KVPY Sample Paper KVPY Stream-SX Model Paper-22

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) \[\frac{1}{\sqrt{2}}\]

Correct Answer: C

Solution :

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 \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{h/\sqrt{2m{{K}_{1}}}}{n/\sqrt{2m{{K}_{2}}}}=\sqrt{\frac{{{K}_{2}}}{{{K}_{1}}}}=\sqrt{2}\]