The potential energy of a particle of mass m is given by $\mathrm{U}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{E}}_{\mathrm{o}}:0\le \mathrm{x}\le 1\\ 0:\mathrm{x}>1\end{array}\right.{\mathrm{\lambda }}_1\mathrm{and}{\mathrm{\lambda }}_2$  are the de-Broglie wavelengths of the particle, when $0\le \mathrm{x}\le 1\mathrm{and}\mathrm{x}>1$ respectively. If the total energy of particle is $2{\mathrm{E}}_{\mathrm{o}},\mathrm{the}\mathrm{ratio}\frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2}\mathrm{will}\mathrm{be}$

• A

  2

• B

  1

• C

  $\sqrt{2}$

• D

  $\frac{1}{\sqrt{2}}$