PS3_key.pdf - Problem set 3 1 For electrons and protons...

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Problem set 3 1. For electrons and protons, kinetic energy E = m v 2 2 = p 2 2 m , and de Broglie wavelength λ = h p . Therefore we get λ = h 2 m E . In[1]:= h = 6.626 × 10 - 34 1.6 × 10 - 19 ; (* in units of eV.s *) me = 0.511 × 10 6 ; (* in units of eV c 2 . hence the rest energy of electrons is me * c 2 = 0.511 * 10 6 eV *) mp = 938 × 10 6 ; (* in units of eV c 2 . hence the rest energy of protons is mp * c 2 = 938 * 10 6 eV *) lambda [ e _ , m _] : = h 2 m e ; Show [ Plot [ lambda [ e, me ] , { e, 0, me * 0.05 }] , AxesLabel { "Energy ( eV ) ", "Wavelength ( m ) " } , PlotLabel "de Broglie wavelength versus energy for electrons" ] Plot [ lambda [ e, mp ] , { e, 0, mp * 0.05 } , AxesLabel { "Energy ( eV ) ", "Wavelength ( m ) " } , PlotLabel "de Broglie wavelength versus energy for protons" ] Out[5]= 5000 10000 15000 20000 25000 Energy ( eV ) 4. × 10 - 20 6. × 10 - 20 8. × 10 - 20 1. × 10 - 19 Wavelength ( m ) de Broglie wavelength versus energy for electrons Out[6]= 1 × 10 7 2 × 10 7 3 × 10 7 4 × 10 7 Energy ( eV ) 2. × 10 - 23 3. × 10 - 23 4. × 10 - 23 5. × 10 - 23 Wavelength ( m ) de Broglie wavelength versus energy for protons

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2. ( a ) Convert the unit of Planck' s constant into the SI base units.
• Fall '18
• Atom, Mass, Electric charge, Fundamental physics concepts

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