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

  • question_answer
    The de-Broglie wavelength \[\lambda \]associated with an electron having kinetic energy E is given by the expression [MP PMT 1990; CPMT 1996]

    A) \[\frac{h}{\sqrt{2mE}}\]             

    B)            \[\frac{2h}{mE}\]

    C)            \[2mhE\]                               

    D)            \[\frac{2\sqrt{2mE}}{h}\]

    Correct Answer: A

    Solution :

               \[\frac{1}{2}m{{v}^{2}}=E\Rightarrow mv=\sqrt{2mE};\ \ \therefore \ \ \lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}\]


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