( ) π π0 ππ 2 π0 π0 π0 2 π π(π0 ) = 2πΆπ β2 β 8πΆπ β2 + 4πΆπ β2 = β2πΆπ β2 ππ 2 indicating a local maxima c) |π(π)|2 = πΆπ β2π/π0 Which has a maxima at r=0. d) 4 β 3 β2π/π 0 β« π π π03 0 using 4 β 3 β2π/π 4 3! 4 6 4 3 0 = π = π0 4 = 3 3β« π π 3 π0 0 π0 2 π0 16 0 2 (π ) 0 As expected, larger than the maximum.