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VMMC VMMC Medical Solved Paper-2005

  • question_answer
    Ultraviolet radiation of 6.2 eV falls on an aluminium foil surface. Work function is 4.2 eV. The K.E. of the fastest electron emitted approximately:

    A) \[3.2\times {{10}^{-21}}J\]                          

    B) \[3.2\times {{10}^{-19}}J\]          

    C) \[3.2\times {{10}^{-15}}J\]          

    D) \[3.2\times {{10}^{-17}}J\]

    Correct Answer: B

    Solution :

    Using the relation \[KE=hv-{{\omega }_{0}}\] \[KE=6.2-4.2\,=2\,eV\] \[=2\times 1.6\times {{10}^{-19}}=3.2\times {{10}^{-19}}\,J\]


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