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Chapter 27 Quantum Physics Answers to Even Numbered Conceptual Questions 4. Measuring the position of a particle implies having photons reflect from it. However, collisions between photons and the particle will alter the velocity of the particle. 6. Light has both wave and particle characteristics. In Young’s double-slit experiment, light behaves as a wave. In the photoelectric effect, it behaves like a particle. Light can be characterized as an electromagnetic wave with a particular wavelength or frequency, yet at the same time, light can be characterized as a stream of photons, each carrying a 10. Ultraviolet light has a shorter wavelength and higher photon energy than visible light. 16. The red beam. Each photon of red light has less energy (longer wavelength) than a photon of blue light, so the red beam must contain more photons to carry the same total energy. 393 394 CHAPTER 27 Problem Solutions 27.3 The wavelength of maximum radiation is given by λm ax = 27.4 0.2898× 10− 2 m ⋅ K = 5.00× 10−7 m = 500 nm 5800 K The energy of a photon having wavelength λ is Eγ = hf = hc λ . Thus, the number of photons delivered by each beam must be: Red Beam: ( 2 500 eV ) ( 690× 10−9 m ) ⎛ 1.60× 10−19 J⎞ Etotal EtotalλR 3 = = nR = ⎜ ⎟ = 1.39× 10 −34 8 1 eV Eγ ,R hc 6. 63 × 10 J ⋅ s 3. 00 × 10 m s ( )( )⎝ ⎠ ( 2 500 eV ) ( 420× 10−9 m ) ⎛ 1.60× 10−19 J⎞ Etotal EtotalλB = = = 845 nB = Eγ ,B hc ( 6.63× 10−34 J⋅ s)( 3.00× 108 m s) ⎜⎝ 1 eV ⎟⎠ Blue Beam: 27.8 The energy entering the eye each second is P = I⋅ A = ( 4.0× 10−11 W m 2 ) ⎡⎢⎣ π4 ( 8.5× 10 −3 m ) 2 ⎤ −15 ⎥⎦ = 2.3× 10 W The energy of a single photon is Eγ = hc λ ( 6.63× 10 = − 34 J⋅ s) ( 3.00 × 108 m s) −9 500 × 10 m = 3.98× 10−19 J so the number of photons entering the eye in ∆t= 1.00 s is −15 ∆E P⋅ ( ∆t) ( 2.3× 10 J s) ( 1.00 s) N = = = = 5.7 × 103 Eγ Eγ 3.98 × 10−19 J Quantum Physics 27.11 395 (a) From the photoelectric effect equation, the work function is φ= hc λ − KEm ax , or ( 6.63× 10 φ= J⋅ s) ( 3.00 × 108 m s) ⎛ ⎞ 1 eV ⎜ ⎟ − 1.31 eV − 19 −9 350× 10 m ⎝ 1.60× 10 J⎠ − 34 φ = 2.24 eV (b) λc = (c) 27.19 fc = hc φ c λc ( 6.63× 10 − 34 = = J⋅ s) ( 3.00 × 108 m s) ⎛ ⎞ 1 eV ⎜ ⎟ = 555 nm − 19 2.24 eV ⎝ 1.60 × 10 J⎠ 3.00 × 108 m s = 5.41× 1014 H z −9 555× 10 m Assuming the electron produces a single photon as it comes to rest, the energy of that photon is Eγ = ( KE )i = eV . The accelerating voltage is then −34 8 hc ( 6.63× 10 J⋅ s)( 3.00 × 10 m s) 1.24× 10−6 V ⋅ m V = = = = e eλ λ (1.60× 10-19 C ) λ Eγ For λ = 1.0 × 10−8 m , V = 1.24× 10−6 V ⋅ m = 1.2× 102 V 1.0× 10−8 m and for λ = 1.0× 10−13 m , V = 27.25 1.24× 10−6 V ⋅ m = 1.2× 107 V 1.0 × 10−13 m The interplanar spacing in the crystal is given by Bragg’s law as d= ( 1)( 0.140 nm ) mλ = = 0.281 nm 2sinθ 2sin14.4° 396 CHAPTER 27 27.26 The scattering angle is given by the Compton shift formula as ⎛ θ = cos−1 ⎜ 1− ⎝ λC = ∆λ ⎞ ⎟ where the Compton wavelength is λC ⎠ h = 0.002 43 nm m ec2 ⎛ 1.50 × 10−3 nm ⎞ Thus, θ = cos−1 ⎜ 1− ⎟ = 67.5° 2.43× 10−3 nm ⎠ ⎝ 27.34 The de Broglie wavelength of a particle of mass m is λ = h p where the momentum is given by p = γ m v = m v 1 − ( v c) . Note that when the particle is not relativistic, then 2 γ ≈ 1, and this relativistic expression for momentum reverts back to the classical expression. (a) For a proton moving at speed v = 2.00 × 104 m s , v << c and γ ≈ 1 so λ= h 6.63× 10−34 J⋅ s = = 1.98× 10−11 m -27 4 m pv ( 1.67 × 10 kg )( 2.00× 10 m s) (b) For a proton moving at speed v = 2.00× 107 m s λ= h γ m pv = h 2 1− ( v c) m pv ( 6.63× 10 J⋅ s) = (1.67 × 10 kg)( 2.00× 10 −34 -27 27.36 7 2 m s) ⎛ 2.00 × 107 m s⎞ −14 1− ⎜ ⎟ = 1.98 × 10 m 8 3. 00 × 10 m s ⎝ ⎠ After falling freely with acceleration ay = − g = −9.80 m s2 for 50.0 m, starting from rest, the speed of the ball will be v = v02 + 2ay ( ∆y) = 0 + 2( −9.80 m s) ( −50.0 m ) = 31.3 m s 2 so the de Broglie wavelength is λ= h h 6.63× 10−34 J⋅ s = = = 1.06× 10−34 m p m v ( 0.200 kg )( 31.3 m s) Quantum Physics 27.43 From the uncertainty principle, the minimum uncertainty in the momentum of the electron is ∆px = h 6.63× 10−34 J⋅ s = = 5.3× 10−25 kg ⋅ m s 4π ( ∆x) 4π ( 0.10× 10−9 m ) so the uncertainty in the speed of the electron is ∆vx = ∆px 5.3× 10−25 kg ⋅ m s = = 5.8× 105 m s or ~ 106 m s −31 m 9.11× 10 kg Thus, if the speed is on the order of the uncertainty in the speed, then v ~ 106 m s 27.45 With ∆x = 5.00 × 10−7 m s, the minimum uncertainty in the speed is ∆vx = ∆px h 6.63× 10−34 J⋅ s ≥ = = 116 m s m e 4π m e ( ∆x) 4π ( 9.11× 10-31 kg )( 5.00× 10-7 m ) 397