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EAMCET Medical EAMCET Medical Solved Paper-2006

  • 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-Plancks constant

    A)  h                                           

    B)  c

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

    D)  \[\frac{1}{c}\]

    Correct Answer: B

    Solution :

                     \[{{\lambda }_{e}}=\frac{h}{mv}\] \[{{\lambda }_{p}}=\frac{h}{mc}\] Given,                   \[{{\lambda }_{c}}={{\lambda }_{p}}\] \[\Rightarrow \]                               \[v=c\] \[\therefore \]  \[\frac{{{E}_{p}}}{{{p}_{e}}}=\frac{hv}{mv}\] \[=\frac{m{{c}^{2}}}{mv}=\frac{{{c}^{2}}}{c}=c\]                                \[(\because \,v=c)\]


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