Chapter 39 - 39.1: a) . s m kg 10 37 . 2 ) m 10 80 . 2 ( )...

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Unformatted text preview: 39.1: a) . s m kg 10 37 . 2 ) m 10 80 . 2 ( ) s J 10 63 . 6 ( 24 10 34 = = = =--- h p p h b) eV. 3 . 19 J 10 08 . 3 ) kg 10 11 . 9 ( 2 ) s m kg 10 37 . 2 ( 2 18 31 2 24 2 = = = =--- m p K 39.2: mE h p h 2 = = m. 10 02 . 7 V) e J 10 60 . 1 ( ) eV 10 20 . 4 ( ) kg 10 64 . 6 ( 2 ) s J 10 63 . 6 ( 15 19 6 27 34---- = = 39. 3: a) m. 10 55 . 1 ) s m 10 70 . 4 ( kg) 10 11 . 9 ( s J 10 63 . 6 10 6 31 34--- = = = v m h e e b) m. 10 46 . 8 m 10 55 . 1 kg 10 67 . 1 kg 10 11 . 9 14 10 27 31---- = = = e p e p m m 39.4: a) keV. 2 . 6 m) 10 20 . ( ) s m 10 00 . 3 ( s) eV 10 136 . 4 ( 9 8 15 = = =-- hc E b) kg) 10 11 . 9 ( 2 )) m 10 20 . ( ) s J 10 626 . 6 (( 2 ) ( 2 31 2 9 34 2 2--- = = = m h m p K eV. 37 J 10 . 6 18 = =- Note that the kinetic energy found this way is much smaller than the rest energy, so the nonrelativistic approximation is appropriate. c) = = = = =---- J 10 3 . 8 kg) 10 64 . 6 ( 2 )) m 10 20 . ( s) J 10 626 . 6 (( 2 ) ( 2 22 27 2 9 34 2 2 m h m p K 5.2 meV. Again, the nonrelativistic approximation is appropriate. 39.5: a) In the Bohr model . 2 nh r mv n = The de Broglie wavelength is m. 10 32 . 3 ) m 10 29 . 5 ( 2 m 10 29 . 5 : 1 for 2 10 11 1 11 1--- = = = = = = = = a r n n r mv h p h n This equals the orbit circumference. b) , 4 4 ) 16 ( 2 16 ) 4 ( : 4 4 2 4 = = = = = a a a r n . m 10 33 . 1 9 4- = The de Broglie wavelength is a quarter of the circumference of the orbit, . 2 4 r 39.6: a) For a nonrelativistic particle, so , 2 2 m p K = . 2 Km h p h = = b) m. 10 34 . 4 ) Kg 10 11 . 9 ( ) J/eV 10 60 . 1 ( ) eV 800 ( 2 ) s J 10 63 . 6 ( 11 31 19 34---- = 39.7: m. 10 90 . 3 ) s m 340 ( ) kg 005 . ( s J 10 63 . 6 34 34-- = = = = mv h p h We should not expect the bullet to exhibit wavelike properties. 39.8: Combining Equations 37.38 and 37.39 gives . 1 2- = mc p a) m. 10 43 . 4 1 ) ( 12 2- =- = = mc h p h (The incorrect nonrelativistic calculation gives m.) 10 05 . 5 12- b) m. 10 07 . 7 1 ) ( 13 2- =- mc h 39.9: a) photon nm . 62 V) e J 10 602 . 1 ( ) eV . 20 ( ) s m 10 998 . 2 ( ) s J 10 626 . 6 ( so 19 8 34 = = = =-- E hc hc E electron = = = =-- V) e J 10 602 . 1 ( eV) . 20 ( ) kg 10 109 . 9 ( 2 2 so ) 2 ( 19 31 2 mE p m p E s m kg 10 416 . 2 24 - nm 274 . = = p h b) photon eV 96 . 4 J 10 946 . 7 19 = = =- hc E electron s m kg 10 650 . 2 so 27 = = =- h p p h eV 10 41 . 2 J 10 856 . 3 ) 2 ( 5 24 2-- = = = m p E c) You should use a probe of wavelength approximately 250 nm. An electron with 250 = nm has much less energy than a photon with 250 = nm, so is less likely to damage the molecule....
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Chapter 39 - 39.1: a) . s m kg 10 37 . 2 ) m 10 80 . 2 ( )...

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