hw6 - 5-2 The issue is: Can we use the simpler classical...

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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 5-5 (a) = h p or p = h = hc c = 1240 eV nm 10 nm ( ) c ( ) = 124 eV c . As K = E " mc 2 = p 2 c 2 + mc 2 ( ) 2 # $ % & ( " mc 2 , we must use the relativistic expression for K , when pc " mc 2 . Here pc = 124 eV << mc 2 = 0.511 MeV , so we can use the classical expression for K , K = p 2 2 m . K = p 2 2 m = p 2 c 2 2 mc 2 = 124 eV ( ) 2 2 0.511 MeV ( ) = 0.150 eV
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(b) Electrons with " = 0.10 nm p = hc c = 12400 eV c as in (a). As pc << mc 2 = 0.511 MeV , use K = p 2 2 m = p 2 c 2 = 12 400 ( ) 2 eV ( ) 2 2 ( ) 0.511 " 10 6 eV ( ) = 150 eV . (c) Electrons with = 10 fm = 10 # 10 $ 15 m , p = hc c = 1.24 # 10 3 MeV c . As pc >> mc 2 = 0.511 MeV , use K = p 2 c 2 + mc 2 ( ) 2 " # $ % & ( mc 2 = pc ( mc 2 = 1240 MeV ( 0.511 MeV = 1239 MeV .
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This note was uploaded on 12/02/2008 for the course PHYS phys 2d taught by Professor Hirsch during the Fall '08 term at UCSD.

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hw6 - 5-2 The issue is: Can we use the simpler classical...

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