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1096_PartUniversity Physics Solution

1096_PartUniversity Physics Solution - 40-6 Chapter 40...

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40-6 Chapter 40 40.26. 0 2 00 2( ) with 16 1 and , L mU E EE TG e G UU κ ⎛⎞ == = ⎜⎟ ⎝⎠ = so 0 22( ) 16 1 . E L Te −− =− = (a) If 61 5 2 7 0 30.0 10 eV, 2.0 10 m, 6.64 10 kg UL m = × and 66 0 1.0 10 eV ( 29.0 10 eV), 0.090. UE E T −= × = × = (b) If 0 10.0 10 eV ( 20.0 10 eV), 0.014. E T −= × = × = 40.27. IDENTIFY and SET UP: The energy levels are given by Eq.(40.26), where . k m ω = EXECUTE: 110 N/m 21.0 rad/s 0.250 kg k m = The ground state energy is given by Eq.(40.26): 34 33 19 15 0 11 (1.055 10 J s)(21.0 rad/s) 1.11 10 J(1 eV/1.602 10 J) 6.93 10 eV 22 E × = × × = × = 1 ; 2 n En =+ = (1 ) 1 1 2 n + + = The energy separation between these adjacent levels is 33 33 14 10 2 2(1.11 10 J) 2.22 10 J 1.39 10 eV nn EE E E + Δ= = = = × = × = × = EVALUATE: These energies are extremely small; quantum effects are not important for this oscillator. 40.28. Let 2, mk δ = = and so 2 d ψ x ψ dx δ and 2 2 (4 2 , d ψ x δδ ) ψ dx and ψ is a solution of Eq.(40.21) if 2 E δ k/m ω . m = = 40.29. IDENTIFY: We can model the molecule as a harmonic oscillator. The energy of the photon is equal to the energy difference between the two levels of the oscillator. SET UP: The energy of a photon is / , Eh fh c γ λ and the energy levels of a harmonic oscillator are given by . n k n m EXECUTE: (a) The photon’s energy is 34 8 6 (6.63 10 J s)(3.00 10 m/s) 0.21 eV 5.8 10 m hc E ×⋅ × = × (b) The transition energy is 1 , k m + − = = which gives 2 . π ck m = = = Solving for , k we get 2 8 2 2 6 26 2 4 4 (3.00 10 m s) (5.6 10 kg) 5,900 N/m. (5.8 10 m) π cm π k ×× = × EVALUATE: This would be a rather strong spring in the physics lab. 40.30. According to Eq.(40.26), the energy released during the transition between two adjacent levels is twice the ground state energy 32 0 2 11.2 eV. ω E −= = = = For a photon of energy E 34 8 19 (6.63 10 J s)(3.00 10 m s) 111 nm. (11.2 eV)(1.60 10 J/eV) ch c f fE ×⋅ × = == = = × 40.31. IDENTIFY and SET UP: Use the energies given in Eq.(40.26) to solve for the amplitude A and maximum speed max v of the oscillator. Use these to estimate x Δ and x p Δ and compute the uncertainty product . x xp ΔΔ EXECUTE: The total energy of a Newtonian oscillator is given by 2 1 2 Ek A = where k is the force constant and A is the amplitude of the oscillator. Set this equal to the energy 1 2 () = of an excited level that has quantum number n , where , k m = and solve for A : 2 kA n = (2 1) n A k + = = The total energy of the Newtonian oscillator can also be written as 2 1 max 2 . Em v = Set this equal to 1 2 = and solve for max : v 2 max mv n = max (2 n v m + = =

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Quantum Mechanics 40-7 Thus the maximum linear momentum of the oscillator is max max (2 1) . pm v nm ω == + = Assume that A represents the uncertainty x Δ in position and that max p is the corresponding uncertainty x p Δ in momentum. Then the uncertainty product is (2 1) 1 (2 (2 1) . x xp n m n n n kk ωω + ⎛⎞ Δ Δ = + =+ ⎜⎟ ′′ ⎝⎠ = = = EVALUATE: For 1 n = this gives 3 , x ΔΔ = = in agreement with the result derived in Section 40.4. The uncertainty product x increases with n .
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1096_PartUniversity Physics Solution - 40-6 Chapter 40...

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