chapter 39 solutions - Solutions to Exercises Chapter 39 12...

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Unformatted text preview: Solutions to Exercises, Chapter 39. 12 th Ed. 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---- ״ = = 5 = ״ . 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) λ 2.37 10 kg m s. λ (2.80 10 m) h h p p--- = = = = ״ (b) 2 24 2 18 31 (2.37 10 kg m s) 3.08 10 J 19.3eV. 2 2(9.11 10 kg) p K m--- = = = = 39.4. λ 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 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 4 so 16 . n r n a r a = = 10 9 2 (16 )/ 4 4(2 ) 4(3.32 10 m) 1.33 10 m a a λ π π-- = = = = B 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. 39.6. (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 a 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....
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This note was uploaded on 04/14/2008 for the course PHYS 2303 taught by Professor Hoffman during the Fall '07 term at University of Texas at Dallas, Richardson.

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chapter 39 solutions - Solutions to Exercises Chapter 39 12...

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