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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is \[1\,\,eV\] for incoming radiation of frequency \[{{v}_{0}}\] and \[3\,\,eV\] for incoming radiation of frequency\[3{{v}_{0}}/2\]. What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency\[9{{v}_{0}}/4\]?

    A) \[3\,\,eV\]                         

    B) \[4.5\,\,eV\]

    C) \[6\,\,eV\]                         

    D) \[9\,\,eV\]

    Correct Answer: C

    Solution :

    \[{{(KE)}_{\max }}=hv-{{\phi }_{0}}\] So\[,\]  \[1eV=h{{v}_{0}}-{{\phi }_{0}}\]                                                ? (i) and        \[3eV=\frac{h{{v}_{0}}}{2}-{{\phi }_{0}}\]                                             ... (ii) \[\Rightarrow \]               \[3eV-1eV=\frac{h{{v}_{0}}}{2}\] or            \[h{{v}_{0}}=4eV\] From Eq. (i),       \[{{\phi }_{0}}=h{{v}_{0}}-1eV\]                 \[=4eV-1eV=3eV\] \[\therefore \]  \[{{(KE)}_{\max }}=h\times \frac{9{{v}_{0}}}{4}-3eV\]                 \[=\frac{9}{4}(4eV)-3eV=6eV\]


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