ComptonEffect 2009-03-08 03:03:52 2009-03-08 03:03:52 Topic Compton scattering equation % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here The Compton effect represents another confirmation of the photon theory, and a refutation of the wave theory. One observes it (Compton, 1924) in the scattering of X-rays by free (or weakly bound) electrons. The wavelength of the scattered radiation exceeds that of the incident radiation. The difference $\triangle \lambda$ varies as a function of the angle $\theta$ between the direction of propagation of the incident radiation and the direction along which one observes the scattered light, according to Compton's formula: \begin{equation} \triangle \lambda = 2 \frac{h}{mc} \sin^2 \frac{\theta}{2} \end{equation} where $m$ is the rest mass of the electron. One notes that $\triangle \lambda$ is independent of the incident wavelength. Compton and Debye have shown that the Compton effect is a simple elastic collision between a photon of the incident light and one of the electrons of the irradiated target. In order to discuss this corpuscular interpretation it is convenient to state a few properties of photons which derive directly from Einstein's hypothesis. Since they possess the velocity $c$,photons are particles of zero mass. The momentum $p$ and the energy $\epsilon$ of a photon are thus connected by the relation \begin{equation} \epsilon = p c \end{equation} Consider a plane, monochromatic light wave $$ exp \left [2 \pi i \left ( \frac{\mathbf{u} \cdot \mathbf{r}}{\lambda} - vt \right ) \right ]$$ $\mathbf{u}$ is a unit vector in the direction of propagation, $\lambda$ is the wavelength, $v$ the frequency: $\lambda v = c$. In accordance with Einstein's hypothesis, this wave represents a stream of photons of energy $hv$. The momentum of these photons is evidently directed along $\mathbf{u}$ and its absolute value, according to (2), is equal to $$p = \frac{h v}{c} = \frac{h}{\lambda}$$ This relation is a special case of the relation of L. de Broglie. It is often convenient to introduce the angular frequency $\omega = 2 \pi v$ and the wave vector $\mathbf{k}=(2\pi/\lambda)\mathbf{u}$ of he plane wave. The connecting relations are then written: \begin{equation} \epsilon = \hbar \omega, \,\,\,\,\,\,\,\, \mathbf{p}=\hbar \mathbf{k} \end{equation} (more to come...) derived from public domain work [1]