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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is (\[c=\]velocity of light, \[h=\]Planck's constant)

    A) \[h\]                                     

    B) \[c\]

    C) \[\frac{1}{h}\]                                  

    D) \[\frac{1}{c}\]

    Correct Answer: B

    Solution :

    We have\[{{\lambda }_{e}}=\frac{h}{mv}\] and        \[{{\lambda }_{p}}=\frac{h}{mc}\] According to the question,                 \[{{\lambda }_{e}}={{\lambda }_{p}}\] \[\therefore \]  \[v=c\]                 \[\frac{{{E}_{p}}}{{{P}_{e}}}=\frac{m{{c}^{2}}}{mv}=\frac{{{c}^{2}}}{c}=c\]


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