Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of photographic film and thus, recorded on the film. Photon absorption will occur if the photon energy equals or exceeds $0.6 \mathrm{eV},$ the smallest amount of energy needed to dissociate an $\mathrm{AgBr}$ molecule in the film. $\left(\mathrm{a}\right)$ What is the greatest wavelength of light that can be recorded by the film? $\left(\mathrm{b}\right)$ In what region of the electromagnetic spectrum is this wavelength located?

In equation $\psi (x,t)=A{e}^{\mathrm{i}(\mathrm{kx}-\mathrm{\omega t})}+B{e}^{-\mathrm{i}(\mathrm{kx}+\mathrm{\omega t})}$ keep both terms, putting $A=B={\psi }_0.$ The equation then describes the superposition of two matter waves of equal amplitude, travelling in opposite directions. (Recall that this is the condition for a standing wave.) $\left(\mathrm{a}\right)$ Show that $\left|\Psi \right(x,t){|}^2$ is then given by,

$\left|\Psi \right(x,t){|}^2=2{\psi }_0^2[1+\cos 2 kx].$

$\left(\mathrm{b}\right)$ Plot this function and demonstrate that it describes the square of the amplitude of a standing matter-wave. $\left(\mathrm{c}\right)$ Show that the nodes of this standing wave are located at $x=(2n+1)\left(\frac{1}{4}\mathrm{\lambda }\right),$ where $n=0,1,2,3, ...$ and $\mathrm{\lambda }$ is the de Broglie wavelength of the particle. $\left(\mathrm{d}\right)$ Write a similar expression for the most probable locations of the particle.