This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHYS126 Supplementary Notes 6 1. De Broglies Hypothesis hf E = and h p = or = E and k p f = 2 h = where s eV s J h . 10 14 . 4 . 10 63 . 6 15 34 = = Example: Find the wavelength of an electron with energy 2 E MeV = Hint: The De Broglie relation / h p = is correct at all energies, but since the energy is relativistic at some case, you will have to use the relation ( 29 ( 29 2 2 2 E pc mc = + to find p . The energy is 2 E MeV = , and for electron 2 0.511 mc MeV = , so we have 1.934 PC MeV = So 15 4.14 10 h eV = , 13 6.42 10 m = 2. Compton's Scattering ) cos 1 (  + = c m h e Example: X rays of wavelength 0.2400nm are ComptonScattered, and the scattered beam is observed at an angle of 60 o C relative to the incident beam. Find a) the wavelength of the scattered X rays, b) the energy of the scattered Xray photons, c) the kinetic energy of the scattered electrons a) ) cos 1 (  + = c m h e nm nm nm 2412 ....
View
Full
Document
 Spring '11
 HoBunCHAN
 Physics, Energy

Click to edit the document details