A deuteron and an alpha particle are accelerated with the same potential. Which one of the two has $\left(\mathrm{i}\right)$ greater value of de Broglie wavelength associated with it and $\left(\mathrm{ii}\right)$ less kinetic energy? Explain.

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.

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12\times {10}^{-15} \mathrm{V} \mathrm{s}$. Calculate the value of Planck’s constant.