JEE PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

The potential energy of a particle of mass $m$ is given by

$U\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{E_0};\;0 \le x \le 1}\\ {0;\;\;\;\;\;\;\;x > 1} \end{array}} \right.$

${\lambda _1}$ 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 $2{E_0},$ the ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ will be