Question

The PE of a particle of mass m is given by $V_{\left(x\right)}=\{\begin{array}{c}{E}_0\\ 0\end{array}$ $\begin{array}{c}0\le x\le 1\\ x>1\end{array}$

${\lambda }_1$ and ${\lambda }_2$ are the de-Broglie wavelengths of the particle, when $0\le x\le 1$ and $x>1$ respectively. The total energy of the particle is $42E_0$. If the ratio $\frac{{\lambda }_1}{{\lambda }_2}$ is $\sqrt{x}$, find $x$.

Answer 1

Kinetic Energy, $KE=2E_0-E_0=E_0$ for $0\le x\le 1$

Wavelength, ${\lambda }_1=\frac{h}{\sqrt{2m{E}_0}}$

$KE=2E_0$ for $x>1\therefore {\lambda }_2=\frac{h}{\sqrt{4m{E}_0}}$

$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{2}$.