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Compton scattering - no solution from quadratic eq? What to do?

  1. Jul 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Incomming photon gives half of its energy to an electron during scatering. After scattering, photon is headed ##\phi=120^\circ## according to the original direction. What is the ##\lambda## of the incomming photon?

    2. Relevant equations
    \begin{align}
    \Delta \lambda &= \frac{h}{m_ec}(1-\cos\phi)\\
    W_{before} &= W_{after}
    \end{align}

    3. The attempt at a solution
    I first calculated the ##\Delta \lambda = \frac{h}{m_e c} (1 - \cos \phi) = 3.65pm## and then wrote the energy conservation equation which i evolved into an quadratic equation:

    \begin{align}
    W_{before} &=W_{after}\\
    W_f + m_ec^2 &= W_f' +W_e\\
    W_f + m_ec^2 &= W_f' +\tfrac{1}{2}W_f\longleftarrow\substack{\text{I used the fact that photon gives}\\\text{half of its energy to an electron} }\\
    m_e c^2 &= W_f' - \tfrac{1}{2}W_f\\
    m_ec^2 &= hc \left(\frac{1}{\lambda'} - \frac{1}{2\lambda}\right)\\
    m_ec^2 &= hc \left(\frac{1}{\lambda + \Delta \lambda} - \frac{1}{2\lambda}\right)\longleftarrow\substack{\text{since i know $\Delta \lambda$ which }\\\text{is defined as $\Delta \lambda = \lambda' - \lambda$}}\\
    \frac{m_ec}{h} &= \frac{1}{\lambda + \Delta \lambda} - \frac{1}{2\lambda}\\
    \frac{m_ec}{h} &= \frac{2\lambda - (\lambda + \Delta \lambda)}{2\lambda (\lambda + \Delta \lambda)}\\
    \frac{2m_ec}{h} &= \frac{\lambda - \Delta \lambda}{\lambda^2 + \Delta \lambda \lambda}\\
    \frac{2m_ec}{h} \lambda^2 + \frac{2m_ec}{h} \Delta \lambda \lambda &= \lambda - \Delta \lambda\\
    \frac{2m_ec}{h} \lambda^2 + \left(\frac{2m_ec}{h} \Delta \lambda - 1 \right)\lambda + \Delta \lambda &=0\\
    \underbrace{\lambda^2}_{A\equiv 1} + \underbrace{\left(\Delta \lambda - \frac{h}{2m_ec} \right)}_{\equiv B}\lambda + \underbrace{\frac{h}{2m_ec}\Delta \lambda}_{\equiv C} &=0\longleftarrow\substack{\text{I finaly got the quadratic equation}\\\text{for whom i define $A$, $B$ and $C$}}\\
    \end{align}

    Now i use the basic quadratic equation formula derived from "compleeting the square", and as you can see the number under the square root is negative. Does this mean there is no solution to this problem? If this is not the case did i miss something?

    \begin{aligned}
    \lambda &= \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\\
    \lambda &= \frac{ \left( \frac{h}{2m_ec} - \Delta \lambda \right) \pm \sqrt{ \left(\Delta \lambda - \frac{h}{2m_ec} \right)^2 - 4\frac{h}{2m_ec}\Delta \lambda} }{ 2 }\\
    \lambda &= \frac{-2.43\times10^{-12}m \pm \sqrt{5.92\times10^{-24}m^2 - 1.78\times10^{-23}m^2}}{2}\\
    \lambda &= \frac{-2.43\times10^{-12}m \pm \sqrt{-1.19\times10^{-23}m^2}}{2}\longleftarrow\substack{\text{I get the negative number}\\\text{under the square root}}\\
    \end{aligned}
     
  2. jcsd
  3. Jul 16, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello.

    Did you leave out the "rest energy" of the electron when you wrote ##W_e = \frac{1}{2}W_f## for the final energy of the electron in the third line of your calculation?

    There might be a simpler approach. The compton formula gives one equation with the two unknowns λ and λ'. See if you can get another equation for these unknown using the fact that the final energy of the photon is 1/2 the initial energy. (There's a simple equation relating energy and λ of a photon.)
     
  4. Jul 17, 2013 #3
    The problem states that photon gives half of its energy to an electron so i just wrote that the energy of an electron after the collision equals the ##\tfrac{1}{2} W_f##. Ouch now i see it. The electron had to have rest energy before and after the scattering - what photon gives to an electron is all kinetic energy!

    Thank you!
     
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