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# Ch39Word - Chapter 39 p 1 CHAPTER 39 – Quantum Mechanics...

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Unformatted text preview: Chapter 39 p. 1 CHAPTER 39 – Quantum Mechanics Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc / λ = (6.63 × 10 –34 J ∙ s)(3 × 10 8 m/s)(10 9 nm/m)/(1.60 × 10 –19 J/eV) λ ; E = (1.24 × 10 3 eV ∙ nm)/ λ . 1. We find the wavelength of the neutron from λ = h / p = h /(2 mK ) 1/2 = (6.63 × 10 –34 J ∙ s)/[2(1.67 × 10 –27 kg)(0.025 eV)(1.6 × 10 –19 J/eV)] 1/2 = 1.81 × 10 –10 m. The peaks of the interference pattern are given by d sin θ = m λ , m = 1, 2, … . and the positions on the screen are y = L tan θ . For small angles, sin θ ˜ tan θ , so we have y = mL λ / d . Thus the separation is ? y = L λ / d = (1.0 m)(1.81 × 10 –10 m)/(1.0 × 10 –3 m) = 1.8 × 10 –7 m . 2. We find the wavelength of the bullet from λ = h / p = h / mv = (6.63 × 10 –34 J ∙ s)/(2.0 × 10 –3 kg)(120 m/s) = 2.8 × 10 –33 m. The half-angle for the central circle of the diffraction pattern is given by sin θ = 1.22 λ / D . For small angles, sin θ ˜ tan θ , so we have r = L tan θ ˜ L sin θ = 1.22 L λ / D ; 0.50 × 10 –2 m = 1.22 L (2.8 × 10 –33 m)/(3.0 × 10 –3 m), which gives L = 4.5 × 10 27 m . Diffraction effects are negligible for macroscopic objects. 3. We find the uncertainty in the momentum: ? p = m ? v = (1.67 × 10 –27 kg)(0.024 × 10 5 m/s) = 4.00 × 10 –24 kg ∙ m/s. We find the uncertainty in the proton’s position from ? x = ˙ /? p = (1.055 × 10 –34 J ∙ s)/(4.00 × 10 –24 kg ∙ m/s) = 2.6 × 10 –11 m. Thus the accuracy of the position is ± 1.3 × 10 –11 m . 4. We find the minimum uncertainty in the energy of the state from ? E = ˙ /? t = (1.055 × 10 –34 J ∙ s)/(10 –8 s) = 1.1 × 10 –26 J = 7 × 10 –8 eV . 5. We find the uncertainty in the momentum: ? p = ˙ /? x = (1.055 × 10 –34 J ∙ s)/(1.6 × 10 –8 m) = 6.59 × 10 –27 kg ∙ m/s. We find the uncertainty in the velocity from ? p = m ? v ; 6.59 × 10 –27 kg ∙ m/s = (9.11 × 10 –31 kg) ? v , which gives ? v = 7.2 × 10 3 m/s ....
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## This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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Ch39Word - Chapter 39 p 1 CHAPTER 39 – Quantum Mechanics...

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