chap39 - a s m kg 10 37 2 m 10 80 2 s J 10 63 6 24 10...

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Unformatted text preview: a) . s m kg 10 37 . 2 ) m 10 80 . 2 ( ) s J 10 63 . 6 ( λ λ 24 10 34 ⋅ × = × ⋅ × = = ⇒ =--- h p p h 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 mE h p h 2 λ = = 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---- × = × × × ⋅ × = a) m. 10 55 . 1 ) s m 10 70 . 4 ( kg) 10 11 . 9 ( s J 10 63 . 6 λ 10 6 31 34--- × = × × ⋅ × = = v m h e e m. 10 46 . 8 m 10 55 . 1 kg 10 67 . 1 kg 10 11 . 9 λ λ 14 10 27 31---- × = × × × = = e p e p m m a) keV. 2 . 6 m) 10 20 . ( ) s m 10 00 . 3 ( s) eV 10 136 . 4 ( λ 9 8 15 = × × ⋅ × = =-- hc E kg) 10 11 . 9 ( 2 )) m 10 20 . ( ) s J 10 626 . 6 (( 2 ) λ ( 2 31 2 9 34 2 2--- × × ⋅ × = = = m h m p K eV. 37 J 10 . 6 18 = × =- at the kinetic energy found this way is much smaller than the rest energy, so the ativistic approximation is appropriate. c) = × = × × ⋅ × = = =---- J 10 3 . 8 kg) 10 64 . 6 ( 2 )) m 10 20 . ( s) J 10 626 . 6 (( 2 ) λ ( 2 22 27 2 9 34 2 2 m h m p K V. Again, the nonrelativistic approximation is appropriate. a) In the Bohr model . 2 π nh r mv n = The de Broglie wavelength is 10 32 . 3 ) m 10 29 . 5 ( 2 λ m 10 29 . 5 : 1 for 2 10 11 1 11 1--- × = × = ⇒ × = = = = = π a r n n r π mv h n uals the orbit circumference. b) , λ 4 4 ) 16 ( 2 λ 16 ) 4 ( : 4 4 2 4 = = ⇒ = = = a π a a r n . m 10 33 . 1 λ 9 4- × = ⇒ Broglie wavelength is a quarter of the circumference of the orbit, . 2 4 πr a) For a nonrelativistic particle, so , 2 2 m p K = . 2 λ 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---- × = × × ⋅ × m. 10 90 . 3 ) s m 340 ( ) kg 005 . ( s J 10 63 . 6 λ 34 34-- × = ⋅ × = = = mv h p h should not expect the bullet to exhibit wavelike properties. Combining Equations 37.38 and 37.39 gives . 1 2- = γ mc p a) m. 10 43 . 4 1 ) ( λ 12 2- × =- γ = = mc h p h (The incorrect nonrelativistic calculation m.) 10 05 . 12- × b) m. 10 07 . 7 1 ) ( 13 2- × =- γ mc h a) photon nm . 62 V) e J 10 602 . 1 ( ) eV . 20 ( ) s m 10 998 . 2 ( ) s J 10 626 . 6 ( λ so λ 19 8 34 = × × ⋅ × = = =-- E hc hc ctron = × × = = =-- V) e J 10 602 . 1 ( eV) . 20 ( ) kg 10 109 . 9 ( 2 2 so ) 2 ( 19 31 2 mE p m p s m kg 10 24 ⋅ ×- nm 274 . = p h photon eV 96 . 4 J 10 946 . 7 λ 19 = × = =- hc E ctron s m kg 10 650 . 2 λ so λ 27 ⋅ × = = =- h p p h eV 10 41 . 2 J 10 856 . 3 ) 2 ( 5 24 2-- × = × = = m p You should use a probe of wavelength approximately 250 nm. An electron with 0nm has much less energy than a photon with 250 λ = nm, so is less likely to damage the le. λ λ m h v mv h = → = 2 2 2 2 λ 2 λ 2 1 2 1 m h m h m mv K = = = ey will not have the same kinetic energy since they have different masses....
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap39 - a s m kg 10 37 2 m 10 80 2 s J 10 63 6 24 10...

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