So electron moving left ($v_{initial}=0.8c$) collides into photon going right. After the collision electron is moving to right ($v_{after}=0.6c$) and the photon is moving to left. What is the wavelength $\lambda_{initial}$ of the photon before collision.

So we are only on the x-axis and $cos\phi=-1$ because there is no scattering $\phi =180^0$. And I think we need to establish the equations for preservation of the linear momentum and energy. Something like this? $$\begin{align} E_{initial}&=E_{final} \\ p_{initial,x} &= p_{final,x} \\ \end{align} $$ Im not sure what are the formulas for linear momentum and energy of photon and electron. Any help? $$\begin{align} E_{initial}&=E_{final} \\ \frac{hc}{\lambda}+\gamma m_e c^2 &= \frac{hc}{\lambda^´}+\gamma^´ m_ec^2 \\ \end{align} $$ Where $p=\frac{h}{\lambda}$ and $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ and $\gamma^´ =\frac{1}{\sqrt{1-\frac{v^´2}{c^2}}}$ ?

And for momentum:$$\begin{align} p_{initial,x} &= p_{final,x} \\ \frac{h}{\lambda}-\gamma m_e v &= \frac{h}{\lambda^´}cos\phi-\gamma^´ m_e v^´ cos\phi \\ \frac{h}{\lambda}-\gamma m_e v &= -\frac{h}{\lambda^´} + \gamma^´ m_e v^´ \\ \end{align}$$

And now I have to equations with 2 unknowns $\lambda$ (what I want to know) and $\lambda^´$. Can someone verify this before I start to solve this system?:D

\begin{cases} \frac{hc}{\lambda}+\gamma m_e c^2 &= \frac{hc}{\lambda^´}+\gamma^´ m_ec^2 \\ \frac{h}{\lambda}-\gamma m_e v &= -\frac{h}{\lambda^´} + \gamma^´ m_e v^´ \\ \end{cases} \begin{cases} \frac{hc}{\lambda}-\frac{hc}{\lambda^´} &= \gamma^´ m_ec^2 - \gamma m_e c^2 \\ \frac{h}{\lambda}+\frac{h}{\lambda^´} &= \gamma^´ m_e v^´ +\gamma m_e v , & \text{let's multiply this with $c$} \\ \end{cases} \begin{cases} \frac{hc}{\lambda}-\frac{hc}{\lambda^´} &= \gamma^´ m_ec^2 - \gamma m_e c^2 \\ \frac{hc}{\lambda}+\frac{hc}{\lambda^´} &= \gamma^´ m_e v^´ c +\gamma m_e v c \\ \end{cases} $$\lambda = \frac{2hc}{\gamma^´ m_ec^2 - \gamma m_e c^2 + \gamma^´ m_e v^´ c +\gamma m_e v c}?$$ $$\lambda = \frac{2\times1.239eV\mu m}{\frac{0.511\times 10^6\frac{eV}{c^2}}{\sqrt{1-\frac{(0.6c)^2}{c^2}}}(c^2+0.6c^2)+\frac{0.511\times 10^6\frac{eV}{c^2}}{\sqrt{1-\frac{(0.8c)^2}{c^2}}}(0.8c^2-c^2)}\approx 3.0\times 10^{-6} \mu m?$$

That's too small I think... i was expecting wavelenght in $nm$. Damn I guess it's wrong :(


1 Answer 1


$p_{photon} = \frac{h}{\lambda}$, with its energy $E_{photon} = pc$

The momentum of an electron is like any other object. Remember to account for relativistic effects!

You now have two equations and two undetermined variables, for example:

$a +b = 1$ and $2a - b = 0$

One way to solve this is to rearrange the first to $a = 1-b$ and insert that into equation 2, solve for b and then find a with either initial equation.

You've botched the vectors in your momentum equation. Pick to the right as positive and to the left as negative. You'll find that pre-collision $p_{photon}$ is positive and $p_{electron}$ is negative and viceversa after the collision.

Try to avoid using eV (and other non-SI units) if you can. I find it's just another way for errors to creep into your calculation.

For one, your expression of relativistic kinetic energy is wrong.

  • $\begingroup$ So the $\vec{p_{photon,i}}+\vec{p_{electron,i}}=\vec{p_{photon,after}}+\vec{p_{electron,after}}$ $\endgroup$
    – ELEC
    Commented Jan 19, 2014 at 11:41
  • $\begingroup$ That's right. You did the same for energy as well, now you'll have to solve those two equations. $\endgroup$
    – Kvothe
    Commented Jan 19, 2014 at 13:49
  • $\begingroup$ I've updated my answer to explain solving multiple equations at once $\endgroup$
    – Kvothe
    Commented Jan 19, 2014 at 15:23
  • $\begingroup$ I've updated my answer again. $\endgroup$
    – Kvothe
    Commented Jan 19, 2014 at 17:22

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