Calculate the de Broglie wavelength of $\left(\mathrm{a}\right)$ a $1.00 \mathrm{keV}$ electron, $\left(\mathrm{b}\right)$ a $1.00 \mathrm{keV}$ photon, and $\left(\mathrm{c}\right)$ a $1.00 \mathrm{keV}$ neutron.

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 $12p?$

The function $\psi \left(x\right)$ displayed in Eq. $38-27$ can describe a free particle for which the potential energy is $U\left(x\right)=0$ in Schrodinger's equation (Eq. $38-19$). Assume now that $U\left(x\right)=U_0=$ a constant in that equation. Show that Eq. $38-27$ is a solution of Schrodinger's equation with

$k=\frac{2\pi }{h}\sqrt{2m\left(E-U_0\right)}$

giving the angular wave number $k$ of the particle.