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1101_PartUniversity Physics Solution

1101_PartUniversity Physics Solution - Quantum Mechanics...

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Quantum Mechanics 40-11 sinh 2 L L e e L κ κ κ = For 1, sinh 2 L e L L κ κ κ W and 1 2 2 0 0 2 2 0 0 0 16 ( ) 1 16 ( ) 16 ( ) L L U e E U E T E U E E U E U e κ κ + = + For 2 2 2 2 0 0 0 1, 16 ( ) L L L E U E U e U e κ κ κ + W 2 0 2 2 0 0 0 16 ( ) 16 1 , L L E U E E E T e U e U U κ κ ⎞⎛ = ⎟⎜ ⎠⎝ which is Eq.(40.21). (b) 0 2 ( ) . L m U E L κ = = So 1 L κ W when L is large (barrier is wide) or 0 U E is large. ( E is small compared to 0 . U ) (c) 0 2 ( ) ; m U E κ κ = = becomes small as E approaches 0 . U For κ small, sinh L L κ κ and 1 1 2 2 2 2 2 0 0 0 2 0 0 2 ( ) 1 1 4 ( ) 4 ( ) U L U m U E L T E U E E U E κ + = + = (using the definition of κ ) Thus 1 2 2 0 2 2 1 4 U L m T E + = 0 U E so 2 0 U E E and 1 2 2 2 1 4 EL m T + = But 2 2 2 , mE k = = so 1 2 1 , 2 kL T + as was to be shown. E VALUATE : When L κ is large Eq.(40.20) applies and T is small. When 0 , E U T does not approach unity. 40.48. (a) 2 1 ( (1 2)) ( (1 2)) , 2 E mv n ω n hf = = + = + = and solving for n , 2 2 30 34 1 1 (1/2)(0.020 kg)(0.360 m/s) 1 2 1.3 10 . 2 (6.63 10 J s)(1.50 Hz) 2 mv n hf = = = × × (b) The difference between energies is 34 34 (6.63 10 J s)(1.50 Hz) 9.95 10 J. ω hf = = × = × = This energy is too small to be detected with current technology 40.49. I DENTIFY and S ET U P : Calculate the angular frequency ω of the pendulum and apply Eq.(40.26) for the energy levels. E XECUTE : 1 2 2 4 s 0.500 s T π π ω π = = = The ground-state energy is 34 1 34 0 1 1 (1.055 10 J s)(4 s ) 6.63 10 J. 2 2 E ω π = = × = × = 34 19 15 0 6.63 10 J(1 eV/1.602 10 J) 4.14 10 eV E = × × = × 1 2 n E n ω = + = 1 1 1 2 n E n ω + = + + = The energy difference between the adjacent energy levels is 1 0 2 n n E E E E ω + Δ = = = = = 33 15 1.33 10 J 8.30 10 eV × = × E VALUATE : These energies are much too small to detect. Quantum effects are not important for ordinary size objects. 40.50. I DENTIFY : We model the electrons in the lattice as a particle in a box. The energy of the photon is equal to the energy difference between the two energy states in the box. S ET U P : The energy of an electron in the n th level is 2 2 2 . 8 n n h E mL = We do not know the initial or final levels, but we do know they differ by 1. The energy of the photon, / , hc λ is equal to the energy difference between the two states. E XECUTE : The energy difference between the levels is 34 8 7 (6.63 10 J s)(3.00 10 m/s) 1.649 10 m hc E λ × × Δ = = = × 18 1.206 10 J. × Using the formula for the energy levels in a box, this energy difference is equal to 2 2 2 2 2 2 ( 1) (2 1) . 8 8 h h E n n n mL mL Δ = =

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40-12 Chapter 40 Solving for n gives 2 18 31 9 2 2 34 2 8 1 (1.206 10 J)8(9.11 10 kg)(0.500 10 m) 1 1 3.
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1101_PartUniversity Physics Solution - Quantum Mechanics...

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