JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Matter Waves

  • question_answer
    The kinetic energy of electron and proton is \[{{10}^{-32}}J\]. Then the relation between their de-Broglie wavelengths is [CPMT 1999]

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

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

    C)            \[{{\lambda }_{p}}={{\lambda }_{e}}\]                         

    D)            \[{{\lambda }_{p}}=2{{\lambda }_{e}}\]

    Correct Answer: A

    Solution :

               By using \[\lambda =\frac{h}{\sqrt{2mE}}\]    E = 10?32 J = Constant for both particles. Hence \[\lambda \propto \frac{1}{\sqrt{m}}\] Since \[{{m}_{p}}>{{m}_{e}}\] so \[{{\lambda }_{p}}<{{\lambda }_{e}}.\]


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