Question

The potential energy of a particle of mass m is given by $V\left(x\right)=\left\{\begin{array}{c}{E}_0\\ 0\end{array}\begin{array}{c}0\le x\le 1\\ x>1\end{array}\right\}$
${\lambda }_1$ and ${\lambda }_2$ are the de-Broglie wavelengths of the particle, when $0\le x\le 1$ and $x>1$ respectively.
If the total energy of particle is ${2E}_0$, find $\left({\lambda }_1/{\lambda }_2\right)$.

Answer 1

Total energy , $E=T+V$ where T=Kinetic energy and V=potential energy.

When $0\le x\le 1$, $2E_0=T+E_0\Rightarrow T=E_0$

de-Broglie wavelength, ${\lambda }_1=\frac{h}{\sqrt{2mT}}=\frac{h}{\sqrt{2m{E}_0}}$

When $x>1,2E_0=T+0$ or $T=2E_0$

now, ${\lambda }_2=\frac{h}{\sqrt{2mT}}=\frac{h}{\sqrt{2m2E_0}}$

Thus, $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{2}$