BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    A photon and an electron possess same de-Broglie wavelength given that \[C=\]speed of light and \[\upsilon =\]space of electron, which of the following relation is correct? (here, \[{{E}_{e}}=K.E\]of electron,\[{{E}_{Ph}}=K.E\] of photon, \[{{P}_{e}}=\] momentum of electron, \[{{P}_{ph}}=\]momentum of photon)

    A)  \[\frac{{{P}_{e}}}{{{P}_{Pn}}}=\frac{C}{2v}\]     

    B)         \[\frac{{{E}_{e}}}{{{E}_{Ph}}}=\frac{C}{2v}\]      

    C) \[\frac{{{E}_{ph}}}{{{E}_{e}}}=\frac{2c}{v}\]        

    D)         \[\frac{{{P}_{e}}}{{{P}_{Ph}}}=\frac{2C}{v}\]

    Correct Answer: C

    Solution :

    We have,  \[{{\lambda }_{Ph}}=\frac{h}{{{P}_{Ph}}}\]and \[{{\lambda }_{e}}=\frac{h}{{{P}_{e}}}\] Given,   \[{{\lambda }_{Ph}}={{\lambda }_{e}}\] \[\therefore \] We get,  \[{{P}_{Ph}}={{P}_{e}}\]                 \[\frac{h}{{{\lambda }_{Ph}}}=mv\] \[\therefore \]  \[\frac{hc}{{{\lambda }_{Ph}}}=mcv=\frac{1}{2}m{{v}^{2}}\left( \frac{2c}{v} \right)\]       or            \[\frac{{{E}_{Ph}}}{{{E}_{e}}}=\frac{2c}{v}\]


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