KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • 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) \[\frac{1}{\sqrt{2}}\]

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

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