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GR0877 Problem 97

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) According to Compton scattering formula, the difference in wavelength of the scattered \lambda' and incident \lambda photon is \displaystyle  \lambda' - \lambda = \frac{h}{mc}(1-\cos{\theta}), where \theta is the scattering angle. In our case, \theta = 90^{\circ} and \displaystyle \lambda' - \lambda = \frac{h}{mc}. But \displaystyle E = \frac{hc}{\lambda}, so \displaystyle E' = \frac{hc}{\lambda'}, where E' is the energy of the scattered photon. Thus, \displaystyle \frac{hc}{E'} - \frac{hc}{E} = \frac{h}{mc}, from which \displaystyle E' = \frac{E \cdot mc^2}{E + mc^2}.

Found a typo? Comment!

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  1. PineApple
    15.10.2011 at 13:12

    2 small typos: you wrote that E=hc/\nu but should be E=hc/\lambda.
    Another typo is in the compton scattering formula: should be h/mc rather than \hbar/mc

  2. Drew
    24.09.2014 at 21:15

    For those that don’t remember (or know anything about) the Compton formula, you can solve with conservation laws. Momentum of the electron will be equal to \(E^2+E_F^2\) (conservation of momentum). Plug into \(E+m=E_f+\sqrt{m^2+p_e^2}\) to get the final answer.

  3. 20.10.2014 at 12:43

    Surely the easiasiasiest way is consider and extreme panorama where
    $\lim_{m \to +\infty} E’ = E$
    cuz the photon shouldn’t lose its energy due to the momentum conservation (it’s an elastic scattering).
    E is the only answer that fits this prediction.

    • 20.10.2014 at 12:49

      OH…IT’S CALLED thomson scattering!
      \frac{h \cdot \nu}{m \cdot c^2} \ll 1

  4. 20.10.2014 at 12:45

    Surely the easiasiasiest way is consider and extreme panorama where
    \[ \lim_{m \to +\infty} E’ = E \]
    cuz the photon shouldn’t lose its energy due to the momentum conservation (it’s an elastic scattering).
    E is the only answer that fits this prediction.

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