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A KE derivation from Compton Effect

  1. Mar 5, 2006 #1

    Pengwuino

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    Gold Member

    I have an odd problem here. I need to show that:

    [tex]KE = \frac{{\frac{{\Delta \lambda }}{\lambda }}}{{1 + (\frac{{\Delta \lambda }}{\lambda })}}hf[/tex]

    I've basically derived [tex] KE = \frac{{hc}}{{\lambda _o }} - \frac{{hc}}{{\lambda '}}[/tex] down to…

    [tex] KE(\frac{{\lambda '}}{{\lambda _o ^2 }}) = (\frac{{\Delta \lambda }}{{\lambda _o }})hf[/tex]

    but I'm not sure how I can turn that [tex] \frac{{\lambda '}}{{\lambda _o ^2 }}[/tex] into a [tex] 1 + (\frac{{\Delta \lambda }}{{\lambda _o }})[/tex]

    Can anyone help?
     
    Last edited: Mar 5, 2006
  2. jcsd
  3. Mar 5, 2006 #2
    [tex]\Delta E=\frac{hc}{\lambda_0}(1-\frac{\lambda_0}{\lambda})[/tex]

    [tex]1-\frac{\lambda_0}{\lambda}=\frac{\lambda-\lambda_0}{\lambda}=\frac{\frac{\lambda-\lambda_0}{\lambda_0}}{1+\frac{\lambda-\lambda_0}{\lambda_0}}[/tex]

    Consider [tex]\frac{\Delta \lambda}{\lambda_0}=\frac{\lambda-\lambda_0}{\lambda_0}[/tex]

    And we have the answer
     
  4. Mar 5, 2006 #3

    Pengwuino

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    Gold Member

    Alright i'll try to get to that myself... is there anything special about the equation they wanted me to find?
     
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