phys for scient & eng serway 5th edition chpt41

Physics for Scientists and Engineers

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Chapter 41 Solutions 41.1 (a) λ == ×⋅ × () = h mv 6 626 10 0 400 34 . . Js 1.67 10 kg m/s -27 992 10 7 . × m (b) For destructive interference in a multiple-slit experiment, dm sin θ =+ 1 2 with m = 0 for the first minimum. Then, = sin . 1 2 0 0284 d y L = tan so yL ° = tan . tan . 10 0 0 0284 m 496 . m m (c) We cannot say the neutron passed through one slit. We can only say it passed through the slits. 41.2 Consider the first bright band away from the center: sin = 600 10 0 400 200 1 1 20 10 81 1 0 . sin tan . . × × −− m = h mv e so h e = and Km v m h m eV e e e e ==== 1 2 2 2 22 2 2 2 V h em e × × × = 2 2 34 2 19 31 10 2 2 6 626 10 2 1 60 10 9 11 10 1 20 10 . .. . C kg m 105 V 41.3 (a) The wavelength of a non-relativistic particle of mass m is given by hp h m K /2 where the kinetic energy K is in joules. If the neutron kinetic energy K n is given in electron volts, its kinetic energy in joules is KK n 160 10 19 . J/eV and the equation for the wavelength becomes × × = h mK K n 2 6 626 10 2 1 67 10 1 60 10 34 27 19 . kg J/eV 287 10 11 . × K n m where K n is expressed in electron volts. (b) If K n 1 00 1000 . keV eV, then
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Chapter 41 Solutions 495 41.4 λ == h p h mK 2 , so K h m = 2 2 2 If the particles are electrons and ~. 01 10 10 nm m = , the kinetic energy in electron volts is K = ×⋅ () × ( ) × = −− 6 626 10 2 9 11 10 10 1 34 2 31 10 2 . . Js kg m eV 1.602 10 J -19 ~10 2 eV 41.5 = h p p h × 6 626 10 100 10 663 10 34 11 23 . . . m kg m/s (a) electrons: K p m e e × = 2 34 2 31 2 2 9 11 10 . . J 15 1 . keV The relativistic answer is more precisely correct: Kp c m c m c ee e =+ −= 22 2 4 12 2 14 9 / /. keV (b) photons: Ep c γ × × = 6 63 10 3 00 10 23 8 .. 124 keV 41.6 The theoretical limit of the electron microscope is the wavelength of the electrons. If K e = 40 0 . keV, then EK m c = 2 551 keV and p c Em c e =−= × × 1 551 511 300 10 160 10 100 110 10 2 2 4 8 16 22 keV keV m/s J keV kg m/s . . . . The electron wavelength, and hence the theoretical limit of the microscope, is then = h p 6 626 10 603 10 34 12 . . 1.10 10 kg m/s m 22 603 . p m 41.7 EKm c e = + = 2 0511 151 . MeV MeV MeV pE m c e 2 4 1 51 0 511 c MeV MeV 2 =− = so pc = 142 . MeV/ = × × × h p hc 6 626 10 3 00 10 1 42 10 1 60 10 874 10 34 8 61 9 13 . . MeV J s m/s J m Suppose the array is like a flat diffraction grating with openings 0 250 n m apart: dm sin θ =  13
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Chapter 41 Solutions 496 41.8 (a) ∆∆ px mvx =≥ h /2 so v h mx ≥= () = 4 2 200 100 π J s 4 kg m .. 0 250 . m/s (b) The duck might move by 025 5 125 m/s s m = . With original position uncertainty of 10 .0 m, we can think of x growing to 100 m m += 225 . m 41.9 For the electron, pmv e == × × −− 9 11 10 500 1 00 10 4 56 10 31 4 32 . kg m/s kg m/s x h p ×⋅ = 4 6 626 10 456 10 34 32 . . Js 4 kg m/s 116 m m For the bullet, × 0 0200 500 1 00 10 1 00 10 43 . kg m/s kg m/s x h p 4 528 10 32 . × m Goal Solution An electron ( m e 911 10 31 k g ) and a bullet (.
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