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BVP Medical BVP Medical Solved Paper-2004

  • question_answer
    A proton and an \[\text{ }\!\!\alpha\!\!\text{ -}\]particle are accelerated through the same potential difference. The ratio of de-Broglie wavelength of proton to the de-Broglie wavelength of alpha particle will be :

    A) \[2.4\times {{10}^{-3}}\]                              

    B)  1 : 2

    C)  2 :1                                       

    D)  1 : 1

    Correct Answer: A

    Solution :

    From the relation, de-Broglie wavelength                 \[\mu F\] \[\mu F\]             \[\Omega \] (when V is same) Hence, \[\frac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\frac{\sqrt{{{m}_{\alpha }}{{q}_{\alpha }}}}{{{m}_{p}}{{q}_{p}}}=\frac{\sqrt{4\times 2}}{1\times 1}=\frac{2\sqrt{2}}{1}\]


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