BCECE Medical BCECE Medical Solved Papers-2006

  • question_answer
    The de-Broglie wavelength of an electron, a-particle and a proton all having the same kinetic energy is respectively given as \[{{\lambda }_{e}},\,{{\lambda }_{\alpha }}\] and \[{{\lambda }_{p}}\]. Then which of the following is not true?

    A)  \[{{\lambda }_{e}}>{{\lambda }_{p}}\]

    B)  \[{{\lambda }_{p}}>{{\lambda }_{\alpha }}\]

    C)  \[{{\lambda }_{e}}>{{\lambda }_{\alpha }}\]

    D)  \[{{\lambda }_{\alpha }}<{{\lambda }_{p}}<{{\lambda }_{e}}\]

    Correct Answer: A

    Solution :

    de-Broglie wavelength                 \[\lambda =\frac{h}{mv}\] or \[\lambda \propto \frac{1}{m}\] \[\therefore \]  \[{{\lambda }_{e}}\propto \frac{1}{{{m}_{e}}},\,{{\lambda }_{\alpha }}\propto \frac{1}{{{m}_{\alpha }}}\] and \[{{\lambda }_{p}}\propto \frac{1}{{{m}_{p}}}\] As we know that \[{{m}_{e}}<{{m}_{p}}<{{m}_{a}}\] So,          \[{{\lambda }_{e}}>{{\lambda }_{p}}>{{\lambda }_{\alpha }}\] or \[{{\lambda }_{e}}>{{\lambda }_{\alpha }}\] or \[{{\lambda }_{p}}>{{\lambda }_{\alpha }}\] or \[{{\lambda }_{e}}>{{\lambda }_{p}}\]


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