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Percent change in photon energy

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A photon of wavelength is incident on an electron at rest. On collision, the photon is scattered at an angle θ with increased wavelength . Suppose θ = pi/2, what is the percent change in photon energy for this collision in (a) a microwave oven with wavelength = 3.0 cm; (b) visible sunlight with wavelength = 5000 Å; (c) X-ray machine with wavelength = 1.0 Å; and (d) ɣ rays with energy 1.0 MeV. [h/mc = 2.426 x 10-12m]

    2. Relevant equations
    wavelength’ - wavelength = h/mc (1 – cosϴ)


    3. The attempt at a solution

    wavelength’ - wavelength = h/mc (1 – cosϴ) = 2.43 x 10-12m (1-cos/2) = 3.76 x 10-4m
    a. 3 x 10-2m / 3.76 x 10-4m = 79.79
    b. 5000 x 10-10m/ 3.76 x 10-4m = 13.30%
    c. 1 x 10-10m / 3.76 x 10-4m = 2.66 x 10-5%
    d. 1 MeV = 1 x 106 eV = 1.6 x 10-13 J/ 6.626 x 10-34 Js = 2.41 x 1020 Hz
    wavelength = c/f = 2.998 x 108 m/s / 2.41 x 1020 Hz = 1.24 x 10-12 m
    1.24 x 10-12 m / 3.76 x 10-4m = 3.30 x 10-7%
    ( is this correct? please help...)
     
  2. jcsd
  3. Aug 4, 2011 #2

    vela

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    What did you do here? What is cos π/2 equal to? Where did 3.76x10-4 m come from?
    No, it's not correct. The problem is asking for the percent change in energy. If E and E' are the energy of the photon before and after the collision, what's the expression for the percentage change in energy?
     
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