PC Chapter 07

PC Chapter 07 - 7 Tunneling Phenomena 7-1 (a) The...

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103 7 Tunneling Phenomena 7-1 (a) The reflection coefficient is the ratio of the reflected intensity to the incident wave intensity, or ( 29 (29 - = + 2 2 1 21 1 i R i . But (29(29 -=- - =- +=+= 22 1 1 1 * 1 112 i ii i , so that = 1 R in this case. (b) To the left of the step the particle is free. The solutions to Schrödinger’s equation are ± ikx e with wavenumber = h 12 2 2 mE k . To the right of the step (29 = U xU and the equation is ( 29 (29 ψ ψ =- h 2 2 dm U Ex dx . With ψ - = kx xe , we find ψ ψ = 2 2 2 d kx dx , so that -  =   h 2 2 mUE k . Substituting = h 2 2 mE k shows that = - 1 E UE or = 1 2 E U . (c) For 10 MeV protons, = 10 MeV E and = 2 938.28 MeV m c . Using ( 29 - == h 15 197.3 MeV fm 1 fm 10 m c , we find ( 29 δ = = h 2 197.3 MeV fm 1 1.44 fm 2 2 938.28 MeV 10 MeV c k mE c . 7-2 (a) To the left of the step the particle is free with kinetic energy E and corresponding wavenumber = h 1 2 2 mE k : ψ - =+ 11 ikx x Ae Be 0 x To the right of the step the kinetic energy is reduced to - EU and the wavenumber is now - = h 2 2 2 mEU k ψ - i x Ce De 0 x
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104 CHAPTER 7 TUNNELING PHENOMENA with = 0 D for waves incident on the step from the left. At = 0 x both ψ and ψ d dx must be continuous: (29 ψ =+= 0 ABC ψ = -= 12 0 d ik A B ik C dx . (b) Eliminating C gives + =- 1 2 k AB k or  - =+   11 22 kk . Thus, ( 29 ( 29 ( 29 ( ( - - = == + + =-= + 2 2 2 2 1 1 4 1 B R A TR (c) As EU , 2 0 k , and 1 R , 0 T (no transmission), in agreement with the result for any energy < . For →∞ E , and 0 R , 1 T (perfect transmission) suggesting correctly that very energetic particles do not see the step and so are unaffected by it. 7-3 With = 25 MeV E and = 20 MeV U , the ratio of wavenumber is ( 29 = = -- 1 2 25 5 2.236 25 20 kE k . Then from Problem 7-2 - + 2 2 51 0.146 R and 1 0.854 . Thus, 14.6% of the incoming particles would be reflected and 85.4% would be transmitted. For electrons with the same energy, the transparency and reflectivity of the step are unchanged. 7-4 The reflection coefficient for this case is given in Problem 7-2 as ( ( ( ( - - = + + 2 2 2 1 1 B R A . The wavenumbers are those for electrons with kinetic energies = 54.0 eV E and - 54.0 eV 10.0 eV 64.0 eV : = - 1 2 54 eV 0.918 6 64 eV k . Then, ( ( - - = + 2 3 2 0.918 6 1 1.80 10 0.918 6 1 R is the fraction of the incident beam that is reflected at the boundary.
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MODERN PHYSICS 105 7-5 (a) The transmission probability according to Equation 7.9 is (29 α  =+  -  2 2 1 1 sinh 4 U L TE EUE with [ ] α - = h 12 2 mUE . For << EU , we find ( 29 α h 2 2 2 2 1 mUL L by hypothesis. Thus, we may write α α≈ 1 sinh 2 L Le . Also -≈ UEU , giving αα +≈ 22 1 1 16 16 LL UU ee T E EE and a probability for transmission α - == 2 16 L E P T Ee U . (b) Numerical Estimates: ( 29 - h 34 1.055 10 Js 1) For - 31 9.11 10 kg m , - - 21 1.60 10 J UE , - = 10 10 m L ; [ ] α - - = h 81 2 5.12 10 m and α - = 2 0.90 L e 2) For - 31 9.11 10 kg m , - - 19 1.60 10 J , - = 10 10 m L ; α - 91 5.12 10 m and α - = 2 0.36 L e 3) For - 27 6.7 10 kg m , - - 13 1.60 10 J , - = 15 10 m L ; α - 1 41 4.4 10 m and α - = 2 0.41 L e 4) For = 8 kg m , -= 1 J , = 0.02 m L ; α - 3 3.8 10 m and α - =≈ 33 2 1.5 10 0 L 7-6 Equation 7.9 gives for the transmission probability α - 2 2 1 1 sinh 4 U L . For 0.1% transmission (29 = 0.001 , and the resulting equation must be solved for E using = 1 nm L and = 5 eV U . We adopt = E x U as the unknown and write α - = =- hh 2 2 2 2 1 m LUE mUL Lx and = -- 2 1 1 U E U E xx . For this case ( 29 ( × h 32 2 2 2 511 10 eV 5.00 eV 1 nm 2 11.46 197.3 eV nm
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PC Chapter 07 - 7 Tunneling Phenomena 7-1 (a) The...

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