Ch39 #45 - 39.45: a) Since K > mc 2 we must use...

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Unformatted text preview: 39.45: a) Since K > mc 2 we must use the relativistic expression for energy. E 2 = p 2c 2 + m 2c 4 but E = K + mc 2 ⇒ ( K + mc 2 ) 2 = p 2c 2 + m 2c 4 [( K + mc 2 ) 2 − m 2 c 4 )]1 2 h hc ⇒λ= = .  ⇒ p = 2 2 c p [( K + mc ) − m 2 c 4 ]1 2 hc h 2 = .  If K = 3mc then λ = 2 2 2 4 12 [(4mc ) − m c ] 15mc  b) i)  m = 9.11 × 10 −31 kg K = 3mc 2 = 3(9.11 × 10 −31 kg)(3.00 × 10 8 m s) 2 = 2.46 × 10 −13 J   = 1.54 MeV. λ=    ii)  m = 1.67 × 10 −27 kg h (6.63 × 10 −13 J ⋅ s) = 15mc 15 (9.11 × 10 −31 kg)(3.00 × 108 m s) = 6.2 × 10 −13 m. K = 3mc 2 = 3(1.67 × 10 −27 kg)(3.00 × 10 8 m s) 2 = 4.51 × 10 −10 J = 2.82 × 103 MeV. h (6.63 × 10 −34 J ⋅ s) λ= = 15mc 15 (1.67 × 10 −27 kg)(3.00 × 108 m s) = 3.42 × 10 −16 m. ...
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This homework help was uploaded on 04/02/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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