**1. The problem statement**

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}

## Best Answer

In my opinion you are using for energy conservation after scattering the wrong Ansatz. The electron which has no energy before scattering has $E_{e1}=m_{e}c^{2}$, but after scattering its energy increases due to the kinetic energy $E_{\text{kin}}=p\cdot c=\frac{1}{2}\frac{h\cdot c}{\lambda}$. So after the scattering process the electrons total energy should be in my opinion $E_{e2}=\sqrt{\left(m_{e}c^{2}\right)^{2}+\left(\frac{1}{2}\frac{h\cdot c}{\lambda}\right)^{2}}$. Of course the photon energy still contribute, so

$\begin{eqnarray}E_{1}&=&E_{2}\\\frac{h\cdot c}{\lambda}+m_{e}c^{2}&=&\frac{h\cdot c}{\lambda^{\prime}\left(<\lambda\right)}+\sqrt{\left(m_{e}c^{2}\right)^{2}+\left(\frac{1}{2}\frac{h\cdot c}{\lambda}\right)^{2}}\end{eqnarray}$.

Additionally I found a well documented problem set including the solution concerning the compton effect, which might help you in case you still got stuck in your calculation: http://physlab.lums.edu.pk/images/c/cb/Solution2_new_modified.pdf .