( ) ! = = = = = = = 0. ev nm. h h hc 5-4. (from Equation 5-2) (a) For an electron:! = = 0. % & (b) For a proton: (c) For an alpha particle:

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1 5-3. ( c) 1 ( 140eV nm) (. )(. ) Ek = evo = = V 940 o = = V 5 m mc e ev 0 04nm 5-4. c = = = mek mc Ek (from Equation 5-) 140eV nm (a) For an electron: = = nm ( )( )( ) 1 /. $ ev. $ ev # % (b) For a roton: 140eV nm = = 4. 7 # 10 1/ 3 $ ( )( # 10 ev )( 4. 5# 10 ev )% 4 nm (c) For an ala article: 140eV nm = =. 14# 10 1/ 9 3 $ ( )( 3. 78# 10 ev )( 4. 5# 10 ev )% 4 nm 5-5. ( ) 1 / = / = / me = c / mc. kt # k $ 1 5 % (from Equation 5-) Mass of N molecule = u MeV / uc = MeV / c = ev / c ( ) eV nm = = 0. 07nm ( )( )( 1 5)( )( 300 ) 1 / #. % ev.. % ev / K K $ = 0. 5nm = m E = Squaring and rearranging, E k k ( c) ( ) ( 140eV nm) ( )(. ) = = = = eV m m c ev nm ( 1)( 0 5 ) ( ) n = D sin # sin = n / D =. nm /. nm sin = 0. 8 = 55

2 5-. c 140eV nm 0. 0nm / me.. k mc Ek ev ev

3 5-1. (a) (b) n n = D sin # D = = nc sin sin mc Ek ( 1)( 140eV nm) ( sin 55. ) ( 5. 11# 10 5 ev )( 50eV ) 1 = = 0 10nm $ % ( 1)( 140eV nm) ( 0 10 ) ( )( 100 ) 1 /. #. % $ n sin = = = D nm ev ev ( ) = sin = / (a) y = y1 + y = 0. 00m cos 8. 0x / m 400t / s m cos 7. x / m 380t / s ( ) ( ) ( ) 1 ( ) 1 = 0. 00m cos $ 8. 0x / m # 7. x / m # ( 400t / s # 380t / s) % 1 1 # cos % ( 8. 0x / m + 7. x / m) $ ( 400t / s + 380t / s) ( = m cos 0. x / m 10t / s cos 7. 8x / m 390t / s ( ) ( ) (b) 390 / s v = = = 50m / s k 7. 8 / m (c) v s 0 / s = = = 50m / s k 0. 4 / m (d) Successive zeros of te enveloe requires tat 0. x / m =, tus 1 # x = = 5 m wit # k = k1 k = 0. 4 m and # x = = 5 m. 0. # k c 140eV nm 5-. (a) = = = = = nm ( )( 5 ) 1 / me. k mc Ek $ ev ev # % d sin = / For first minimum (see Figure 5-17). (b) nm d = = = nm slit searation sin sin5 0. 5cm sin 5 = 0. 5 cm / L were L = distance to detector lane L = = 5. 74cm sin5

4 5-3. (a) Te article is found wit equal robability in any interval in a force-free region. Terefore, te robability of finding te article in any interval x is roortional to x. Tus, te robability of finding te sere exactly in te middle, i.e., wit x = 0 is zero. (b) Te robability of finding te sere somewere witin 4.9cm to 5.1cm is roortional to x =0.cm. Because tere is a force free lengt L = 48cm available to te sere and te robability of finding it somewere in L is unity, ten te robability tat it will be found in x = 0.cm between 4.9cm and 5.1cm (or any interval of equal size) is: P x ( )( cm) = 1/ = Because te article must be in te box # * # sin ( / ) L dx = 1 = A x L dx = Let / ; 0 0 ; and ( / ) u = x L x = u = x = L u = dx = L du, so we ave A L / sin udu A L / sin udu ( ) ( ) = = u sinu / sin = / $ = / / = / = 1 4 ( # ( L ) A ) udu ( L ) A % ( L ) A ( ) ( LA ) 0 = = ( ) 1 / A / L A / L 5-5. (a) At 0 : ( 0, 0) 0 0 x = Pdx = dx = Ae dx = A dx L (b) At (c) At / 4 1/ 4 x = : Pdx = Ae dx = Ae dx = 0. 1A dx 4 / 4 1 x = : Pdx = Ae dx = Ae dx = 0. 14A dx (d) Te electron will most likely be found at x = 0, were Pdx is largest.

5 5-3. Because c f for oton, c / f c / f c / E, so c 140eV nm 5 E ev nm 5 E ev 7 and ev s / m 8 c 3 10 m / s For electron: ev s x m 4. / ev s m Notice tat for te electron is 1000 times larger tan for te oton.

6 J s # # $ % # $ # = $ 10 s E t E / t. 10 ev 7 19 (. J / ev ) Te size of te object needs to be of te order of te wavelengt of te 10MeV neutron. = / = / mu. and u are found from: ( ) Ek = mnc 1 or 1 = 10MeV / 939MeV ( ) 1 / = / 939 = = 1/ 1 u / c or u = 0. 14c c 140eV nm Ten, = = = = fm mu # mc ( u / c) $ #( )( 939 % 10 ev )( 0. 14) $ Nuclei are of tis order of size and could be used to sow te wave caracter of 10MeV neutrons E t # t = E ev s # t = = ev 4 s (a) For a roton or neutron: x and = m v assuming te article seed to be non-relativistic J s # = = = $ m# x kg 10 m 34 7 v m / s 0. 1c 7 15 (. )( ) (nonrelativistic) 1 ( kg)( m / s) 13 (b) Ek # mv = = J = 5. 1MeV (c) Given te roton or neutron velocity in (a), we exect te electron to be relativistic, in wic case, E = mc ( ) k 1 and

7 = # mv $ v # x m x For te relativistic electron we assume v c # 10 J s $ = = 193 mc% x 9 11# 10 kg 3 00# 10 m s 10 m (. )(. / )( ) ( ) ( )( ) ( ) Ek = mc 1 = 9. 11# 10 kg # 10 m / s 19 = 1. 58# 10 J = 98MeV