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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 electronvolts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc / = (6.63 10 34 J s)(3 10 8 m/s)(10 9 nm/m)/(1.60 10 19 J/eV) ; E = (1.24 10 3 eV nm)/ . 1. We find the wavelength of the neutron from = h / p = h /(2 mK ) 1/2 = ( 6 . 6 3 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 = m , m = 1, 2, . and the positions on the screen are y = L tan . For small angles, sin tan , so we have y = mL / d . Thus the separation is ? y = L / 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 = h / p = h / mv = ( 6 . 6 3 10 34 J s)/(2.0 10 3 kg)(120 m/s) = 2.8 10 33 m. The halfangle for the central circle of the diffraction pattern is given by sin = 1.22 / D . For small angles, sin tan , so we have r = L tan L sin = 1.22 L / 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 protons 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 . 6. ( a ) We find the wavelength of the bullet from = h / p = h / mv = ( 6 . 6 3 10 34 J s)/(12 10 3 kg)(150 m/s) = 3.7 10 34 m . ( b ) We find the uncertainty in the momentum component perpendicular to the motion: ? p y = /? y = (1.055 10 34 J s)/(0.55 10 2 m) = 1.9 10 32 kg m/s . ( c ) We find the possible uncertainty in the yposition at the target from y / L = ? v y / v x = ? p y / p x ; Chapter 39 p. 2 y /(300 m) = (1.9 10 32 kg m/s)/(12 10 3 kg)(150 m/s), which gives y = 3.2 10 30 m ....
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 Spring '09
 Fielding
 Physics, Energy, Photon

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