Question

The potential energy of a particle of mass m is given by V(x) =$\{\begin{array}{c}{E}_0,0\le x\le 1\\ 0,x>1\end{array}$

${\lambda }_1$ and ${\lambda }_2$ are the de-Broglie wavelengths of the particle, when $0\le x\le 1$ and x > 1respectively. If the total energy of particle is $2E_0$, then $\frac{{\lambda }_1}{{\lambda }_2}$ is

• A. 2
• B. $\sqrt{2}$
• C. 1
• D. 4

Answer 1

The correct option is D

4

Kinetic energy + Potential energy = Total energy De-Broglie wavelength $\lambda =\frac{h}{\sqrt{2mK}}$