$X$ rays of wavelength $8.50 \mathrm{pm}$ are directed in the positive direction of an $x$-axis onto a target containing loosely bound electrons. For Compton scattering from one of those electrons, at an angle of $180°$, what are $\left(\mathrm{a}\right)$ the Compton shift, $\left(\mathrm{b}\right)$ the corresponding change in photon energy, $\left(\mathrm{c}\right)$ the kinetic energy of the recoiling electron. and $\left(\mathrm{d}\right)$ the angle between the positive direction of the $x$-axis and the electron's direction of motion?

Show that when a photon of energy $E$ is scattered from a free electron at rest, the maximum kinetic energy of the recoiling electron is given by $K_{\max }=\frac{E^2}{E+{\displaystyle \frac{m{c}^2}{2}}}$.

Figure shows a case in which the momentum component $p_{\mathrm{x}}$ of a particle is fixed, so that, $\mathrm{\Delta }{p}_{\mathrm{x}}=0;$ then, from Heisenberg's uncertainty principle, the position $x$ of the particle is completely unknown. From the same principle, it follows that the opposite is also true, that is, if the position of a particle is exactly known $(\mathrm{\Delta }x=0),$ the uncertainty in its momentum is infinite.

Consider an intermediate case. In which the position of a particle is measured, not to infinite precision, but to within a distance of $\lambda /2\pi ,$ where $\lambda$ is the particle's de Broglie wavelength. Show that the uncertainty in the (simultaneously measured) momentum component is then equal to the component itself; that is, $\mathrm{\Delta }{p}_{\mathrm{x}}=p.$ Under these circumstances, would a measured momentum of zero surprises you? What about a measured momentum of $0.5p?$ Of $2p?$ Of $12p?$