• # question_answer The potential energy of a particle of mass m is given by U\left( x \right)=\left\{ \begin{align} & {{E}_{0}};\,0\le x\le 1 \\ & 0;\,\,\,\,\,\,\,\,\,x>1 \\ \end{align} \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 A) 2 B) 1 C) $\sqrt{2}$         D) $\frac{1}{\sqrt{2}}$

 For $0\le x\le 1,{{K}_{1}}=E-U=2{{E}_{0}}-{{E}_{0}}={{E}_{0}}$ For $x>1,{{K}_{2}}=E-U=2{{E}_{0}}-0=2{{E}_{0}}$ Now $\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{h/\sqrt{2m{{K}_{1}}}}{n/\sqrt{2m{{K}_{2}}}}=\sqrt{\frac{{{K}_{2}}}{{{K}_{1}}}}=\sqrt{2}$