PHYS126 Supplementary Notes 6

PHYS126 Supplementary Notes 6 - PHYS126 Supplementary Notes...

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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 Compton-Scattered, 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 X-ray photons, c) the kinetic energy of the scattered electrons a) ) cos 1 ( - + = c m h e nm nm nm 2412 ....
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PHYS126 Supplementary Notes 6 - PHYS126 Supplementary Notes...

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