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

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T HE W AVE N ATURE OF P ARTICLES 39.1. IDENTIFY and SET UP: hh pm v λ == . For an electron, 31 9.11 10 kg m . For a proton, 27 1.67 10 kg m . EXECUTE: (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) ×⋅ × = ×× (b) is proportional to 1 m , so 31 10 14 e pe 27 p 9.11 10 kg (1.55 10 m) 8.46 10 m 1.67 10 kg m m λλ −− ⎛⎞ × × = × ⎜⎟ × ⎝⎠ . 39.2. IDENTIFY and SET UP: For a photon, hc E = . For an electron or proton, h p = and 2 2 p E m = , so 2 2 2 h E m = . EXECUTE: (a) 15 8 9 (4.136 10 eV s)(3.00 10 m/s) 6.2 keV 0.20 10 m hc E × = × (b) 2 23 4 18 29 3 1 6.63 10 J s 1 6.03 10 J 38 eV 2 2(9.11 10 kg) h E m = × = (c) 31 e 27 p 9.11 10 kg (38 eV) 0.021 eV 1.67 10 kg m EE m × = × EVALUATE: 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) p p = = × × (b) 22 4 2 18 31 (2.37 10 kg m s) 3.08 10 J 19.3 eV. ( 9 . 1 1 1 0 k g ) p K m = × = × 39.4. λ 2 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. IDENTIFY and SET UP: The de Broglie wavelength is . v In the Bohr model, (/2) , n mvr n h π = so /(2 ). n mv nh r = Combine these two expressions and obtain an equation for in terms of n . Then . nn rr h nh n ππ EXECUTE: (a) For 10 11 0 1, 2 with 0.529 10 m, so nr r a λπ = = × 10 10 2 (0.529 10 m) 3.32 10 m = × 1 2; r = the de Broglie wavelength equals the circumference of the orbit. (b) For 4 4, 2 / 4. 2 04 0 so 16 . n rn a r a 10 9 00 2 (16 )/4 4(2 ) 4(3.32 10 m) 1.33 10 m aa = × = × 4 2/ 4 ; r = the de Broglie wavelength is 4 n = times the circumference of the orbit. EVALUATE: 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 hh 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. IDENTIFY: A person walking through a door is like a particle going through a slit and hence should exhibit wave properties. SET UP: The de Broglie wavelength of the person is = h/mv . EXECUTE: (a) Assume m = 75 kg and v = 1.0 m/s. = h/mv = (6.626 × 10 –34 J s)/[(75 kg)(1.0 m/s)] = 8.8 × 10 –36 m EVALUATE: (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. pm c γ =− (a) 21 2 () 1 4 . 4 3 1 0 m . h hmc p λγ −= × (The incorrect nonrelativistic calculation gives 12 5.05 10 m.) × (b) 3 1 7 . 0 7 1 0 m . γ × 39.9. IDENTIFY and SET UP: A photon has zero mass and its energy and wavelength are related by Eq.(38.2). An electron has mass. Its energy is related to its momentum by 2 /2 Ep m = and its wavelength is related to its momentum by Eq.(39.1).
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39_InstructorSolutions - THE WAVE NATURE OF PARTICLES 39 hc...

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