RAJASTHAN ­ PET Rajasthan PET Solved Paper-2010

  • question_answer
    The de- Broglie wavelength of an electron having kinetic energy E is given by the expression

    A)  \[\frac{h}{\sqrt{2mE}}\]

    B)  \[\frac{2h}{mE}\]

    C)  \[2mhE\]

    D)  \[\frac{2\sqrt{2mE}}{h}\]

    Correct Answer: A

    Solution :

     The kinetic energy \[\frac{1}{2}m{{v}^{2}}=E\] \[mv=\sqrt{2mE}\] \[\therefore \] \[\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}}\]


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