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 . . J s 1.67 10 kg m/s -27 9 92 10 7 . × m (b) For destructive interference in a multiple-slit experiment, d m sin θ λ = + 1 2 with m = 0 for the first minimum. Then, θ λ = = ° sin . 1 2 0 0284 d y L = tan θ so y L = = ( ) ° ( ) = tan . tan . θ 10 0 0 0284 m 4 96 . mm (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: d m sin θ λ = 6 00 10 0 400 200 1 1 20 10 8 1 10 . sin tan . . × ( ) = = × m m λ λ = h m v e so m v h e = λ and K m v m v m h m e V e e e e = = = = ( ) 1 2 2 2 2 2 2 2 2 λ 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 λ . . . . J s C kg m 105 V 41.3 (a) The wavelength of a non-relativistic particle of mass m is given by λ = = h p h mK / 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 K K n = × ( ) 1 60 10 19 . J/eV and the equation for the wavelength becomes λ = = × × ( ) × ( ) = h mK K n 2 6 626 10 2 1 67 10 1 60 10 34 27 19 . . . J s kg J/eV 2 87 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 λ If the particles are electrons and λ ~ . 0 1 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 . . J s kg m eV 1.602 10 J -19 ~ 10 2 eV 41.5 λ = h p p h = = × × = × λ 6 626 10 1 00 10 6 63 10 34 11 23 . . . J s m kg m/s (a) electrons: K p m e e = = × ( ) × ( ) = 2 34 2 31 2 6 63 10 2 9 11 10 . . J s J 15 1 . keV The relativistic answer is more precisely correct: K p c m c m c e e e = + ( ) = 2 2 2 4 1 2 2 14 9 / /. keV (b) photons: E pc 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 E K m c e e = + = 2 551 keV and p c E m c e = = ( ) ( ) × × = × 1 551 511 3 00 10 1 60 10 1 00 1 10 10 2 2 4 2 2 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 λ = = × × = × = h p 6 626 10 6 03 10 34 12 . . J s 1.10 10 kg m/s m 22 6 03 . pm 41.7 E K m c e = + = + = 2 1 00 0 511 1 51 . . . MeV MeV MeV p E m c e 2 2 2 4 2 2 1 51 0 511 c MeV MeV 2 = = ( ) ( ) . . so p c = 1 42 . MeV/ λ = = = × ( ) × ( ) × ( ) × ( ) = × h p hc 1 42 6 626 10 3 00 10 1 42 10 1 60 10 8 74 10 34 8 6 19 13 . . . . . . MeV J s m/s J m Suppose the array is like a flat diffraction grating with openings 0 250 . nm apart: d m sin θ λ = 13 5 8 74 10
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Chapter 41 Solutions 496 41.8 (a) ∆ ∆ ∆ ∆ p x m v x = h /2 so v h m x = ( ) ( ) = 4 2 2 00 1 00 π π π J s 4 kg m . . 0 250 . m/s (b) The duck might move by 0 25 5 1 25 . . m/s s m ( ) ( ) = . With original position uncertainty of 1 0 . 0 m , we can think of x growing to 1 00 1 25 . . m m + = 2 25 . m 41.9 For the electron, p m v 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 4 56 10 34 32 π π .
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