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JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Self Evaluation Test - Dual Nature of Radiation and Matter

  • question_answer
    Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V. The de-Broglie wavelengths are \[{{\lambda }_{e}}\] and \[{{\lambda }_{p}}\] for electrons and protons respectively. The ratio of \[\frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}\]is given by: (given \[{{m}_{e}}\]is mass of electron and \[{{m}_{p}}\]is mass of proton)

    A) \[\sqrt{\frac{{{m}_{p}}}{{{m}_{e}}}}\]

    B) \[\sqrt{\frac{{{m}_{e}}}{{{m}_{p}}}}\]

    C) \[\frac{1}{2}\sqrt{\frac{{{m}_{e}}}{{{m}_{p}}}}\]

    D) \[2\sqrt{\frac{{{m}_{e}}}{{{m}_{p}}}}\]

    Correct Answer: D

    Solution :

    [d] Energy in joule (E)= charge \[\times \] potential diff. in volt. \[{{E}_{electron}}={{q}_{e}}V\text{ and }{{\text{E}}_{proton}}={{q}_{p}}4V\] De-Broglie wavelength \[{{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}eV}}\text{and }{{\lambda }_{p}}=\frac{h}{\sqrt{2{{m}_{p}}e4V}}\] \[(\therefore {{q}_{e}}={{q}_{p}})\] \[\therefore \frac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\frac{\frac{h}{\sqrt{2{{m}_{e}}eV}}}{\frac{h}{\sqrt{2{{m}_{p}}e4V}}}=\sqrt{\frac{2{{m}_{p}}e4V}{2{{m}_{e}}eV}}=2\sqrt{\frac{{{m}_{p}}}{{{m}_{e}}}}\]


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