p39_037

# p39_037 - E =(1240 eV Â nm/Î where E is the energy and Î is...

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37. (a) Since the mass of an electron is m =9 . 109 × 10 31 kg, its Compton wavelength is λ C = h mc = 6 . 626 × 10 34 J · s (9 . 109 × 10 31 kg)(2 . 998 × 10 8 m / s) =2 . 426 × 10 12 m=2 . 43 pm . (b) Since the mass of a proton is m =1 . 673 × 10 27 kg, its Compton wavelength is λ C = 6 . 626 × 10 34 J · s (1 . 673 × 10 27 kg)(2 . 998 × 10 8 m / s) =1 . 321 × 10 15 m=1 . 32 fm . (c) We use the formula developed in Exercise 3:
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Unformatted text preview: E = (1240 eV Â· nm) /Î» , where E is the energy and Î» is the wavelength. Thus for the electron, E = (1240 eV Â· nm) / (2 . 426 Ã— 10 âˆ’ 3 nm) = 5 . 11 Ã— 10 5 eV = . 511 MeV. (d) For the proton, E = (1240 eV Â· nm) / (1 . 321 Ã— 10 âˆ’ 6 nm) = 9 . 39 Ã— 10 8 eV = 939 MeV....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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