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39_InstructorSolutions

39_InstructorSolutions - THE WAVE NATURE OF PARTICLES 39 hc...

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T HE W AVE N ATURE OF P ARTICLES 39.1. I DENTIFY and S ET U P : h h p mv λ = = . For an electron, 31 9.11 10 kg m = × . For a proton, 27 1.67 10 kg m = × . E XECUTE : (a) 34 10 31 6 6.63 10 J s 1.55 10 m 0.155 nm (9.11 10 kg)(4.70 10 m/s) λ × = = × = × × (b) λ is proportional to 1 m , so 31 10 14 e p e 27 p 9.11 10 kg (1.55 10 m) 8.46 10 m 1.67 10 kg m m λ λ × = = × = × × . 39.2. I DENTIFY and S ET U P : For a photon, hc E λ = . For an electron or proton, h p λ = and 2 2 p E m = , so 2 2 2 h E m λ = . E XECUTE : (a) 15 8 9 (4.136 10 eV s)(3.00 10 m/s) 6.2 keV 0.20 10 m hc E λ × × = = = × (b) 2 2 34 18 2 9 31 6.63 10 J s 1 6.03 10 J 38 eV 2 0.20 10 m 2(9.11 10 kg) h E m λ × = = = × = × × (c) 31 e p e 27 p 9.11 10 kg (38 eV) 0.021 eV 1.67 10 kg m E E m × = = = × E VALUATE : For a given wavelength a photon has much more energy than an electron, which in turn has more energy than a proton. 39.3. (a) 34 24 10 (6.63 10 J s) 2.37 10 kg m s. (2.80 10 m) h h p p × = = = = × × (b) 2 24 2 18 31 (2.37 10 kg m s) 3.08 10 J 19.3 eV. 2 2(9.11 10 kg) p K m × = = = × = × 39.4. 2 h h p mE = = 34 15 27 6 19 (6.63 10 J s) 7.02 10 m. 2(6.64 10 kg) (4.20 10 eV) (1.60 10 J eV) × = = × × × × 39.5. I DENTIFY and S ET U P : The de Broglie wavelength is . h h p mv λ = = In the Bohr model, ( / 2 ), n mvr n h π = so /(2 ). n mv nh r π = Combine these two expressions and obtain an equation for λ in terms of n . Then 2 2 . n n r r h nh n π π λ = = E XECUTE : (a) For 10 1 1 0 1, 2 with 0.529 10 m, so n r r a λ π = = = = × 10 10 2 (0.529 10 m) 3.32 10 m λ π = × = × 1 2 ; r λ π = the de Broglie wavelength equals the circumference of the orbit. (b) For 4 4, 2 /4. n r λ π = = 2 0 4 0 so 16 . n r n a r a = = 10 9 0 0 2 (16 )/4 4(2 ) 4(3.32 10 m) 1.33 10 m a a λ π π = = = × = × 4 2 /4; r λ π = the de Broglie wavelength is 1 1 4 n = times the circumference of the orbit. E VALUATE : As n increases the momentum of the electron increases and its de Broglie wavelength decreases. For any n , the circumference of the orbits equals an integer number of de Broglie wavelengths. 39
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39-2 Chapter 39 39.6. (a) For a nonrelativistic particle, 2 , so 2 p K m = . 2 h h p Km λ = = (b) 34 -19 -31 11 (6.63 10 J s) 2(800 eV)(1.60 10 J/eV)(9.11 10 kg) 4.34 10 m. × × × = × 39.7. I DENTIFY : A person walking through a door is like a particle going through a slit and hence should exhibit wave properties. S ET U P : The de Broglie wavelength of the person is λ = h/mv . E XECUTE : (a) Assume m = 75 kg and v = 1.0 m/s. λ = h/mv = (6.626 × 10 –34 J s)/[(75 kg)(1.0 m/s)] = 8.8 × 10 –36 m E VALUATE : (b) A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too small to show wave behavior through a “slit” that is about 10 35 times as wide as the wavelength. Hence ordinary objects do not show wave behavior in everyday life. 39.8. Combining Equations 37.38 and 37.39 gives 2 1. p mc γ = (a) 2 12 ( ) 1 4.43 10 m.
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