I was working through a question to work out the wavelength of emitted photons when an electron moving in circular motion collides with a stationary positron, like this:
The following was given in the question: \begin{align} v_{e} &= 6.0\times10^{7} \tag{velocity of the electron} \\ m_{e} &= 9.11\times10^{-31} \tag{rest mass of an electron} \\ h &= 6.63\times10^{-34}\text{Js} \tag{Planck constant} \\ c &= 3.0\times10^{8}\text{ms}^{-2} \tag{Speed of light} \\ \end{align} This was my approach to the question: \begin{align} \lambda &= \frac{h}{\textbf{p}} \tag{de Broglie Wavelength} \\ \Sigma\textbf{p}_{\text{before annihilation}} &= \Sigma\textbf{p}_{\text{after annihilation}} \tag{conservation of momentum}\\ \textbf{p}_{\text{before annihilation}} &= \textbf{p}_{\text{e}} \\ \textbf{p}_{\text{after annihilation}} &= 2\times\textbf{p}_{\text{photon}} \\ \textbf{p}_{\text{photon}} &= \frac{1}{2}\times\textbf{p}_{\text{e}} \\ \textbf{p}_{\text{photon}} &= \frac{1}{2}\times m_{\text{e}}\textbf{v}_{\text{e}} \\ \textbf{p}_{\text{photon}} &= \frac{1}{2}\times(9.11\times10^{-31})(6.0\times10^{7}) \\ \lambda &= \frac{6.63\times10^{-34}}{\frac{1}{2}\times(9.11\times10^{-31})(6.0\times10^{7})} \\ \lambda &= 2.43\times10^{-11} \end{align} My way assumed the equality of de broglie wavelength and em radiation wavelength - which I'm sure is right, but the mark scheme uses the following method and gets a different answer: \begin{align} E = mc^2 &= \frac{hc}{\lambda} \\ mc^2 &= \frac{hc}{\lambda} \\ \lambda &= \frac{h}{mc} \\ \lambda &= \frac{6.63\times10^{-34}}{(9.11\times10^{-31})(3\times10^8)} \\ \lambda &= 2.43\times10^{-12} \end{align}
Why is there such a large difference between my $\lambda$ and theirs? Have I made an incorrect assumption about where the momentum goes?
Is it a coincidence my value is the similar, only off by an order of magnitude?
EDIT: I just noticed an additional clause in the question:
calculate the wavelength ... assuming the kinetic energy of the electron is negligible
(Does this impact their/my reasoning?)
