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CHAPTER 44 - 2

# CHAPTER 44 - 2 - f(26MHz)=(1(1/2,whichgivesf =f =13MHz d d...

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v av = v N d N 0 n n = v max n 3/ 2 N 1/ 2 d N 0 n = 2 3 v max . f d /(26 MHz) = (1)(1/2), which gives  f d  =  f α  =       13 MHz . 7. The size of a nucleon is d  ˜ 2(1.2 × 10 –15  m) = 2.4 × 10 –15  m. Because 30 MeV  «   mc 2 , we find the momentum from K  =  p 2 /2 m , so the wavelength is λ  =  h / p  =  h /(2 mK ) 1/2 . For the  α  particle we have λ α   = (6.63 × 10 –34  J     s)/[2(4)(1.67 × 10 –27  kg)(30 MeV)(1.60 × 10 –13  J/MeV)] 1/2   =       2.6 × 10 –15  m ˜ size of nucleon . For the proton we have λ p = (6.63 × 10 –34  J     s)/[2(1.67 × 10 –27  kg)(30 MeV)(1.60 × 10 –13  J/MeV)] 1/2   =       5.2 × 10 –15  m ˜ 2(size of nucleon). Thus the       α  particle is better . 8. The alternating frequency means the proton is accelerated by the voltage twice in each revolution.  The number of revolutions is n  =  E /2 eV  = (25 MeV)(10 3  keV/MeV)/2(1 e)(55 kV) =      2.3 × 10 2  rev .

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CHAPTER 44 - 2 - f(26MHz)=(1(1/2,whichgivesf =f =13MHz d d...

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