# Difference between DeBroglie wavelength and wavelength of radiation

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?)

• I have doubts on your treatment. The electron comes with a linear momentum $\vec P_e$ which is tangential to the circle at the point where annihilation occurs. Right? Well, as the positron was at rest, the total linear momentum before the collision is $\vec P_e$ and this shouls also be the total linear momentum of the emitted photons. Right? Then, $\sum \vec P_{photon} \ne 0$, the linear momenta of the two photons won't be equal. The photon emitted in the same direction of movement as the electron moved, will have a bigger linear momentum exactly by $\vec P_e$. So, first improve your calculus. – Sofia Feb 2 '15 at 15:14
• I mean, if one photon moves in the direction of movement of the electron, and one photon moves in opposite direction, the former photon has a bigger linear momentum. For making decisions you need the two equations, of momentum conservation and of energy conservation. Of course, if you'll find that the photons should have also a component of linear momentum perpendicular to the electron velocity, these components will be equal in abs. value and opposite in sign. – Sofia Feb 2 '15 at 15:18
• Another point to add for this problem is that the electron is relativistic so the momentum of the electron is given by $p=\gamma m v$ where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ although this is kind of irrelevant as the arguments made by @Sofia I think are true. – Chris2807 Feb 2 '15 at 15:30
• @Ell it won't help you much. Linear momentum conservation is saint. You have to use it, and write your equations, as I suggested. We cannot know in which directions the emitted photons move, but for sure at least one of them has a linear momentum component along the direction of movement of the electron. Now, if the kinetic energy of the electron is small comparing with the rest mass, that doesn't absolve you from linear momentum conservation. – Sofia Feb 2 '15 at 15:38