The potential energy of a particle of mass m is given by \[U(x)=\left\{ \begin{align} & {{E}_{0}};\ \ \ 0\le x\le 1 \\ & \,0\ ;\ \ \ \,\,\,\,x>1 \\ \end{align} \right.\]l1 and l2 are the de-Broglie wavelengths of the particle, when 0 £ x £ 1 and x > 1 respectively. If the total energy of particle is 2E0, the ratio \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\] will be [Based on IIT-JEE (Mains) 2005]

The potential energy of a particle of mass m is given by \[U(x)=\left\

1.	The potential energy of a particle of mass m is given by \[U(x)=\left\{ \begin{align} & {{E}_{0}};\ \ \ 0\le x\le 1 \\ & \,0\ ;\ \ \ \,\,\,\,x>1 \\ \end{align} \right.\]l1 and l2 are the de-Broglie wavelengths of the particle, when 0 £ x £ 1 and x > 1 respectively. If the total energy of particle is 2E0, the ratio \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\] will be [Based on IIT-JEE (Mains) 2005]
A.	2
B.	1
C.	\[\sqrt{2}\]
D.	\[\frac{1}{\sqrt{2}}\]
Answer» D. \[\frac{1}{\sqrt{2}}\]