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PC Chapter 05

# PC Chapter 05 - 5 Matter Waves 5-1 = h h 6.63 10 34 Js =...

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67 5 Matter Waves 5-1 ( 29 λ - - - × = = = = × × 34 6 13 27 6.63 10 Js 10 m s 3.97 10 m 1.67 10 kg h h p mv 5-2 The issue is: Can we use the simpler classical expression ( 29 = 1 2 2 p mK instead of the exact relativistic expression ( 29 + = 2 1 2 2 1 mc K K p c ? As the relativistic expression reduces to ( 29 = 1 2 2 p mK for << 2 2 K mc , we can use the classical expression whenever << 1 MeV K because 2 mc for the electron is 0.511 MeV. (a) Here << 50 eV 1 MeV , so ( 29 = 1 2 2 p mK ( 29 ( 29 ( 29 ( 29( 29( 29 [ ] ( 29 ( 29 ( 29( 29 λ = = = = = × 2 1 2 1 2 0.511 MeV 1 2 6 2 2 0.511 MeV 50 eV 2 50 eV 1 240 eV nm 0.173 nm 2 0.511 10 50 eV c h h hc p (b) As << 50 eV 1 MeV , ( 29 = 1 2 2 p mK ( 29 ( 29 ( 29 λ - = = × × 2 3 1 2 3 0.511 MeV 5.49 10 nm 2 50 10 eV c hc . As this is clearly a worse approximation than in (a) to be on the safe side use the relativistic expression for p : ( 29 + = 2 1 2 2 1 mc K p K c so ( 29 ( 29 ( 29 ( 29( 29 λ - = = = + × + × × = × = 1 2 1 2 2 2 2 3 3 6 3 1 240 eV nm 2 50 10 2 50 10 0.511 10 eV 5.36 10 nm 0.005 36 nm h hc p K Kmc

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68 CHAPTER 5 MATTER WAVES 5-3 ( 29 λ - - × = = = = × 34 36 6.63 10 Js 5 m s 1.79 10 m 74 kg h h p mv 5-4 Taking λ = 0.1 nm and using λ = = h p mv , we get ( 29 λ - - - × = = × = × × 34 9 6 31 6.63 10 Js 0.1 10 m 7.28 10 m s 9.11 10 kg h v m . As << v c , it is okay to use = p mv instead of γ = p mv . 5-5 (a) λ = h p or ( 29( 29 λ λ = = = = 1 240 eV nm 124 eV 10 nm h hc p c c c . As ( 29 = - = + - 1 2 2 2 2 2 2 2 K E mc p c mc mc , we must use the relativistic expression for K , when 2 pc mc . Here = << = 2 124 eV 0.511 MeV pc mc , so we can use the classical expression for K , = 2 2 p K m . ( 29 ( 29 = = = = 2 2 2 2 2 124 eV 0.150 eV 2 2 0.511 MeV 2 p p c K m mc (b) Electrons with λ λ = = = 12 400 eV 0.10 nm hc p c c as in (a). As << = 2 0.511 MeV pc mc , use ( 29 ( 29 ( 29 ( 29 = = = = × 2 2 2 2 2 6 12 400 eV 150 eV 2 2 0.511 10 eV p K p c m . (c) Electrons with λ - = = × 15 10 fm 10 10 m , λ × = = 3 1.24 10 MeV hc p c c . As = 2 0.511 MeV pc mc , use ( 29 = + - = - = - = 1 2 2 2 2 2 2 2 1 240 MeV 0.511 MeV 1 239 MeV K p c mc mc pc mc . For alphas with = 2 3 726 MeV mc : (a) p still is 124 eV c . As << 3 726 MeV pc , we use = 2 2 p K m : ( 29 ( 29 ( 29 - = = = × 2 2 2 6 2 124 eV 2.06 10 eV 2 3 726 MeV 2 p c K mc .
MODERN PHYSICS 69 (b) For alphas with λ = 0.10 nm , = 12 400 eV p c . As << = 2 3 726 MeV pc mc , ( 29 ( 29 ( 29 = = = = 2 2 2 2 2 12 400 eV 0.020 6 eV 2 2 3 726 MeV 2 p p c K m mc . (c) × = 3 1.24 10 MeV p c and = = 2 1 240 MeV~ 3 726 MeV pc mc . We use ( 29 ( 29 ( 29 = + - = + - = 1 2 1 2 2 2 2 2 2 2 2 1 240 MeV 3 726 MeV 3 726 MeV 201 MeV. K p c mc mc 5-6 From Problem 5-2, a 50 keV electron has λ - = × 3 5.36 10 nm . A 50 keV proton has = << = 2 50 keV 2 1 877 MeV K mc so we use ( 29 = 1 2 2 p mK : ( 29 ( 29 ( 29 ( 29( 29 ( 29 [ ] ( 29 ( 29( 29 λ - = = = = = × × × 2 1 2 1 2 938.3 MeV 4 1 2 3 3 2 938.3 MeV 50 keV 2 50 keV 1 240 eV nm 1.28 10 nm 2 938.3 10 eV 50 10 eV c h h hc p 5-7 A 10 MeV proton has = << = 2 10 MeV 2 1 877 MeV K mc so we can use the classical expression ( 29 = 1 2 2 p mK . (See Problem 5-2) ( 29( 29( 29 [ ] ( 29( 29( 29( 29 λ - = = = = = × 15 1 2 1 2 2 1 240 MeV fm 9.05 fm 9.05 10 m 2 938.3 MeV 10 MeV 2 938.3 10 MeV h hc p 5-8 ( 29 ( 29 ( 29 λ - = = = = 1 2 1 2 1 2 1 2 2 2 2 h h h h V p mK meV me ( 29 λ λ - - - - - - × = × × × × × = 34 1 2 1 2 31 19 1 2 9 2 1 2 1 2 6.626 10 Js 2 9.105 10 kg 1.602 10 C 1.226 10 kg m V V sC 5-9 = 0.20 kg m : = 2 2 mv mgh : ( 29 = 1 2 2 v gh ( 29 ( 29 ( 29( 29 [ ] λ - - = = = = × = = = × 1 2 1 2 34 34 2 0.20 2 9.80 50 6.261 kg m s 6.626 10 J s 1.06 10 m 6.261 kg m s p mv m gh h p

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PC Chapter 05 - 5 Matter Waves 5-1 = h h 6.63 10 34 Js =...

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