HW 6 - 5 -2 12 The issue is: Can we u se the simpler cla...

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5-2 The issue is: Can we use the simpler classical expression p = 2 mK ( ) 1 2 instead of the exact relativistic expression p = K 1 + 2 mc 2 K ( ) c ? As the relativistic expression reduces to p = 2 mK ( ) 1 2 for K << 2 mc 2 , we can use the classical expression whenever K << 1 MeV because mc 2 for the electron is 0.511 MeV. (a) Here 50 eV << 1 MeV , so p = 2 mK ( ) 1 2 " = h p = h 2 ( ) 0.511 MeV c 2 ( ) 50 eV ( ) [ ] = hc 2 ( ) 0.511 MeV ( ) 50 eV ( ) [ ] = 1240 eV nm 2 ( ) 0.511 # 10 6 ( ) 50 ( ) eV ( ) 2 [ ] = 0.173 nm (b) As 50 eV << 1 MeV , p = 2 mK ( ) 1 2 = hc 2 ( ) 0.511 MeV c 2 ( ) 50 # 10 3 eV ( ) [ ] 1 2 = 5.49 # 10 $ 3 nm . As this is clearly a worse approximation than in (a) to be on the safe side use the relativistic expression for p : p = K 1 + 2 mc 2 K ( ) c so = h p = hc K 2 + 2 Kmc 2 ( ) = 50 # 10 3 ( ) 2 + 2 ( ) 50 # 10 3 ( ) 0.511 # 10 6 eV ( ) $ % & ( ) = 5.36 # 10 * 3 nm = 0.005 36 nm 5-7 A 10 MeV proton has K = 10 MeV << 2 mc 2 = 1877 MeV so we can use the classical expression p = 2 mK ( ) 1 2 . (See Problem 5-2) = h p = hc 2 ( ) 938.3 MeV ( ) 10 MeV ( ) [ ] 1 2 = 1240 MeV fm 2 ( ) 938.3 ( ) 10 ( ) MeV ( ) 2 [ ] 1 2 = 9.05 fm = 9.05 # 10 $ 15 m 5-8 = h p = h 2 mK ( ) = h 2 meV ( ) = h 2 me ( ) # $ % % & ( ( V ) = 6.626 # 10 $ 34 Js 2 # 9.105 # 10 $ 31 kg # 1.602 # 10 $ 19 C ( ) % & ( ) * * * V $ = 1.226 # 10 $ 9 kg 1 2 m 2 sC % & ( ) * * V $ 5-10 As = 2 a 0 = 2 0.052 9 ( ) nm = 0.105 8 nm the energy of the electron is nonrelativistic, so we can use
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p = h " with K = p 2 2 m ; K = h 2 2 m 2 = 6.626 # 10
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This note was uploaded on 06/02/2011 for the course PHYS 2D taught by Professor Hirsch during the Fall '08 term at UCSD.

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HW 6 - 5 -2 12 The issue is: Can we u se the simpler cla...

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